# Diagonalization eigenvalues

1. Jul 17, 2011

### hpayandah

1. The problem statement, all variables and given/known data

I think my teacher made a mistake in his homework answer. I need to verify this for practice. The answer I got is below. The answer the teacher has is in the pdf.

2. Relevant equations

3. The attempt at a solution

So there is two eigenvalues= 4 and 2
but the eigenvalue 2 has 2 eigenvectors [-1 1 0]T and [0 0 1]T but my teacher has only one [-1 1 0]T. That's why he says A is not diagonalizable. Do you think it's correct?

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2. Jul 17, 2011

### NEGATIVE_40

Re: Diagonalization

I get the same result as your teacher for $$\lambda = 2$$

$$A= \begin{bmatrix} 3 & 1 & 0\\ 0 & 2 & 1\\ 1 & 1 & 3 \end{bmatrix}$$

so for lamda = 2,
$$(A-2I)\vec{v}=\vec{0}$$
$$\begin{bmatrix} 1 & 1 & 0\\ 0 & 0 & 1\\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix}$$
I've personally always found it easier to not do row operations here and just jump straight in. From that you get
$$v_3 = 0$$ and $$v_1 = -v_2$$
so therefore $$\vec{v} = \begin{bmatrix} 1\\ -1\\ 0 \end{bmatrix}$$

Since it is a 3x3 matrix, it needs 3 eigenvectors to be diagonalizable.

Hope this helps.

3. Jul 18, 2011

### hpayandah

Re: Diagonalization

Hi, thanks for replying. Attached is how I got my vectors, do you think my steps are correct.

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4. Jul 18, 2011

### vela

Staff Emeritus
Re: Diagonalization

No. Why do you think (0, 0, 1) is an eigenvector? You seem to just pull that out of thin air.

5. Jul 18, 2011

### NEGATIVE_40

Re: Diagonalization

You seem to have made a mistake in the step
$$(A-2I)\vec{v}=\vec{0}$$
Why is your second row 1 0 1 instead of 0 0 1 ?
Have you said $$v_2= \delta$$ ? I'm not sure if I have read that correctly. If it is a delta, you can't say that unless the whole row equals zero. i.e.
$$\begin{bmatrix} 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \end{bmatrix} = \begin{bmatrix} 0 \end{bmatrix}$$