Diagonalization of Eigenvalues: A Mistake in Homework Answer?

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Homework Help Overview

The discussion revolves around the diagonalization of a matrix based on its eigenvalues and eigenvectors. The original poster questions the correctness of their teacher's answer regarding the diagonalizability of a matrix with eigenvalues 4 and 2, specifically focusing on the eigenvectors associated with the eigenvalue 2.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster presents their findings on eigenvalues and eigenvectors, questioning the teacher's assertion about diagonalizability. Some participants verify the teacher's result and discuss the implications of having fewer eigenvectors than the dimension of the matrix. Others challenge the validity of certain eigenvectors and the steps taken to derive them.

Discussion Status

Contextual Notes

Participants are working with a 3x3 matrix and discussing the requirement for three linearly independent eigenvectors for diagonalizability. There is mention of a pdf with the teacher's answer, which is not present in the discussion.

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.
 

Attachments

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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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