Solving Linear Algebra Final Questions with Help

In summary, the final exam given by the professor was a take-home exam intended as a learning tool. Two weeks before the exam, the professor gave the students a packet of notes and said class was over. The professor is not well-respected in the department and the whole semester was essentially self-taught. The final exam was not too difficult but the format was strange, and the professor's tests are generally disliked. The students were asked to type up their answers, including matrices, which some were unsure how to do. The exam had several questions regarding eigenvalues and eigenvectors, orthogonality, and diagonalization. The students were also asked to find an orthogonal matrix for a given matrix and to determine the dimensions of eig
  • #1
Gale
684
2
ok, so here's the deal. The final was a take home, and an intended "learning tool." meaning, 2 weeks ago he threw us a packet of notes, said class was over, and good luck with the final. I hate this prof... somehow he's well respected in the dept... anyways, he doesn't teach us at all, this whole semesmter has basically been self taught, but whatever... the final's not too hard... just... weird... i hate his tests...

So, i'll write the questions, write what i have, and hopefully you all will help.
Also, he wants us to type these all up, and I'm not sure how to go about that... i don't know how to type matrices in word... so any advice as to how to do that is appreciated as well...


Q1) explain why, if a is an eigenvalue of the nxn matrix A, ie, a root of f(t)= lA-tIl there is an [tex] X \neq 0[/tex] in [tex] R_{n} [/tex] with AX= aX.

A) is a is an eigenvector of A then by definition it has a corresponding eigenvector [tex]\xi[/tex], not equal to zero, where [tex] A\xi= a\xi[/tex]
therefore [tex] X=\xi[/tex] and X is an eigenvector of A with a coresponding eigenvalue of a.

(i'm not sure if that proves "exactly when" do i need to show the reverse or something?)



Q2) Show that if U is an nxn matrix, then (UX,UY) = (X,Y) exactly when the columns of U form an orthagonal set.

A) [tex] (UX,UY)= (UX)^T (UY) = X^T U^T UY = X^T (U^T U) Y= X^T Y = (X,Y) [/tex]
where [tex] U^T U= I[/tex] because U is orthagonal, and therefore symmetric.



Q3) Let C be an orthagonal nxn matrix (as in 2) and A an nxn matrix. Show that A is symmetric exactly when [tex] C^-1 AC[/tex] is symmetric.

A) [tex] (C^{-1} AC)^T = (C^{-1} (AC))^T = (AC)^T (C^{-1})^T = C^T A^T (C^{-1})^T = C^{-1} A^T C[/tex] because [tex] C^T = C^{-1}[/tex] because U is orthagonal. therefore. [tex] (C^{-1} AC)^T [/tex] is symmetric when [tex] A= A^T [/tex], when A is symmetric..



Q4) Let A be an upper triangular matrix with all diagonal entries equal. Show that if A is not a diagonal matrix, then A cannot be diagonalized.

A) A = aI + B... (i don't know what diagonal means... so I'm not really sure what to do here... help)



Q5) Show that A is a symmetric nxn matrix and Y and X are eigenvectors of A belonging to different eigevalues, then (X,Y)=0

A) [tex] AX=\lambda X[/tex] and [tex] AY=\mu Y[/tex] where [tex] \lambda \neq \mu [/tex]

[tex]\lambda (X,Y) = (\lambda X,Y) = (AX,Y) = (X,AY) = (X,\mu Y) = \mu (X,Y) [/tex] and since [tex]\mu \neq \lambda[/tex] then we must have [tex] (X,Y)=0 [/tex]

(ok, i pretty much copied that directly from those notes he gave us... not sure if its right... or if it is, why he'd put it on the final... the guy makes no sense... )



Q6) Let A be the matrix
[tex]\left(\begin{array}{cccc}0&0&0&-1\\0&1&\sqrt{2}&0\\0&\sqrt{2}&1&0\\-1&0&0&-1\end{array}\right) [/tex]

find an orthagonal matrix U so that [tex] U^{-1} AU[/tex] is a diagonal matrix.

A) ok... this one took a page and a half of work just to find one eigenvector... i have to find 4 i believe. The first eigenvector i got was [tex]\left(\begin{array}{cccc}0\\1\\1\\0\end{array}\right) [/tex] if someone could check that... that'd be cool. After i find the eigenvectors i... umm... have no idea what i do actually... combine them into one matrix? is that U? not too sure... anyways, if its necessary i'll show more of my work, I'm just not having fun will all this tex right now...



