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We are given the matrix:

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

I want to find the eigenvalues

I did so by solving

$$|\lambda I - A| = 0 $$

\begin{align*} d &= \left|\begin{array}{ccc} \lambda -1 & -2 & 0 \\ -2 & \lambda -1 & 0 \\ 0 & 0 & \lambda - 1 \\\end{array}\right| \\ \end{align*}

$$\rightarrow -\lambda^3 - 3\lambda^2 - \lambda + 3 = (\lambda^2-1)(\lambda-3) = 0$$

This yields 2 eigenvalue, $\lambda = 1$ and $\lambda = 3$

Im not sure how to get my eigen vectors from there.

I think I have to replace lamda in the matrix I found earlier, to get new values.

I tryed with $\lambda = 1$

It gives me : $\begin{bmatrix}0 & -2 & 0\\ -2 & 0 & 0\\0 & 0 & 0\\\end{bmatrix}$

Which I re-arranged (swap $L1 L2$ and / both by -2) : $\begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\0 & 0 & 0\\\end{bmatrix}$

Now I get $v3 = t , v2 = 0, v1 = 0$

It gives me the vector $\begin{bmatrix}0 & 0 & 1\\\end{bmatrix}$

With $\lambda = 3$

It gives me : $\begin{bmatrix}2 & -2 & 0\\ -2 & 2 & 0\\0 & 0 & 2\\\end{bmatrix}$

Which I re-arranged (swap $L2+ L1$ and / all by 2 and swap L2-3) : $\begin{bmatrix}1 & -1 & 0\\ 0 & 0 & 1\\0 & 0 & 0\\\end{bmatrix}$

Now I get $v3 = 0 , v2 = t, v1 = v2 = t$

It gives me the vector $\begin{bmatrix}1 & 1 & 0\\\end{bmatrix}$

Here are my questions :

A) Are my eigen values correct

B) Are my eigen vectors correct

C) How do I go from there and diagonalize the matrix A ?

Can I actually do it with only 2 eigen vectors ?

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  • $\begingroup$ If you're certain of the characteristic polynomial (I didn't check) then there are three eigenvalues: $\;\pm1, 3\;$ $\endgroup$ – DonAntonio Dec 3 '13 at 21:55
  • $\begingroup$ Absolutly ! You're right. there goes my third eigen value and vector. $\endgroup$ – Dave Dec 3 '13 at 21:56
  • $\begingroup$ Ok, I checked: yes, the char. pol. is right and it has three real roots. $\endgroup$ – DonAntonio Dec 3 '13 at 21:56
  • $\begingroup$ Correct, and you left out $\lambda = -1$ which will give you a third eigenvector. To answer your general question, you cannot diagonalize a $3\times 3$ matrix unless you have three eigenvectors. $\endgroup$ – Doc Dec 3 '13 at 21:56
  • $\begingroup$ And might I add that it is very refreshing to see a person like you who shows all attempts to solve a problem, rather than just ask for a solution. Kudos. $\endgroup$ – Doc Dec 3 '13 at 21:58
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We are given:

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

The eigenvalues are found by solving $|A - \lambda I\ | = 0$, which yields:

$$-\lambda^3 + 3 \lambda^2 + \lambda -3 = -(\lambda-3) (\lambda-1) (\lambda+1)$$

So, we have three distinct eigenvalues, which means we can diagonalize this system.

To find the eigenvectors, we solve $[A - \lambda_i I]v_i = 0$, yielding (we just RREF this system):

  • $\lambda_1 = -1, v_1 = (-1,1,0)$
  • $\lambda_2 = 1, v_2 = (0,0,1)$
  • $\lambda_3 = 3, v_3 = (1,1,0)$

To diagonalize the matrix, we have:

$$J = P^{-1} A P$$

$P$ is formed using the column vectors $[v_1 | v_2 | v_3]$.

You should get:

$$J = \begin{bmatrix}-1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 3\\\end{bmatrix}$$

I purposely left the details out so you can work through them.

Notice that since we had three distinct eigenvalues, we could have written the diagonal matrix straight off.

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  • $\begingroup$ Great job! +1 $\uparrow$ $\endgroup$ – Namaste Dec 5 '13 at 0:20

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