1. ## Diagonalize the matrix

$\textsf{ Diagonalize the matrix:}$
$$A=\left[\begin{array}{rrrr} 2& 2& -1\\1& 3& -1\\-1& -2& 2 \end{array}\right]$$
$\textit{ a. Charateristic equation:}$
$$A-\lambda I= \left[\begin{array}{rrrr} 2& 2& -1\\1& 3& -1\\-1& -2& 2 \end{array}\right]-\left[\begin{array}{rrrr} \lambda& 0& 0\\0& \lambda& 0\\0& 0& \lambda \end{array}\right]$$
$$-\lambda^3+7\lambda^2-\lambda+5$$
$\textit{ b.$D=$,$P=$}$
$$v_1=\begin{bmatrix}-1\\-1\\1\end{bmatrix} v_2=\begin{bmatrix}1\\0\\1\end{bmatrix} v_3=\begin{bmatrix}-2\\1\\0\end{bmatrix}$$
$$\therefore P=(v_1,v_2,v_3)\begin{bmatrix} \,\, 1&1&0\\-1&0&1\\ \,\,1&1&0 \end{bmatrix}$$
$\textit{ c. Verify}\\$
ok I got the Characteristic equation and $v_1,v_2,v_3$ from W|A
but didn't understand the example?
And didn't know how to verify it

2. ## Re: Diagonalize the matrix

you haven't done any of it.

come up with matrices, $P, ~D$ such that

$PDP^{-1} = A$

$P$ is the matrix of normalized eigenvectors as columns

$D$ is a diagonal matrix of associated eigenvalues, i.e. element $d_{1,1}$ is the eigenvalue corresponding to column 1 of P, etc.

3. ## Re: Diagonalize the matrix

Originally Posted by bigwave
$\textsf{ Diagonalize the matrix:}$
$$A=\left[\begin{array}{rrrr} 2& 2& -1\\1& 3& -1\\-1& -2& 2 \end{array}\right]$$
$\textit{ a. Charateristic equation:}$
$$A-\lambda I= \left[\begin{array}{rrrr} 2& 2& -1\\1& 3& -1\\-1& -2& 2 \end{array}\right]-\left[\begin{array}{rrrr} \lambda& 0& 0\\0& \lambda& 0\\0& 0& \lambda \end{array}\right]$$
$$-\lambda^3+7\lambda^2-\lambda+5$$
$\textit{ b.$D=$,$P=$}$
$$v_1=\begin{bmatrix}-1\\-1\\1\end{bmatrix} v_2=\begin{bmatrix}1\\0\\1\end{bmatrix} v_3=\begin{bmatrix}-2\\1\\0\end{bmatrix}$$
$$\therefore P=(v_1,v_2,v_3)\begin{bmatrix} \,\, 1&1&0\\-1&0&1\\ \,\,1&1&0 \end{bmatrix}$$
$\textit{ c. Verify}\\$
ok I got the Characteristic equation and $v_1,v_2,v_3$ from W|A
but didn't understand the example?
And didn't know how to verify it
Look At This.

You may get some very useful verification.

4. ## Re: Diagonalize the matrix

The point is that a 3 by 3 matrix, A, is diagonalizable if and only all 3 eigenvectors are independent so form a basis for $R^3$. In that case, A is similar to the diagonal matrix having the eigenvalues of A on its diagonal. When you were finding the eigenvectors of A didn't you find the corresponding eigenvalues? Saying that A is similar to a diagonal matrix, D, means that there exist an invertible matrix, P, Such that $PAP^{-1}= D$. And P can be constructed by taking the eigenvectors of A as columns.

5. ## Re: Diagonalize the matrix

The point is that a 3 by 3 matrix, A, is diagonalizable if and only all 3 eigenvectors are independent so form a basis for $R^3$. In that case, A is similar to the diagonal matrix having the eigenvalues of A on its diagonal. When you were finding the eigenvectors of A didn't you find the corresponding eigenvalues? Saying that A is similar to a diagonal matrix, D, means that there exist an invertible matrix, P, Such that PAP^{-1}= D. And P can be constructed by taking the eigenvectors of A as columns.

To "verify" this, actually do that multiplication.

6. ## Re: Diagonalize the matrix

I'm going to interject a quick question here.

In my work in Physics I usually don't bother with finding P, I simply find the eigenvalues and immediately write the diagonal matrix. Am I losing any generality by doing it this way?

-Dan

7. ## Re: Diagonalize the matrix

The characteristic polynomial you listed is wrong. $\displaystyle -\lambda^3+7\lambda^2-11\lambda+5 = -(\lambda-5)(\lambda-1)^2$ is the characteristic polynomial.
So, how did you find the correct eigenvectors? Setting your characteristic polynomial equal to zero would give only one real root, not three.

8. ## Re: Diagonalize the matrix

Originally Posted by topsquark
I'm going to interject a quick question here.

In my work in Physics I usually don't bother with finding P, I simply find the eigenvalues and immediately write the diagonal matrix. Am I losing any generality by doing it this way?

-Dan
If you find n distinct eigenvalues for a nxn matrix then you can immediately write down the diagonal matrix.

What do you do when you have fewer than n eigenvalues?

9. ## Re: Diagonalize the matrix

Actually the number of distinct eigenvalues is irrelevant. It is the number of independent eigenvectors that is important. Yes, if you have n distinct eigenvalues, for an n by n matrix, the matrix is diagonalizable because you necessarily have n independent eigenvectors- eigenvectors corresponding to distinct eigenvalues are independent. But if some or even all of the n eigenvalues are the same, there still might be n independent eigenvectors.

10. ## Re: Diagonalize the matrix

Originally Posted by Idea
If you find n distinct eigenvalues for a nxn matrix then you can immediately write down the diagonal matrix.

What do you do when you have fewer than n eigenvalues?
The usual. I construct the space with the degenerate eigenvalues and take an appropriate linear combinations.

-Dan

11. ## Re: Diagonalize the matrix

Once again, if an n by n matrix has fewer than n eigenvalues it might still have n independent eigenvectors and so be diagonalizable just as I said before. If there are fewer than n independent eigenvectors, then the matrix is not diagonalizable. One can, in that case, find the "Jordan normal form" where we have "block submatrices" with copies of the single eigenvalue on the diagonal and "1"s just above the diagonal. For example, if a matrix has
'1' as a double eigenvalue, 2 as a triple eigenvalue, and 3 as a single eigenvalue (so 6 eigenvalues and the matrix is 6 by 6) [b]and there exist only 1 eigenvector for each eigenvalue (not counting dependent vectors) then the Jordan Normal form is
$\displaystyle \begin{bmatrix}1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 3\end{bmatrix}$.

However if there exist two independent eigenvectors with eigenvalue 1 and 3 with eigenvalue 2, then the matrix is diagonalizable, to matrix
$\displaystyle \begin{bmatrix}1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 3 \end{bmatrix}$.

If there are other "combinations" of number of eigenvector, if, say, the was only one eigenvector corresponding to eigenvalue 1 and 2 corresponding too eigenvalue 2, then the matrix would be
$\displaystyle \begin{bmatrix}1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 3 \end{bmatrix}$.