Q7) Let A be the matrix
[tex]\left(\begin{array}{ccc}1&0&-1\\0&2&0\\-1&0&1\end{array}\right) [/tex]

i) find the distinct eigenvalues [tex]\lambda[/tex] of A and their multiplicities
ii) Determine the dimensions of the eigenspaces [tex] N(A- \lambda I)[/tex]
iii) find orthonormal bases of these eigenspaces
iv) Combine these bases into one orthonormal basis B of [tex]R^3[/tex] and verify that the matrix of A relative to B is a diagonal matrix with entries the eigenvalues of A, each repeated as many times as its multiplicity

A) ok [tex] \lambda_{1} = 0 \lambda_{2,3} = 2[/tex]
my first eigenvector is [tex] \xi_{1} =\left(\begin{array}{cc}1\\0\\1\end{array}\right) [/tex] so the dimension of that is 1?
the other two eigenvectors I'm not sure about. I get
[tex]\left(\begin{array}{cc}-1&-1\\0&0\\1&1\end{array}\right) [/tex] I'm not even sure if that's the right notation... or if the numbers are right, cause i had an undefined variable. but i wrote it this way so it'd have a dimension of 2... cause our teacher said there's was one dim1 one dim2

after that, I'm not really sure what to do, because we never went over orthonormal anything... and i don't know how to find bases well... its a mess... i'll have to sit with my notes for a while in order to understand this at all

but anyways, thanks for any help, or just checking my answers. And if you have an idea as to how i can type this all up, that'd be awesome. (maybe i can even print my work straight off pf? cause i don't know how else to type matrices...)

~gale~
 
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  • #2
my copy of word has a program called equation editor that has matrices. but i somehow feel constrained from helping you with your final.

but in the first question you are confused as to what eh is asking. i.e. you are taking a different definition of eigenvalue from the one given in thew problem.
'
probably the problem should ask: "show that any root of the characteristic polynomial is an eigenvalue."

the way you are interporeting it it is asking the tautological question: " show that any eigenvalue is an eigenvalue." that is pointless.


in question 2 you are wrong that an orthoigonal matrix is symmetric but your solution is correct otherwise, since Atrans A = I is true for an orthogonal matrix.

in question 4, ooops it loooks as if i gave an incorrect answer elsewhere about this, so maybe i am confused here,

but anyway, for this you need to nkow that a matric is "diagonal" if the only possible non zero entries are on the main diagonal, and this can be achieved if and only if each root of the characteristic polynomial r has algebraic multiplicity as a root, equal to the dimension of the nullspace of A-rI.


it also sounds as if you need to read the book a little bit more, if you do not know this fundamental stuff. i am going to leave you to do that now.
 
  • #3
heres a t shirt i wore to class last week: (imagine it centered and in a fancy font)

And God said:
LET ALL SYMMETRIC MATRICES
BE “ORTHOGONALLY DIAGONALIZABLE”
(and vice versa)
to wit:

A = At
IF and ONLY IF there is an ORTHONORMAL BASIS of EIGENVECTORS for A
 
  • #4
hmm well i realized my typo in problem two, so ya. problem one... i'll look at again, cause you were right, i misinterpreted, problem 4, i figured out what diagonal meant on my own. And I'm working on it. mostly though, i can't get 10 and 11. in problem 10 i keep getting zero vectors for eigenvectors. 11 i don't understand well, and will have to read more, yes.

By the way, i don't have a book. He doesn't issue books for the class. He hands us his notes, which are all implicity written, with nearly no examples or plain english, and then in class all he does is rewrite them on the board. Its been a very, VERY hard class to learn anything from. but thanks for adding to my frustration... I'm well aware of my weak grasp of linear algebra, and I've done my best.

grr...
 
  • #5
have you noticed you have a tendency to blame other people for your difficulties, first your teacher, now me?

if you need a book, the one by ruslan sharipov mentioned in another thread here is free and excellent.

the very short one on the following webpage is also free.

http://www.math.uga.edu/~roy/


by the way 10 and 11 do not seem to be there.
 
