Tricky one - solve for k - linear algebra

  • #1

Homework Statement



A=
1 5 2
1 3 k
2 1 1

Solve for k if A has the eigenvector

3
2
5



The Attempt at a Solution


I first tried to put in eigenvalues L1, L2 and L3

2-L1 [3-L1, k ; -5 , -2-L1] and so on...

and was going to isolate k but it doesnt´make any sense to me.

Can you help?
 

Answers and Replies

  • #2
tiny-tim
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Hi sciencegirl1! :smile:
A=
1 5 2
1 3 k
2 1 1

Solve for k if A has the eigenvector

3
2
5

doesn't look right :confused:

are you sure you've copied the question correctly?
 
  • #3
If x is an eigenvector, Ax=qx, for some constant q, which when multiplied out, will give you three equations in two variables. Unfortunately, when I tried to work it out, I got 23=3q and 13=5q, which leads me to believe that [3, 5, 2] isn't an eigenvector of this matrix.
 
  • #4
2 1 -1
1 3 k
-1 -5 -2

-5
-2
3

You´re right. This wasn´t the right A. Sorry. This one is correct :)
 
  • #5
tiny-tim
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Homework Helper
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2 1 -1
1 3 k
-1 -5 -2

-5
-2
3

You´re right. This wasn´t the right A. Sorry. This one is correct :)

D'uh! :rolleyes:

ok … now just do the obvious

multiply the vector by the matrix … what do you get? :smile:
 
  • #6
thanks tiny tim you´re a live saver

when I multiply it I get
-15
-11+3k
9

Then I don´t know what to do :(
 
  • #7
35,125
6,872
What's the definition of eigenvector? Knowing that will help you understand the results you got.
 
  • #8
i don´t know...
can you help me solve this?
I don´t have that much time left.
 
  • #9
35,125
6,872

Homework Statement



A=
1 5 2
1 3 k
2 1 1

Solve for k if A has the eigenvector

3
2
5



The Attempt at a Solution


I first tried to put in eigenvalues L1, L2 and L3

2-L1 [3-L1, k ; -5 , -2-L1] and so on...

and was going to isolate k but it doesnt´make any sense to me. The expression you wrote is complete gibberish.
Right. It doesn't make any sense.
Can you help?

You have a matrix A, and you're told that a given vector is an eigenvector. An eigenvectors is a special vector with regard to a particular matrix because the product Ax is a vector that is parallel to x. Any other vector makes a product, Ax that is not the same direction as x. For an eigenvector x, which is not allowed to be the zero vector, Ax = [itex]\lambda[/itex].

Historically the Greek letter [itex]\lambda[/itex] (lambda) has been used to represent an eigenvalue.

What you need to do is to solve for k in the equation A[3 2 5]^T = [itex]\lambda[/itex][3 2 5]^T. The exponent T means "transpose."

That's equivalent to (A - [itex]\lambda[/itex]I)[3 2 5]^T = [0 0 0]^T.

Surely you can do that...
 
  • #10
tiny-tim
Science Advisor
Homework Helper
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thanks tiny tim you´re a live saver

when I multiply it I get
-15
-11+3k
9

Then I don´t know what to do :(

Hi sciencegirl1! :smile:

I can't help thinking that you try to make everything too difficult :cry:

go back to basics

Q: what is the definition of an eigenvector? A: if you multiply it by the matrix, the result is an exact multiple of the original

sooooo :wink: … what value of k makes -15 -11+3k 9 an exact multiple of -5 -2 3 ? :smile:
 
  • #11
So I do -15/3=-5
9/3=3
and -11+5k=-2 and therefore k=5/3 ??
 
  • #12
tiny-tim
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So I do -15/3=-5
9/3=3
and -11+5k=-2 and therefore k=5/3 ??

oh, its 5k now, is it? :rofl:

well, that does make it a lot easier :smile:


nooo, it's not -2 (and how do you get 5/3 out of that anyway? :confused:)

Hint: you'll make less mistakes if you do it in stages …

so: stage 1: what is the eigenvalue? :wink:
 
  • #13
can you show me how you do it?
 
  • #14
tiny-tim
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can you show me how you do it?

Nope! :smile:

You try it … what is the eigenvalue?
 
  • #15
35,125
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To repeat what I said in post 9,
That's equivalent to (A - [itex]\lambda[/itex]I)[3 2 5]^T = [0 0 0]^T.
We could show you how we do it, but we aren't going to. We're trying to help you do your work, and understand it.
 
  • #16
3 2 5]^T
why do you use this vector. The vector is -5 -2 3

and why use T?

