Q:

Finding kernel and image of a linear transformation, given eigenvalues and eigenvectors?

Problem: Consider the linear transformation $T : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ with eigenvalues $-1, 0$ and $1$ and the corresponding eigenvectors $v_1, v_2, v_3$. Determine $\text{ker}(T)$ and $\text{Im}(T)$.
Attempt at solution: In general we have $D = P^{-1} A P$, where $D$ is a diagonal matrix with the eigenvalues, $P$ is the matrix with columns the corresponding eigenvectors, and $A$ is the matrixrepresentation of $T$ with respect to some basis. Hence $A = PDP^{-1}$. Now the kernel of $T$ is the nullspace of the matrix $A$, while the image of $T$ is the columnspace of $A$. I'm not sure how to solve this problem, because we are not given any specific eigenvectors. So how could we ever find $P^{-1}$? Since $0$ is an eigenvalue of $T$, $A$ is singular. But what does that say about the nullspace and columnspace of $A$?
Any help would be appreciated.

A:

Here's a possible strategy:
Note that the dimension of $\operatorname{im}(T)$ equals the rank of $A$ and hence the rank of $D$ since conjugate matices have same rank. What is the rank of $D$? Hence, what is the dimension of $\operatorname{im}(T)$? Given the eigenvectors corresponding to the distinct non-zero eigenvalues, what are obvious independent elements of $\operatorname{im}(T)$?
Further, how are the dimensions of $\operatorname{im}(T)$ and $\operatorname{ker}(T)$ related? Hence, given that $0$ is an eigenvalue, what is $\operatorname{ker}(T)$?