  • #6
my professor sucks... you suck...
it frustrates me... meh...

anyways, by 10 and 11 i meant 6 and 7, they're 10 and 11 on my paper, but we only had to chose so many problems... whatever, 6 and 7 are giving me problems. specifically 10. Those were the two "learning" problems. He didn't teach us anything about them, but figured he'd hand us the notes, and if we could figure them out, we'd learn... whatever...

the semester is bout over, i don't think i'll get a textbook now actually. thanks though... again... for the excellent help.

i'll post my work for 4 a bit later. if anyone could figure out the eigenvectors for 10, or tell me if I'm doing 11 right, that'd be cool. thanks...
 
  • #7
have you noticed i am the only person helping you here? your people skills could use work. solve your own problems.
 
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  • #8
ok ok, I'm sorry... PMS...
but thanks. I've just been working on this stuff for too many hours and i think you were a bit unfair...

anyways, it'd be a huge help if someone could show me how to get something decent for an answer to problem 10. I also worked more on 11, but i don't know if what i posted here was right to begin with...
 
  • #9
mathwonk said:
have you noticed i am the only person helping you here? your people skills could use work. solve your own problems.

You are not the only one helping her, you're the only one helping her on PF. Gale is doing her best with what she has. I think your accusations are extremely unfair and your posts unnecessarily condescending. If you don't want to help, just don't respond.
 
  • #10
Gale17 said:
Q1) explain why, if a is an eigenvalue of the nxn matrix A, ie, a root of f(t)= lA-tIl there is an [tex] X \neq 0[/tex] in [tex] R_{n} [/tex] with AX= aX.

If some matrix is invertible, then it returns a nonzero vector for every vector of any basis. Conversely, if a matrix is singular, then there is some basis for which there is some vector which is sent to zero by the matrix.

Can you see how to use that to answer the question?

A) is a is an eigenvector of A then by definition it has a corresponding eigenvector [tex]\xi[/tex], not equal to zero, where [tex] A\xi= a\xi[/tex]
therefore [tex] X=\xi[/tex] and X is an eigenvector of A with a coresponding eigenvalue of a.

(i'm not sure if that proves "exactly when" do i need to show the reverse or something?)
there are some problems with your answer. Firstly, I assume it was a typo when you said "a is an eigenvector". In fact, a is an eigenvalue. We don't know if this guy has an eigenvector yet (remember that there may be more eigenvalues than eigenvectors, so we can't be sure that the matrix has any eigenvectors. that's what we have to prove).

secondly, the definition of eigenvalue that we're using is: root of the secular polynomial, not solution to the eigenvalue equation. so we do not have an eigenvector "by definition" as you claim.



Q2) Show that if U is an nxn matrix, then (UX,UY) = (X,Y) exactly when the columns of U form an orthagonal set.
"exactly when" means you have to show two things: first show that if (UX,UY)=(X,Y), then the columns of U are orthogonal (spelling!). then you have to show that if the columns of U form an orthogonal set, then (UX,UY)=(X,Y).

A) [tex] (UX,UY)= (UX)^T (UY) = X^T U^T UY = X^T (U^T U) Y= X^T Y = (X,Y) [/tex]
where [tex] U^T U= I[/tex] because U is orthagonal, and therefore symmetric.
you have shown that U^TU=1 implies (UX,UY)=(X,Y), which is true, and useful, but not quite the answer that we need.

recognize that the i-th column vector of U is given by U(ei) where ei is the i-th vector in the (orthonormal) basis. and orthonormal bases satisfy (ei,ej)=1 if i=j, 0 if i !=j

remember to do the implication in both directions


Q3) Let C be an orthagonal nxn matrix (as in 2) and A an nxn matrix. Show that A is symmetric exactly when [tex] C^-1 AC[/tex] is symmetric.

A) [tex] (C^{-1} AC)^T = (C^{-1} (AC))^T = (AC)^T (C^{-1})^T = C^T A^T (C^{-1})^T = C^{-1} A^T C[/tex] because [tex] C^T = C^{-1}[/tex] because U is orthagonal. therefore. [tex] (C^{-1} AC)^T [/tex] is symmetric when [tex] A= A^T [/tex], when A is symmetric..
Yes, very good. Remember that "exactly when" means that you have to prove two implications, although in this case, I think every step of the argument you made is reversible. still, make sure you know what is meant by that.



Q4) Let A be an upper triangular matrix with all diagonal entries equal. Show that if A is not a diagonal matrix, then A cannot be diagonalized.

