Why is this matrix diagonalizable?

Question

$$ A= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$

I know that is symmetric, but to diagonalize it we want to invert the matrix for $$P^{-1}AP=D$$ But we can't find the determinant of matrix A because $$\frac{1}{det(A)}$$ the denominator is zero.

Why is matrix A diagonalizable?

Answer 1

It is already a diagonal matrix. So P is the identity matrix.

Answer 2

We have $P^{-1}AP=D$ with $D=A$ and $P=I_3$

$A$ is a diagonal matrix, hence diagonalizable.

Answer 3

Let $v$ a vector in $\mathbb{R}^3$, then

$$ A v = 0 v $$

So $0$ is an eigenvalue of $A$. The eigenvalues of $A$ are all zeros, while the eigenvectors are arbitrary, you can use the vectors of the base so that $P = I_3$