Eigenvectors

This is driving me mad!

I have to find the eigenvectors of the matrix A=[1, 0 : 1, 2 ]

So first I find the eigenvalues by solving the det (A - kI) = 0
Where k = lamda , the eigenvalue and I is the identity matrix.

So I end up with [ 1-k , 0 : 1 , 2-k] which gives the characteristic polynomial as:
k^2 -2k +2 = 0

So k=2 ,1

Then To find eigenvectors I solve (A-kI)v = 0

Where v is the eigenvector.

So if I sub my eigenvalues back into the A-kI matrix and then solve I SHOULD get my eigenvectors. BUT my matrix becomes:

for k=1

[0,0 : 1,1] [v1:v2] = [0:0]

Giving:

0x + 0y =0

which to me gives the eigenvector [1:1] not to sure why just thats what others have looked like but the actual answers are:

[0:1] and 1/sqrt(2) *[1:-1]

HOW?

$\begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \end{pmatrix}$

Eigenvectors that satisfy the equation are $\begin{pmatrix} 0 \\ y \end{pmatrix}$ and $k \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ where k is an arbitrary constant. Of course, $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ is a trivial solution.

I'm sorry I have no idea what you mean here.

(A-kI)v=0

Expanding and rearranging we have:

Av = kv where k is an eigenvalue and v is the eigenvector corresponding to the eigenvalue.

v = $\begin{pmatrix} x \\ y \end{pmatrix}$ where we want to know what x and y are right?

We get the two equations:

x = 2x
x+2y=2y

Solve, you'll find that x must be 0 and y can be anything.

Do the same for

x = x

x+ 2y = y ( y=-x)

If you chose x=800 then y=-800 or $800 \begin{pmatrix} 1 \\ -1 \end{pmatrix}$


$\dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ are both normalised eigenvectors, that might be specified in the question.

Did you get different eigenvalues than I did, or did you just pick two for example? Mine where 2 and 1 which for k=1 gave me the equations : 0x+0y = x and x+y=y

Yeah, I got k=2 and 1 as well

I might be taking your matrix A to be different though

A = $\begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix}$

or

A = $\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}$? Are you multiplying matrix A with $\begin{pmatrix} x \\ y \end{pmatrix}$ correctly? where are the two 0s coming from in the first equation?

$A=\begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix}$

and then when I put the value 1 back into the matrix as k I get
$\begin{pmatrix} 1-1 & 0 \\ 1 & 2-1 \end{pmatrix}$

giving:

$\begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix}$

This is correct right?

You've got to add latex] or tex] at the start of latex statements and when you're done, finish with /latex] or /tex] (all the latex commands start with "[")

Yeah, I see what you've done now, so the equations you have now are

$\begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} =0$
giving x +y = 0 i.e. x = -y. You can now choose any value for x or y and the corresponding y or x value. The easiest being 1 giving $v_1 = \begin{pmatrix} 1\\ -1 \end{pmatrix}$ They have then normalised this vector.

For the eigenvalue k=2 we have $\begin{pmatrix} -1 & 0 \\ 1 & 0 \end{pmatrix}$ giving -x=0 and x=0 and no conditions for y i.e. y can be anything but x must be 0. And so we have $v_2 = \begin{pmatrix} 0 \\ y \end{pmatrix}$ The most straightforward being y=1 as y=0 gives the trivial zero vector. With y =1, this is already a normalised eigenvector.

Ah thank you, your explanation of the first answer makes perfect sense to me. So basically I want to simplify the equations I get into x=y and then just give a value to x and see what the corresponding y value is. How does this work if I have 2 equations, or will they both be multiples of eachother? ie x=y and 4x=4y kind of thing?
The answer given in my notes for the first one is, as you showed me [1:-1] however its multiplied by 1/sqrt(2) why is this .... and for k=2 it has been given as [0:1] as you said, thank you

Yeah, if there are two equations then you just solve for x and y like you would for standard simultaneous equations, if one of the equations was x=y then yes the other could be a multiple of it.

It's multiplied by 1/sqrt2 to 'normalise' the eigenvector. A normalised vector is a vector of unit magnitude. You normalise a vector $\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ ... \\ x_n \end{pmatrix}$ by dividing each element by the magnitude of the vector, which is $\sqrt{(x_1)^2 + (x_2)^2 + (x_3)^2 + ... + (x_n)^2}$

So the eigenvector $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ has a magnitude of $\sqrt{(1)^2 + (-1)^2}$ Dividing each element of the eigenvector by this you get $\dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix}$

Thank you so so much for all that help! I missed the lecture on it and have been trying to learn it on my own but with your help I not understand it. When would I know to normalise the vector or not? It doesn't say in the question, would I just always do it?

Thank you