Diagonalization of Eigenvalues: A Mistake in Homework Answer?

hpayandah
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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?
 

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

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

so for lamda = 2,
(A-2I)\vec{v}=\vec{0}
\begin{bmatrix}<br /> 1 &amp; 1 &amp; 0\\ <br /> 0 &amp; 0 &amp; 1\\ <br /> 1 &amp; 1 &amp; 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}<br />
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}<br />

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

Hope this helps.
 


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

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


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 &amp; 0 &amp; 0<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> v_1<br /> \end{bmatrix}<br /> = \begin{bmatrix}<br /> 0<br /> \end{bmatrix}<br />
 
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