diagonalized matrices

**Jane99** · October 13th 2010, 06:19 PM

Hi guys, I'm hoping you can help me. I'm sure I've done something wrong, though I can't figure out what. The question I have to answer is this:

Find invertible U and diagonal D such that $A = UDU^{-1}$ . A is the matrix:

2, -4, 2
-4, 2, -2
2, -2, -1

I have found the three eigenvalues -2, -2, 7. I understand that these values, along the diagonal of a matrix otherwise filled with zeros, make up the D matrix.

I have also found the eigenvectors that go with the eigenvalues. Respectively they are [0,1,2], [1,0,-2], [1,-1,1/2]. I understand that these three vectors together make up U.

My problem is how to fit it all together. I don't understand what order I have to put the vectors in within U, or the values in within D, to prove that $A = UDU^{-1}$ . I've tried to brute force it but I'm not getting anywhere and it's making me think I've done everything wrong.

Can anyone explain my mistake, please?

**lvleph** · October 13th 2010, 06:23 PM

Define
$D = \begin{bmatrix} \lambda_1 & 0 & 0 & \cdots & 0 \\ 0 & \lambda_2 & 0 & \cdots & \vdots \\ 0 & 0 & \lambda_3 & \ddots & \vdots \\ \vdots & \cdots & \ddots & \ddots & 0 \\ 0 & \cdots &\cdots &0& \lambda_n\end{bmatrix}$
and then
$U = \begin{bmatrix} v_1 & v_2 & v_3 & \cdots & v_n\end{bmatrix}$
where $\lambda_i$ is the ith eigenvalue and $v_i$ is the corresponding eigenvector. This is called the Jordan Decomposition of A.

**Jane99** · October 13th 2010, 06:29 PM

Yes, but I don't understand which eigenvalue is lambda1, which is lambda2 etc. I'm sure I don't just pick randomly?

**lvleph** · October 13th 2010, 06:31 PM

Well as long as you are consistent it it doesn't matter (number of eigenvalues is meaningless anyway). You just have to make sure that column 1 of matrix U corresponds to the eigenvector for the first eigenvalue in D, etc.

**lvleph** · October 13th 2010, 06:49 PM

So look at it this way
$AU = UD$
$A \begin{bmatrix}v_1 & v_2 & v_3 & \cdots & v_n\end{bmatrix} = \begin{bmatrix}v_1 & v_2 & v_3 & \cdots & v_n\end{bmatrix} \begin{bmatrix} \lambda_1 & 0 & 0 & \cdots & 0 \\ 0 & \lambda_2 & 0 & \cdots & \vdots \\ 0 & 0 & \lambda_3 & \ddots & \vdots \\ \vdots & \cdots & \ddots & \ddots & 0 \\ 0 & \cdots &\cdots &0& \lambda_n\end{bmatrix}$
Which leads to
$\begin{bmatrix}\lambda_1 v_1 & \lambda_2 v_2 & \lambda_3 v_3 & \cdots & \lambda_n v_n\end{bmatrix} = \begin{bmatrix}\lambda_1 v_1 & \lambda_2 v_2 & \lambda_3 v_3 & \cdots & \lambda_n v_n\end{bmatrix}$ .
So if you rearrange things you get the same result, but in a different order.

**Jane99** · October 13th 2010, 06:58 PM

Ok, thanks. It's sounding like I did everything right then got cold feet at the end!

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