# A conceptual doubt about characteristic polynomial and eigenvalues

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!

• 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... Oct 15 '20 at 6:53
• Where is this $11$ coming from? Oct 15 '20 at 7:12
• Sorry! You got the reason I did badly my computes. Sorry! Oct 15 '20 at 7:16

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.
• 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. Oct 15 '20 at 11:43