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Is it possible? I know that for a matrix to be diagonalizable it needs sufficient linearly independent eigenvectors to make up invertible matrix P, but does that mean B itself must also be invertible?

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    $\begingroup$ Yes it is. For example take the matrix $[1 \ 0; 0 \ 0]$. $\endgroup$ – AnyAD Dec 12 '18 at 3:48
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A matrix $A$ is said to be diagonalizable, if there exists a diagonal matrix $D$ and an invertible matrix $P$ such that $A = PDP^{-1}$.

Note that $A$ is invertible if and only if $D$ is invertible. Therefore, the question boils down to : are all diagonal matrices invertible?

The answer is no, simply because a diagonal matrix is invertible if and only if every diagonal element is invertible. But if some diagonal elements are zero, then the diagonal matrix will not be invertible. Therefore, neither will $A$.

For example, consider the $2 \times 2$ matrix $E_{11}$, having a $1$ at the position $11$ and zero elsewhere. Then, $E_{11}$ is a diagonal non-invertible matrix, as the diagonal entry $(E_{11})_{22} = 0$.

Now, $A = D$, or $A=PDP^{-1}$ for any invertible $P$ serves as a counterexample.

Also, note that the zero matrix is diagonalizable, in fact diagonal, but not invertible. This would be the most obvious example.

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Let $$B= \begin{bmatrix} 0 & 0 \\ 0 & 0\end{bmatrix}.$$

It is a diagonal matrix and it is diagonalizable, we can let $P$ be any non-singular matrices, but it is not invertible.

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Yes it is possible. Say we have a diagonal matrix $D=\mathrm {diag}(2,0)$, then $D$ is not invertible. Let $P$ be some invertible matrix, say $$ P = \begin{bmatrix} a & b \\ -b & a \end{bmatrix} $$ where $a^2 +b^2 \neq 0$, then $A = PDP^{-1}$ works.

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