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Diagonalization eigenvalues 
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#1
Jul1711, 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? 


#2
Jul1711, 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] (A2I)\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. 


#3
Jul1811, 12:18 AM

P: 18

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



#4
Jul1811, 04:33 AM

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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.



#5
Jul1811, 06:37 AM

P: 23

You seem to have made a mistake in the step
[tex] (A2I)\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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