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Linear Algebra final

  1. May 11, 2005 #1
    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 dunno 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 dunno 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 thats 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 dunno 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 dunno how else to type matrices....)

    Last edited: May 11, 2005
  2. jcsd
  3. May 11, 2005 #2


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    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.
  4. May 11, 2005 #3


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    heres a t shirt i wore to class last week: (imagine it centered and in a fancy font)

    And God said:
    (and vice versa)
    to wit:

    A = At
  5. May 11, 2005 #4
    hmm well i realised 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.

  6. May 11, 2005 #5


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    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.


    by the way 10 and 11 do not seem to be there.
  7. May 11, 2005 #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...
  8. May 11, 2005 #7


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    have you noticed i am the only person helping you here? your people skills could use work. solve your own problems.
    Last edited: May 11, 2005
  9. May 12, 2005 #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...
  10. May 12, 2005 #9


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    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.
  11. May 12, 2005 #10
    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?

    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.

    "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).

    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

    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.

    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.


    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!

    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.

    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.

    what is an undefined variable? What happened? You should definitely be able to get 2 eigenvectors here.

    once you have the 3 eigenvectors, it'll be easy to make them orthonormal, they're almost orthonormal already. and then you're done.

    type it in Latex! msword sucks for math documents.
  12. May 12, 2005 #11


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    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.
    Last edited: May 12, 2005
  13. May 12, 2005 #12
    So here's what i've redone:

    umm.... not really. if A has an eigenvalue, a, does that mean A's invertible? i guess if thats true, then i sorta 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 thats 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 gonna 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?

    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 thats the orthonormal bases? i guess thats part iii) and then i combine those... and i do what? i dunno 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.
    Last edited: May 12, 2005
  14. May 12, 2005 #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 definetly condescending, and i'm admittedly already very frustrated. it certainly doesn't help me to be antagonistic. i dunno 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
    Last edited: May 12, 2005
  15. May 12, 2005 #14


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    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?
  16. May 12, 2005 #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.

    the diags of D are all a. i assume it tells me that A must be diagonal... seeing as thats what i'm trying to prove... but i don't quite get it. lets 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?
  17. May 12, 2005 #16


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    You know how to do this. Think about what D looks like and how this is related to the first quote in this post.
  18. May 12, 2005 #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...
  19. May 12, 2005 #18
    Gale, here is another decent free "e-book" on linear algebra.

    http://home.comcast.net/~bigboa/linear.htm [Broken]
    Last edited by a moderator: May 2, 2017
  20. May 14, 2005 #19


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    ". 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)!
  21. May 14, 2005 #20

    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.
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