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Let be the matrix:\begin{align*} A=\begin{pmatrix} -3 & -1 & 2\\ 2 & 1 & -1\\ -3 & -1 & 2 \end{pmatrix} \end{align*}

Prove that $\lambda=0$ is the unique eigenvalue of A and find an eigenvector associated $w_1 \in Ker(A)$. Also prove that Dim(Ker(A))=1.

To try to prove it, what I did is to calculate the characteristic polynomial of $A$, and I got this:

$det (A-tI)=11(t+3)(t-1)(t-2) \Longrightarrow $ the eigenvalues should be $-3, 1$ and $2$. Neverthless I have to prove that the unique eigenvalue of A is zero, so I suspect that I'm understanding something conceptually bad, so what am I understanding or doing wrong? Intuitively, I think it might be related with the fact that the columns of A are linear combinations of each other, but I don't know how it impacts. I would really appreciate your pacience and help!

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    $\begingroup$ Your computation of the characteristic polynomial is wrong. For starters, minus the LHS should be a monic polynomial in $t$, but that doesn’t hold for the RHS... $\endgroup$
    – Mindlack
    Oct 15 '20 at 6:53
  • $\begingroup$ Where is this $11$ coming from? $\endgroup$
    – Berci
    Oct 15 '20 at 7:12
  • $\begingroup$ Sorry! You got the reason I did badly my computes. Sorry! $\endgroup$
    – luisegf
    Oct 15 '20 at 7:16
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I got the characteristic polynomial to be -$\lambda^{3}$. Obviously the only solution to $-\lambda^{3} = 0$ is when $\lambda = 0$. Make sure you're doing your determinant right. Look up the cofactor expansion formula and try again.

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  • $\begingroup$ Well, not every vector except the zero vector is an eigenvector, take $(1,0,0)$ for example. The eigenspace has obviously dimension $1$. As the sum of the first and third column is the negative second column, $(1,-1,1)$ is a basis of the eigenspace. $\endgroup$ Oct 15 '20 at 11:43

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