A) A = aI + B... (i don't know what diagonal means... so I'm not really sure what to do here... help)
Diagonal means having only entries on the diagonal of the matrix. Like, the matrix

1 0 0
0 2 0
0 0 3

is diagonal, but the matrix

1 2 3
0 4 0
0 0 5

is not, because that 2 and 3 aren't on the diagonal.

If a matrix can be diagonalized (meaning there is some change of basis which brings the matrix to diagonal form), then the diagonal entries will be the eigenvalues of the matrix. Use that fact, along with the fact that the determinant of an upper triangular matrix is the product of its diagonal elements, and you can get this one.


Q5) Show that A is a symmetric nxn matrix and Y and X are eigenvectors of A belonging to different eigevalues, then (X,Y)=0

A) [tex] AX=\lambda X[/tex] and [tex] AY=\mu Y[/tex] where [tex] \lambda \neq \mu [/tex]

[tex]\lambda (X,Y) = (\lambda X,Y) = (AX,Y) = (X,AY) = (X,\mu Y) = \mu (X,Y) [/tex] and since [tex]\mu \neq \lambda[/tex] then we must have [tex] (X,Y)=0 [/tex]
yeah

(ok, i pretty much copied that directly from those notes he gave us... not sure if its right... or if it is, why he'd put it on the final... the guy makes no sense... )



Q6) Let A be the matrix
[tex]\left(\begin{array}{cccc}0&0&0&-1\\0&1&\sqrt{2}&0\\0&\sqrt{2}&1&0\\-1&0&0&-1\end{array}\right) [/tex]

find an orthagonal matrix U so that [tex] U^{-1} AU[/tex] is a diagonal matrix.

A) ok... this one took a page and a half of work just to find one eigenvector... i have to find 4 i believe. The first eigenvector i got was [tex]\left(\begin{array}{cccc}0\\1\\1\\0\end{array}\right) [/tex] if someone could check that... that'd be cool. After i find the eigenvectors i... umm... have no idea what i do actually... combine them into one matrix? is that U? not too sure... anyways, if its necessary i'll show more of my work, I'm just not having fun will all this tex right now...
Hoh boy. Diagonalizing matrices can be a bunch of calculation. Tell you what I'll do: I'll plug this guy into mathematica, and see what the answer is and tell you if you're right. You'll still have to do the calculation yourself. By the way, that calculation is as follows:

1. get the secular polynomilal det(M-x)=0, and find its roots.
2. for each root, find the nullity of M-xi
3. choose an orthonormal basis for each null space.
4. those vectors go in the columns of U
5. and the eigenvalues go in the diagonal of the resulting matrix

and the answer is... yes! the vector you list is indeed one of the eigenvectors, corresponding to eigenvalue 1+sqrt(2). 3 more to go!


Q7) Let A be the matrix
[tex]\left(\begin{array}{ccc}1&0&-1\\0&2&0\\-1&0&1\end{array}\right) [/tex]

i) find the distinct eigenvalues [tex]\lambda[/tex] of A and their multiplicities
ii) Determine the dimensions of the eigenspaces [tex] N(A- \lambda I)[/tex]
iii) find orthonormal bases of these eigenspaces
iv) Combine these bases into one orthonormal basis B of [tex]R^3[/tex] and verify that the matrix of A relative to B is a diagonal matrix with entries the eigenvalues of A, each repeated as many times as its multiplicity
Again? didn't we just diagonalize a matrix in the last problem? Ugh. Well this one is only 3x3, so it won't be as bad, so I guess I'll do it by hand, so I can check you step by step.

OK, actually, it wasn't bad. lots of zeros.

A) ok [tex] \lambda_{1} = 0 \lambda_{2,3} = 2[/tex]
yes.
my first eigenvector is [tex] \xi_{1} =\left(\begin{array}{cc}1\\0\\1\end{array}\right) [/tex] so the dimension of that is 1?
yes
the other two eigenvectors I'm not sure about. I get
[tex]\left(\begin{array}{cc}-1&-1\\0&0\\1&1\end{array}\right) [/tex] I'm not even sure if that's the right notation...
that notation is ****ed up. is that supposed to be a 3x2 matrix or is it supposed to be 2 vectors? anyway, the column of that thing is the correct answer for one of the eigenvectors corresponding to eigenvalue 2. there is another one.

or if the numbers are right, cause i had an undefined variable. but i wrote it this way so it'd have a dimension of 2... cause our teacher said there's was one dim1 one dim2
what is an undefined variable? What happened? You should definitely be able to get 2 eigenvectors here.

after that, I'm not really sure what to do, because we never went over orthonormal anything... and i don't know how to find bases well... its a mess... i'll have to sit with my notes for a while in order to understand this at all
once you have the 3 eigenvectors, it'll be easy to make them orthonormal, they're almost orthonormal already. and then you're done.

but anyways, thanks for any help, or just checking my answers. And if you have an idea as to how i can type this all up, that'd be awesome. (maybe i can even print my work straight off pf? cause i don't know how else to type matrices...)

