Question

) a, p and d are n×n matrices. check the true statements below:
a. if a is diagonalizable, then a has n distinct eigenvalues.
b. if ap=pd, with d diagonal, then the nonzero columns of p must be eigenvectors of
a.
c. if a is invertible, then a is diagonalizable.
d. a is diagonalizable if a has n distinct eigenvectors.

Answer 1

a. False. Consider the identity matrix, which is diagonalizable (it's already diagonal) but all its eigenvalues are the same (1).

b. True. Suppose $\mathbf P$ is the matrix of the eigenvectors of $\mathbf A$, and $\mathbf D$ is the diagonal matrix of the eigenvalues of $\mathbf A$:


$\mathbf P=\begin{bmatrix}\mathbf v_1&\cdots&\mathbf v_n\end{bmatrix}$

$\mathbf D=\begin{bmatrix}\lambda_1&&\\&\ddots&\\&&\lambda_n\end{bmatrix}$

Then

$\mathbf{AP}=\begin{bmatrix}\mathbf{Av}_1&\cdots&\mathbf{Av}_n\end{bmatrix}=\begin{bmatrix}\lambda_1\mathbf v_1&\cdots&\lambda_n\mathbf v_n\end{bmatrix}=\mathbf{PD}$

In other words, the columns of $\mathbf{AP}$ are $\mathbf{Av}_i$, which are identically $\lambda_i\mathbf v_i$, and these are the columns of $\mathbf{PD}$.

c. False. A counterexample is the matrix

$\begin{bmatrix}1&1\\0&1\end{bmatrix}$

which is nonsingular, but it has only one eigenvalue.

d. False. Consider the matrix

$\begin{bmatrix}0&1\\0&0\end{bmatrix}$

with eigenvalue $\lambda=0$ and eigenvector $\begin{bmatrix}k&0\end{bmatrix}^\top$, where $k\in\mathbb R$. But the matrix can't be diagonalized.