Can you just show me how to do this simply?
 
  • #17
Avodyne
Science Advisor
1,396
88
Some are using the vector and/or matrix you gave originally, not the corrected ones which are now buried in the middle of many posts.

You are given

[tex]
A=
\begin{pmatrix}
2 & 1 & -1 \\
1 & 3 & k \\
-1 & -5 & -2
\end{pmatrix}
\ \ \ \ \ \
x=
\begin{pmatrix}
-5 \\
-2 \\
3
\end{pmatrix}
[/tex]

and are told that [itex]x[/itex] is an eigenvalue of [itex]A[/itex], which means that

[tex]Ax = \lambda x[/tex]

where on the left, we matrix-multiply the matrix [itex]A[/itex] times the column-vector [itex]x[/itex] to get a new column vector; some of the components of this column vector will depend on [itex]k[/itex]. On the right, we multiply each component of [itex]x[/itex] by the number [itex]\lambda[/itex] (which we don't know yet). We equate each component of the left side with the corresponding component of the right side. This gives us three equations in two unknowns ([itex]k[/itex] and [itex]\lambda[/itex]). Presumably, the equations will be compatible; that is, there will be a solution. So, pick any two equations, solve them for [itex]k[/itex] and [itex]\lambda[/itex], and then, if you like, check that the third equation is also satisfied.
 
  • #18
I do know that Ax= [-15, -11+3k, 9]

Then I have to find \\lamda * [-5, -2, 3] = [-15, -11+3k, 9]

therefore

-5 \\lamda -2\\lamda + 3\\lamda =-15
and = 9
and = -11+3k

???
 
  • #19
You are trying to make the individual vector components match up, the sum of the components is fairly meaningless here.
 
  • #20
Avodyne
Science Advisor
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No, each component of each side is equal.

First of all,

[tex]\lambda\times
\begin{pmatrix}
-5 \\
-2 \\
3
\end{pmatrix}
= \begin{pmatrix}
-5\lambda \\
-2\lambda \\
3\lambda
\end{pmatrix}
[/tex]

Then, the left side equals the right side:

[tex]
\begin{pmatrix}
-15 \\
-11+3k \\
9
\end{pmatrix} =
\begin{pmatrix}
-5\lambda \\
-2\lambda \\
3\lambda
\end{pmatrix}[/tex]

This gives us three equations, one for each component:

[tex]-15 = -5\lambda[/tex]

[tex]-11+3k = -2\lambda[/tex]

[tex]9 = 3\lambda[/tex]
 
  • #21
lamda is 3 and then you find k?

But in problems like this, do I always know that there is only one eigenvalue?
 
  • #22
tiny-tim
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Homework Helper
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But in problems like this, do I always know that there is only one eigenvalue?

You do if the question tells you :smile:
Solve for k if A has the eigenvector

3
2
5

(and of course each eigenvector has only one eigenvalue)
 
  • #23
tiny-tim I do not recommend you becomming a teacher.
 
  • #24
1,851
7
tiny-tim I do not recommend you becomming a teacher.

Tiny tim is already a teacher :biggrin:

Reading this tread, I got the impression that your attitude toward learning has to be improved. It is now too "high school like". You had difficulties solving this problem and then the first thing you should do is if that difficulty has anything to do with not understanding the theory.

You did do that correctly. But then when you arrived at the conclusion that you have difficulties with the concepts of eigenvalues and eigenvectors, you should have forgotten about this problem for the moment and study the theory.

The goal of studying is to get a deep understanding of the theory, not per se to be able to solve a few problems your Prof. has set, especially not if you have difficulties with the theory.
 
  • #25
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6,872
tiny-tim I do not recommend you becomming a teacher.

And, in line with Count Iblis's comments, I would recommend that you become a student, which implies being able to study and work with some independence.

A lot of people have taken the time to help you with this problem--not do it for you, but help you understand what it is that you are trying to do. It didn't help that your first post had the wrong matrix and an incorrect eigenvector.

At each step in the process, we have tried to get you to carry out the next step, but instead what we saw all too often were replies from you such as these:
Then I don´t know what to do :(

i don´t know...
can you help me solve this?
I don´t have that much time left.

can you show me how you do it?

Can you just show me how to do this simply?

???

We're not here to do your work for you. That's a stated policy of this forum. However, most of us are more than willing to help you, provided that we see you making an honest effort at working the problem, and that includes looking up and trying to understand the basic terms, such as eigenvalue and eigenvector.

Mark
 
Last edited:

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