~gale~
type it in Latex! msword sucks for math documents.
 
  • #11
space tiger, when you ask for free advice, sometimes you get true advice that you don't want. especially likely if you start out trashing people and continue by saying they "suck".

i believe discontinuing such behavior would indeed be helpful.

by the way it is your remarks about gale that are condescending. i think gale is capable of better.
 
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  • #12
So here's what I've redone:

Q1)
Don Aman said:
If some matrix is invertible, then it returns a nonzero vector for every vector of any basis. Conversely, if a matrix is singular, then there is some basis for which there is some vector which is sent to zero by the matrix.

Can you see how to use that to answer the question?

umm... not really. if A has an eigenvalue, a, does that mean A's invertible? i guess if that's true, then i sort of see what you're saying...

I'm looking back through my notes... it says this "Let T:V-->V be a linear transformation. A non-zero vector X in V is called an eigenvector of T with eigenvalue [tex]\lambda[/tex] ( scalar) if [tex] TX=\lambda X[/tex] The eigen value [tex]\lambda[/tex] may be zero, but by definition, an eigenvector is not zero. We say that [tex]\lambda[/tex] is an eigenvalue of T if there is a non-zero vextor X in C with [tex] TX=\lambda X[/tex]"

So that's where i think i was initially going off of... trying to find something like what you're talking about... i find this...
"If [tex] P(t) = a_{0} + a_{1}t+...+a_{r}t^r[/tex] is a polynomial in the variable t, and X is an eigenvector of T with eigenvalue [tex]\lambda[/tex], then X is an eigenvector of [tex] P(t) = a_{0}I + a_{1}T+...+a_{r}T^r[/tex] with eigenvalue [tex]P(\lambda)[/tex], and if T is invertible then [tex]\lambda\neq0[/tex] and X is an eigenvector of [tex]t^{-1}[/tex] with eigenvalue [tex]\frac{1}{\lambda}[/tex]

the non zero vectors in the nullspace [tex]N(T-\lambda I)[/tex] are the eigenvectors of T with the eigenvalue [tex]\lambda[/tex]. But we know that [tex]N(T-\lambda I)[/tex] contains a non-zero vector exactly when [tex]|T-\lambda I|=0[/tex] But we also know that if V has a dimension n, then [tex]|T-\lambda I|[/tex] is a polynomial of degree n in the unknown [tex]\lambda[/tex], called the characteristic polynomial of T."

i'm going to sit and try to make sense of that for a while. I'm really not sure what I'm sposed to do.

Q2 I'm still thinking about question one... i kinda made some sense of what you said... but i'll get back to you on that.

Q3 i think I'm all set on this one, i redid everything backwards, we're good.

Q4 ok, here i go...
Since A is upper trianglular, you can write it as A=aI+B, if B is a strictly upper triangular martix. We know that [tex]C^{-1}AC[/tex] will be a diagonal matrix. so, [tex] D=C^{-1}AC=aI+C^{-1}BC[/tex] (ok, so i don't actually know why we know that fist thing has to be diagonal, and I'm not sure why the aI doesn't get multiplied by C and its inverse... but my prof said this was all right...) so, then [tex] D-aI=C^{-1}BC[/tex]. A diagonal matrix summed with another diagonal matrix will still be a diagonal matrix, therefore, [tex]C^{-1}BC[/tex]. must also be diagonal. now, because B is strictly upper triangular we know that [tex]B^n=0[/tex] therefore [tex]C^{-1}B^nC=(C^{-1}BC)^n=0=D-aI[/tex] therefore [tex]D=aI=C^{-1}AC[/tex]
...ok so now this is supposed to prove something? i think i lost myself...

Q5 we're good...

