how do we find eigenvectors of any matrix where detA =0?

- anonymous

how do we find eigenvectors of any matrix where detA =0?

- katieb

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- anonymous

i mean det (A -(lamda)I)= 0 say for example A = (1 2;-1 4)

- anonymous

i know the eigen values are lamda = 3 and lamda =2 but how to find eigenvectors

- anonymous

@KingGeorge

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## More answers

- KingGeorge

If you have a matrix with a determinant of 0, I'm not sure what to do. However, for your example, let \[A=\begin{bmatrix} 1&2\\-1&4\end{bmatrix}\]Then we look at the determinant of \(A-\lambda I_2\). So we have \[\det\left(\begin{bmatrix} 1&2\\-1&4\end{bmatrix}-\begin{bmatrix} \lambda&0\\0&\lambda\end{bmatrix}\right) \\
=\det\left(\begin{bmatrix} 1-\lambda&2\\-1&4-\lambda\end{bmatrix}\right) \\
=(1-\lambda)(4-\lambda)-(2)(-1)\]And set this equal to 0.

- anonymous

yeah i know how to find the eigenvalues completely like i am good at that but it asks what are the eigen vectors thats where i get confused

- KingGeorge

To be honest, I'm not too great at eigenvectors either.

- anonymous

hmm true they are a pain. like i am reading the theorem you have to sub it back into (A - (lamda)I)x =0 and then solve for x using those lamdas but i reduce my matrix down for both the lamdas and i get x1 = x2 but wth does that mean

- anonymous

You've pretty much solved one of the eigenvectors already. You just simplified it too much.
x1 = x2 means x1 - x2 = 0. What values of x1 and x2 will solve that equation? Eigenvectors are not unique, so there's infinitely many that will work in this case. Then, your eigenvector is [x1 x2]. Note: Eigenvalues are unique, so your eigenvector only corresponds to that eigenvalue, not both. You'll have to plug in the next eigenvalue to get your other eigenvector.

- KingGeorge

So if we have \(\lambda=2,3\), then we have \(A=2I_2 x\) or \(A=3I_2 x\). Let's start with 3. That gives us\[\begin{bmatrix} 1&2\\-1&4\end{bmatrix}\begin{bmatrix} x_1\\x_2\end{bmatrix}=3\begin{bmatrix} x_1\\x_2\end{bmatrix}\]Solving for this, we get that \(x_1=x_2\). So I guess that means that your eigenvector for \(\lambda=3\) is just any vector where both \(x_1\) and \(x_2\) are the same.

- KingGeorge

If you solve for \(\lambda=2\) instead, you get that \(x_1=2x_2\). So in this case, your eigenvector for \(\lambda=2\) is any vector where \(x_2\) is twice \(x_1\).

- anonymous

hmm ya thats what it said but i wasnt getting the right answer do they use any x values like we can use any number right?

- KingGeorge

You should be able to use any numbers you want, unless I'm grossly misunderstanding eigenvectors.

- anonymous

Right. So, in the example where lamda is equal to 3, you have x1 - x2 = 0. So, [x1 x2] could be [1, 1] or [3,3] or [sqrt(2),sqrt(2)]. The simpler the better though, in my opinion.

- anonymous

true for some reason they have [1/sqrt2 1/sqrt2] thats where i was going crazy in how they determined that

- KingGeorge

They wanted a normalized vector apparently (length of 1). Anyways, I've got to get going. Good luck.

- anonymous

ooo alright thanks a lot though

- anonymous

but @DanielxAK to get a normalized eigen vector we find the main eigenvector and normalize it?

- anonymous

Yes, if x is your vector, then to normalize it, you divide it by the norm:
x/||x||. The norm of x (||x||) is just the distance equation. (So, sqrt(x1^2 + x2^2) = ||x||)
Also, note that the normalized eigenvector is unique. That's probably one reason why they're asking you to normalize it.

- anonymous

oo true probably thanks a lot man, just another question do you know anything about symmetric matrices?

- anonymous

A bit. What's your question?

- anonymous

how do we show something is symmetric like i know A^T = A but its talking about some orthagonal stuff where A = CDC^T

- anonymous

Can you state the whole question? Is it asking you to show A^T = A when A = CDC^T? I'm guessing C is an orthogonal matrix and D is a diagonal matrix?

- anonymous

yeahh it isnt a question my TA just said just know something about that because it may be on the test

- anonymous

is this like a spectral decomposition?

- anonymous

Yes, sort of. It's a very specific type of decomposition. Matrices that have certain properties are easier to decompose. Decomposition is important in computing solutions to matrices. Some decompositions are more ideal than others. Anyway,
If A = CDC*, then A* = (CDC*)* = (C*)*D*C*= CD*C* = CDC* = A. (D* = D since D is diagonal).
Does that make sense?

- anonymous

hmm not quite

- anonymous

i didnt know (C*DC)* = (C*)*D*C* like i thought it was (AB)* = B*A* ?

- anonymous

Correct. But you have three matrices instead of two. So, in the case of two matrices, you're reversing the order. The same is true for three.

- anonymous

hmm true do you mind showing me an example of spectral decomposition if you dont mind ?

- anonymous

i have A = ( 3 6 -1; 6 9 4; -1 4 3)

- anonymous

I'm not going to work it out by hand. That'd be quite a bit of work for a 3x3 matrix. I did it on Matlab really quick though. So, you'd have it of the form:
A = C*D*C'
Where, roughly,
C = [ -0.4551 0.5802 -0.6754
-0.8460 -0.0451 0.5313
-0.2778 -0.8132 -0.5114]
D = [ 13.5417 0 0
0 3.9353 0
0 0 -2.4770]

- anonymous

ok sounds good and C= the normalized matrix for A?

- anonymous

C is your unitary (or orthogonal) matrix. Spectral decomposition is also known as eigenvalue decomposition. If you worked out the eigenvalues of that matrix, you would see that diagonal of D makes up your eigenvalues and each column of C makes up your normalized eigenvectors (which correspond to its eigenvalue).
For example, v =
-0.4551
-0.8460
-0.2778
corresponds to the eigenvalue 13.5417.

- anonymous

ahh okayy i have to keep on practicing thenn i apparently have to do this by hand tommorow lol

- anonymous

For a 3x3? I hope not. That's prone to error if done by hand. A 2x2 isn't too bad, but it gets much worse past that.

- anonymous

hmm truee just one final question, i have found the eigenvalues its just the vectors that throw me off

- anonymous

I have A = (3 1 1; 1 0 2; 1 2 0)

- anonymous

and i found the eigenvalues to be lamda1=1, lamda2=4, lamda3 = -2

- anonymous

and the question states that. Find the normalized eigenvectors and use them as columns in orthagonal matrix C

- anonymous

Alright. First, build your matrix D using your eigenvalues since you know those. Then, compute the eigenvector for each eigenvalue. Normalize the eigenvectors. Then, to build your C, match the column of C with the diagonal of D. So, the eigenvalue of your first diagonal of D should match your eigenvector which is the first column of C.

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