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Diagonalization eigenvalues

by hpayandah
Tags: algebra, diagonalization, eigenvalue, linear
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hpayandah
#1
Jul17-11, 06:14 PM
P: 18
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
A7-Answers-2.jpg  
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NEGATIVE_40
#2
Jul17-11, 08:40 PM
P: 23
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
#3
Jul18-11, 12:18 AM
P: 18
Hi, thanks for replying. Attached is how I got my vectors, do you think my steps are correct.
Attached Thumbnails
Untitled-1.jpg  

vela
#4
Jul18-11, 04:33 AM
Emeritus
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HW Helper
Thanks
PF Gold
P: 11,690
Diagonalization eigenvalues

No. Why do you think (0, 0, 1) is an eigenvector? You seem to just pull that out of thin air.
NEGATIVE_40
#5
Jul18-11, 06:37 AM
P: 23
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]


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