Let $V$ be a vector space, $\dim(V)=5$.
Let be $T:v \rightarrow V$ a linear operator that verifies:
- $T$ is not invertible
- $\dim(\operatorname{Ker}(T+4I))=2$
- $\dim(\operatorname{Im}(T-I))=3$
The associated matrix of $T$ in any basis of $V$ is denoted $A$.
Check the correct option:
A) $A$ is diagonalizable and $\operatorname{tr}(A)=-3$
B) $A$ is diagonalizable and $\operatorname{tr}(A)=-6$
C) $A$ is diagonalizable and $\operatorname{tr}(A)=-5$
D) $A$ is not diagonalizable and $\operatorname{tr}(A)=0$
Well, I don't really know where to begin. I suspect the solution has to do with a lot of properties and so on but I'm not seeing it. Any kind of help will be appreciated.
The given information is sufficient for finding the dimensions of all (three) eigenspaces of $T$. Note that $v \in \ker(T+4I)$ just means $Tv = -4Iv = -4v$, so $\ker(T+4I)$ is the eigenspace of $T$ belonging to the eigenvalue $-4$. Since its dimension is $2$ we can find two linearly independent eigenvectors $v_1,v_2$ for the eigenvalue $-4$.
Of course we also have that $\ker(T-I)$ is the eigenspace of $T$ belonging to the eigenvalue $1$, and by rank-kernel-theorem $$ \dim(\ker(T-I)) = \dim(V)-\dim(Im(T-I))= 5-3 = 2. $$ Then we can also find two linearly independent eigenvectors $v_3,v_4$ belonging to eigenvalue $1$.
Finally, $T$ is not invertible, implying the existence of $0 \ne v_5 \in V$ s.t. $Tv_5=0$. Which is simply an eigenvector with corresponding eigenvalue $0$.
Then $(v_1,v_2,v_3,v_4,v_5)$ form a basis of $V$ and the associated matrix of $T$ with respect to this basis is $diag(-4,-4,1,1,0)$.
So $A$ is diagonalisable, and trace is invariant under change of basis, so $trc(A)=-4-4+1+1=-6$.