Diagonalization

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

Please refer to attached pdf

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?

Attached Thumbnails

 

NEGATIVE_40

I get the same result as your teacher for [tex] \lambda = 2 [/tex]

[tex] A= \begin{bmatrix}
3 & 1 & 0\\
0 & 2 & 1\\
1 & 1 & 3
\end{bmatrix}
[/tex]

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

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.

Attached Thumbnails

 

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
[tex] (A-2I)\vec{v}=\vec{0} [/tex]
Why is your second row 1 0 1 instead of 0 0 1 ?
Have you said [tex] v_2= \delta [/tex] ? 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.
[tex] \begin{bmatrix}
0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
v_1
\end{bmatrix}
= \begin{bmatrix}
0
\end{bmatrix}
[/tex]