Let $T: \mathcal{P}_3 \to \mathcal{P}_3: a + bX + cX^2 \mapsto a + (2b + c)X + (b+2c)X^2$ be a linear map. I have to find the eigenvalues and the corresponding eigenspaces of $T$. First, I determined the matrix of $T$ with respect to the ordered base $\{1, X, X^2\}$, namely $$A = \begin{pmatrix}1&0&0 \\ 0&2&1 \\ 0&1&2\end{pmatrix}$$ Then I got $3$ (multiplicity $1$) and $1$ (multiplicity $2$) as eigenvalues. For the eigenspace of $3$, I have $V_3 = S[(0,1,1)]$ and for the eigenspace of $1$, I have $V_1 = S[(0,-1,1)]$. Well, now my question is if this is correct. When I fill in $3$ in the matrix $A - 3I_3$, I get the rank $2$, and in my book of linear algebra, I am given generally: $\dim V_\mu = n - \operatorname{rang}(A-\mu I_n)$. So for the eigenvalue $3$, I have no issues. But the dimension of $V_1$ is $1$, and for the rank of $A - 1I_3$ I got $1$, so according to that general formula, I must get $2$ as the dimension of $V_1$.
Thank you for your help...
Yes, the eigenvalues are 1, with multiplicity 2, and 3. The eigenvectors corresponding to eigenvalue 3 are multiples of $\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}$ so that eigenspace has dimension 1 as it should. The eigenvectors corresponding to eigenvalue 1 are linear combinations of $\begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}$ and $\begin{pmatrix}0 \\ 1 \\ -1 \end{pmatrix}$ so that eigenspace has dimension 2.