Diagonalization of Matrices: Confusion about Eigenvalues and Eigenvectors

[bon]

Homework Statement

Ok so

I have to construct a real symmetric matrix R whose eigenvalues are 2,1,-2 and who corresponding normalized eigenvectors are bla bla bla..

So let the matrix of eigenvalues down diagonal be E and matrix of eigen vectors be V

Is R = VEV^T or R = V^TEV??

How am i meant to know..? Different books are saying different things!

In my notes it says that for unitary U, A = Udagger A' U where A' is the diagonalised matrix, but elsewhere it says the opposite! ahhhh

Homework Equations

The Attempt at a Solution

 

[fzero]

Since $$VEV^T$$ is real and symmetric, it's actually equal to $$V^TEV$$. [STRIKE]The point is that if a matrix S diagonalizes a matrix M, then so does S^-1.[/STRIKE]

Edit: Last sentence is untrue.

 

[bon]

The point is that if a matrix S diagonalizes a matrix M, then so does S^-1.

Thanks. Is this true for any matrix S and M?

Can you recommend and good books on this/websites?

Thanks again!

 

[fzero]

Thanks. Is this true for any matrix S and M?

Sorry, I was actually wrong about that. You need to assemble the eigenvectors into the matrix S in a particular way, and that will determine whether S^-1 M S or S M S^-1 is diagonal in the most general case.

Can you recommend and good books on this/websites?

Thanks again!

I'm not so familiar with current texts, but I found this thread:

https://www.physicsforums.com/showthread.php?t=429607&highlight=linear+algebra+text

which has some online texts.

 

[HallsofIvy]

If E is a diagonal matrix with eigenvalues on the main diagonal and V is the matrix with corresponding eigenvectors as columns, then $R= VEV^{-1}$ is a matrix having those eigenvalues and eigenvectors. If you choose the eigenvectors to be "orthonormal" (each has length 1 and is perpendicular to the others) then $V^{-1}= V^T$ so $R= VEV^T$.

 

[bon]

Sorry, I was actually wrong about that. You need to assemble the eigenvectors into the matrix S in a particular way, and that will determine whether S^-1 M S or S M S^-1 is diagonal in the most general case.
.

I'm trying to understand this a bit better..

so let M' be the diagonalised matrix of eigenvectors.. obviously A^T = A if it is diagonal

if the eigenvectors that make up M are orthonormal, then surely either S^-1MS or SMS^-1 will work..

If the eigenvectors are not orthonormal, which one will be the one that gives the diagonalized version of M and why?

Thanks

 

[bon]

If E is a diagonal matrix with eigenvalues on the main diagonal and V is the matrix with corresponding eigenvectors as columns, then $R= VEV^{-1}$ is a matrix having those eigenvalues and eigenvectors. If you choose the eigenvectors to be "orthonormal" (each has length 1 and is perpendicular to the others) then $V^{-1}= V^T$ so $R= VEV^T$.

Thanks. Sorry, I thought I'd just quote you too in case you are able to help with my above question..

Thanks.

 

[fzero]

I'm trying to understand this a bit better..

so let M' be the diagonalised matrix of eigenvectors.. obviously A^T = A if it is diagonal

if the eigenvectors that make up M are orthonormal, then surely either S^-1MS or SMS^-1 will work..

If the eigenvectors are not orthonormal, which one will be the one that gives the diagonalized version of M and why?

Thanks

I'll try to explain, but I'm not sure that I remember a really clean way to get this. Let $$\{ \mathbf{v}_i \}$$ be an orthonormal basis for M with associated eigenvalues $$\lambda_i$$. Let S be the matrix that has the $$\mathbf{v}_i$$ as its columns, so

$$S = \begin{pmatrix} \mathbf{v}_1 & \mathbf{v}_2 & \cdots & \mathbf{v}_n \end{pmatrix}.$$

By construction$$S^{-1} = S^T = \begin{pmatrix} \mathbf{v}_1^T \\ \mathbf{v}_2^T \\ \vdots \\ \mathbf{v}_n^T \end{pmatrix},$$

which is the matrix with the $$\mathbf{v}_i^T$$ as its rows.

Then

$$M S = \begin{pmatrix}\lambda_1 \mathbf{v}_1 & \lambda_2\mathbf{v}_2 & \cdots &\lambda_n \mathbf{v}_n \end{pmatrix}$$

and

$$S^T M S = \text{diag}~(\lambda_1, \ldots, \lambda_n) \equiv \Lambda.$$

From this we conclude that, with the given choice of S,

$$M = S \Lambda S^T,$$

while

$$S^T \Lambda S = M^T.$$

In the case of your original question, R was symmetric, so these were equal. Also note that $$S M S^T$$ has no obvious special form.

 

[bon]

thanks a lot!