Q6 Ok, so here are all the eigenvectors...
[tex]\left(\begin{array}{c}0\\1\\1\\0\end{array}\right) \left(\begin{array}{c}0\\1\\-1\\0\end{array}\right) \left(\begin{array}{c}1\\0\\0\\\frac{-1+\sqrt{5}}{2}\end{array}\right)\left(\begin{array}{c}1\\0\\0\\\frac{-1-\sqrt{5}}{2}\end{array}\right) [/tex]

ok, so now i make them unit vectors, yes? umm... i can't find anything about doing that in my notes... so i'll assume its just pythagoras. so my new results... (this gets messy, but i did it so you could check my work, more easily...)
[tex]\left(\begin{array}{c}0\\\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\\0\end{array}\right \left(\begin{array}{c}0\\\frac{1}{\sqrt{2}}\\\frac{-1}{\sqrt{2}}\\0\end{array}\right) \left(\begin{array}{c}\frac{1}{\sqrt{1+(\frac{-1+\sqrt{5}}{2})^2}}\\0\\0\\\frac{\frac{-1+\sqrt{5}}{2}}{\sqrt{1+(\frac{-1+\sqrt{5}}{2})^2}}\end{array}\right) \left(\begin{array}{c}\frac{1}{\sqrt{1+(\frac{-1-\sqrt{5}}{2})^2}}\\0\\0\\\frac{\frac{-1-\sqrt{5}}{2}}{\sqrt{1+(\frac{-1-\sqrt{5}}{2})^2}}\end{array}\right)[/tex]

OOOk... so i have all that now, i just put them all in a matrix together, call it U, and I'm done?

Q7
ok, so ya, the way i did that last matrix was wrong. I wasn't sure how to find the last eigenvector... i guess i just take that last one, and find one orthagonal to it? So then the last vector i had, and the new one i'll get form together to create an eigenspace of dimension 2. ALRIGHT... hmm.. well, i still don't think i understand parts iii) and iv) then. i take my eigenvectors and make them unit vectors, and that's the orthonormal bases? i guess that's part iii) and then i combine those... and i do what? i don't know what it means by combine them, and i really don't get the rest of part iv). he wouldn't even explain it in class cause he says "thats where it really tests your understanding..." ugh. i obviously don't understand, but I'm trying and augh.
 
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  • #13
By the way, thanks a ton Don Aman, really, thanks. and thanks spacetiger. Math wonk... if you want to help, that'd be awesome. you were definately condescending, and I'm admittedly already very frustrated. it certainly doesn't help me to be antagonistic. i don't know what you mean by "i think gale is capable of better," but i think I'm trying pretty hard actually.

ANYways, how do i use Latex? isn't that an online thing?

heh, awesome, mines not even showing up in my last post... arg
 
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  • #14
Gale17 said:
ANYways, how do i use Latex? isn't that an online thing?

Try a package like MikTex.


I'm hesitant to offer much help on a take home exam, so I'll just a couple of things:

Q1)"if A has an eigenvalue, a, does that mean A's invertible?"

No it doesn't. You should be able to find a non-invertible matrix that has an eigenvalue.

Think about how a non-zero vector in [tex]N(T-\lambda I)[/tex] relates to this question (you should also know why there is one).

Q4) "(ok, so i don't actually know why we know that fist thing has to be diagonal, and I'm not sure why the aI doesn't get multiplied by C and its inverse... but my prof said this was all right...)"

It does, perform the multiplication [tex]C^{-1}(aI+B)C[/tex] carefully.

What are the possibilities for the diagonal entries of D? Combined with [tex]D=C^{-1}AC[/tex] what does this tell you about A?
 
  • #15
Thanks i'll go about downloading that thing in a bit...
Oh, and btw, my professor encourages us to get help, we just have to list our resources. I'm afraid i won't list off everyone's name who helps me, but i'll write down physicsforums. He doesn't teach us like... anything... of course its expected that we go get help. Normally, i work with my partner, but she's been really busy, and we haven't been able to meet up, so we're going solo... but only help me as much is comfortable... I'm happy with whatever i can get.

