Does eigenvalue 2 have two eigenvectors, making the matrix diagonalizable?

[hpayandah]

Homework Statement

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.

Homework Equations

Please refer to attached pdf

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?

 

[NEGATIVE_40]

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

$$A= \begin{bmatrix}<br /> 3 & 1 & 0\\ <br /> 0 & 2 & 1\\ <br /> 1 & 1 & 3<br /> \end{bmatrix}$$

so for lamda = 2,
$$(A-2I)\vec{v}=\vec{0}$$
$$\begin{bmatrix}<br /> 1 & 1 & 0\\ <br /> 0 & 0 & 1\\ <br /> 1 & 1 & 1<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> v_1\\ <br /> v_2\\ <br /> v_3<br /> \end{bmatrix}<br /> = \begin{bmatrix}<br /> 0\\<br /> 0\\<br /> 0\\<br /> \end{bmatrix}$$
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 $$\vec{v} = \begin{bmatrix}<br /> 1\\ <br /> -1\\ <br /> 0<br /> \end{bmatrix}$$

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

Hope this helps.

 

[hpayandah]

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

 

[vela]

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

 

[NEGATIVE_40]

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.
$$\begin{bmatrix}<br /> 0 & 0 & 0<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> v_1<br /> \end{bmatrix}<br /> = \begin{bmatrix}<br /> 0<br /> \end{bmatrix}$$