) 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.

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

b. True. Suppose [tex]\mathbf P[/tex] is the matrix of the eigenvectors of [tex]\mathbf A[/tex], and [tex]\mathbf D[/tex] is the diagonal matrix of the eigenvalues of [tex]\mathbf A[/tex]:


[tex]\mathbf P=\begin{bmatrix}\mathbf v_1&\cdots&\mathbf v_n\end{bmatrix}[/tex]

[tex]\mathbf D=\begin{bmatrix}\lambda_1&&\\&\ddots&\\&&\lambda_n\end{bmatrix}[/tex]

Then

[tex]\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}[/tex]

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

c. False. A counterexample is the matrix

[tex]\begin{bmatrix}1&1\\0&1\end{bmatrix}[/tex]

which is nonsingular, but it has only one eigenvalue.

d. False. Consider the matrix

[tex]\begin{bmatrix}0&1\\0&0\end{bmatrix}[/tex]

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