Anyways... that first problem is supposed to be simple, and it probably is... but its just not clicking for me at all...

question 4:
i get that multiplication thing... god i was dense... that was simple... the scalar a commutes, as does I... somehow, i just didn't get that.

shmoe said:
What are the possibilities for the diagonal entries of D? Combined with [tex]D=C^{-1}AC[/tex] what does this tell you about A?

the diags of D are all a. i assume it tells me that A must be diagonal... seeing as that's what I'm trying to prove... but i don't quite get it. let's see [tex]CDC^{-1}=A[/tex] if D's a bunch of zeros off the diagonal... they'll stay zeros through the multiplication, and therefore A must be diagonal to begin with?
 
  • #16
Gale17 said:
question 4:
i get that multiplication thing... god i was dense... that was simple... the scalar a commutes, as does I... somehow, i just didn't get that.


Gale17 said:
the diags of D are all a. i assume it tells me that A must be diagonal... seeing as that's what I'm trying to prove... but i don't quite get it. let's see [tex]CDC^{-1}=A[/tex] if D's a bunch of zeros off the diagonal... they'll stay zeros through the multiplication, and therefore A must be diagonal to begin with?

You know how to do this. Think about what D looks like and how this is related to the first quote in this post.
 
  • #17
ok, i have this written in my notes for diff eq, i just realized its the answer for problem one... (ok ok, i didn't realize it so much as someone was quickly explaining problem one to me, and it sounded too familiar...) i can't quite make sense of it though...

[tex]\lambda[/tex] is an eigenvalue with eigenvector [tex]\xi\neq0[/tex] if [tex] A\xi=\lambda\xi[/tex]
to find eigenvalue, use the fact that matrix is not invertible if the determinent equals 0.
we have [tex]A\xi-\lambda\xa=0[/tex] which goes to [tex](A-\lambda I)\xi=0[/tex]
so if [tex]|A-\lambda I|\neq0[/tex] then [tex]A-\lambdaI [/tex] is invertible. So then, [tex]\xi(A-\lambda I)(A-\lambda I)^{-1}=0(A-\lambda I)^{-1}[/tex]
and... [tex] I\xi=0 ; \xi=0[/tex] which contradicts our initial statement so, [tex]|A-\lambda I|=0[/tex]

ok, i follow the math just fine. what I've said is... um... if there's a value a such that the determinant is zero... then there must be a non zero vector \xi or X...? I'm just not quite getting it...
 
  • #18
Gale, here is another decent free "e-book" on linear algebra.

http://home.comcast.net/~bigboa/linear.htm
 
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  • #19
". I hate this prof... somehow he's well respected in the dept... anyways, he doesn't teach us at all, this whole semesmter has basically been self taught"

Actually, I suspect it is not at all unusual for a professor who expects students to think for themselves and recognizes that the most important thing he can teach them is how to learn on their own is both well respected by his colleagues and hated by his student (or at least by the poorer students)!
 
  • #20
HallsofIvy said:
". I hate this prof... somehow he's well respected in the dept... anyways, he doesn't teach us at all, this whole semesmter has basically been self taught"

Actually, I suspect it is not at all unusual for a professor who expects students to think for themselves and recognizes that the most important thing he can teach them is how to learn on their own is both well respected by his colleagues and hated by his student (or at least by the poorer students)!


aye... i don't want to get into a convo about my professor... but look... I'm a smart girl, and well enough equipped to study things on my own. i pay to have a teacher though... and i expected one. Instead, i got a verbal copy of my notes. that's just not teaching. mostly, i think he's just way too old to still be a professor. if we asked a question, he would just get really scattered and confused, and then just rewrite the notes on the board.

anyways, for anyone who cares. I finished up the exam already. I'm not sure how well i did cause i had trouble typing it. but thanks for any help.
 
  • #21
i don't remember who the guy's name was (but his name started with M..) - he is a prof of either math or physics and he would sit in the back of the class and let his students ask the question and answer to them

a whole 'technique' that was adopted by universities and other elite institutions was named after him - which in essence is practiced today, because his students actually went on to become professors and famous scientists who discovered things.. anyone knows his name?
 
  • #22
RL Moore perhaps?
 
  • #23
yes! (min limit filler here)
 
  • #24
The Moore Method and other things...

Gale,

This is coming from a student who has sat in your chair (although not as frusterated) and understands, but has a little more time under her belt.

First of all, yes, I've been there with the take-home tests thing and it sucks. But you probably wouldn't want a sit-in exam from the same type of person.

Professors who teach in my department have an augmented Moore method way of teaching, and if you had the actual R.L. Moore for math, you probably wouldn't like math at all.

It isn't meant to make you frusterated, although at times it can be that way. But it's supposed to stretch your mind and make you think about things in a different way.

The actual method, if I understand it correctly, works a bit like this:

- Supply definitions to students (No Books)
- See what theorems they can prove by themselves without my (professor's) help (No Collaboration)

Sounds fun, doesn't it? (read: sarcasm)

The reason Moore's students were so successful is because they ended up being top researchers who didn't need others to hold their hand.

I had an interesting conversation with my advisor last week about what the difference between undergraduate and graduate education is.

His answer: the way education is perceived.

Yes, you are paying for a teacher, heck, at my school it's $11,000/semester for resident students.

You should demand your monies worth, but at the same time, if you are putting something into this, shouldn't you demand the BEST education possible for your money?

That would mean not only asking your professor to teach you the course, but to well-prepare you for the future, including your future in mathematics/physics/sciences in general.

Which means putting in 100% to the course, looking for extra resources early on and going to your profs. office hours, no matter how unapproachable he is. Also seeking out the help of others, like you did here. But, beggers can't be choosers and you have to take the bad with the good.

Standard Linear Algebra texts have all the information you are looking for, and if you don't want to ask another student to borrow theirs or look up free texts online, it's hard to give you definitions or say this proof works for XYZ reasons online because it's hard to gauge your level and understanding.

The people here aren't trying to hurt you, and I would hope that (a) your studies in mathematics continue and (b) you possibly take a course in Linear Algebra with another professor or take an advanced/graduate course in it later.

When confronted with a problem, solve it like a problem solver. Don't just give up, but work on it in another way.

Other good resources:

The Wolfram Research, Math World Site:
http://www.mathworld.com

Ask Dr. Math:
http://www.mathforum.com

Hope some of that sticks,
Vanes.
 
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  • #25
did i miss something here? gale was working on her _final exam_, right? at my school we have take home exams too, but they are all bound by an "honor principle", meaning even if they are open resource the internet is usually not allowed but in some cases it is, but no other person is ever permitted. is there no academic dishonesty here? is this not cheating?
 
  • #26
wintercarver said:
did i miss something here? gale was working on her _final exam_, right? at my school we have take home exams too, but they are all bound by an "honor principle", meaning even if they are open resource the internet is usually not allowed but in some cases it is, but no other person is ever permitted. is there no academic dishonesty here? is this not cheating?

we were allowed to work on it with other people. we were encouraged to go out and find any other resources. internet is certainly allowed as far as i knew. no dishonesty, no cheating. I've really more character than that...
 
  • #27
Gale17 said:
we were allowed to work on it with other people. we were encouraged to go out and find any other resources. internet is certainly allowed as far as i knew. no dishonesty, no cheating. I've really more character than that...

alright, i didn't mean to imply anything, just clarifying. i am still somewhat startled by the amount of help you got prior to this further explanation though. maybe other people are more aware of your integrity than i am and knew everything from the get go :)
 

Related to Solving Linear Algebra Final Questions with Help

1. How do I approach solving linear algebra final questions?

When solving linear algebra final questions, it's important to first review and understand the concepts and equations involved. Make sure to practice solving similar problems beforehand, and break down the question into smaller, manageable steps. It can also be helpful to work backwards from the answer to check your work.

2. Can I use a calculator or computer program to solve linear algebra questions?

It depends on the specific guidelines provided by your instructor. Some exams may allow the use of a calculator or computer program, while others may require you to solve the problems by hand. Make sure to check with your instructor beforehand to avoid any potential issues during the exam.

3. How can I check my answers when solving linear algebra questions?

One way to check your answers is to substitute your values back into the original equations and see if they satisfy the equations. You can also use a graphing calculator or computer program to graph the equations and visually check if they intersect at the points you calculated.

4. What are some common mistakes to avoid when solving linear algebra questions?

Some common mistakes to avoid when solving linear algebra questions include not properly understanding the question, making errors in calculations, and not simplifying or reducing equations. It's also important to check for extraneous solutions and make sure your answers make sense in the context of the problem.

5. How can I prepare for solving linear algebra questions on my final exam?

To prepare for solving linear algebra questions on your final exam, it's important to review and practice the concepts and equations covered throughout the course. Make sure to also review any past exams or quizzes to familiarize yourself with the types of questions that may be asked. You can also seek help from a tutor or study with classmates to reinforce your understanding of the material.

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