Diagonalizability of a matrix $A$.

Question

Let $A=\begin{bmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&0&1&1\\ 0&0&0&1\\ \end{bmatrix}$.

Then is A diagonalizable?

My attempt: Matrix A is a Jordan matrix consisting of Jordan blocks corresponding to eigen values $0$ and $1$. Clearly nullity of A is $2$ and hence geometric multiplicity of $0$ is $2.$ Also geometric and algebraic multiplicity of $1$ is $2$. Now as A.M=G.M. corresponding to each e.v. hence A is diagonalizable. Is the above reasoning correct?

One more question: If one of the block of a block diagonal matrix is not diagonalizable then is matrix not diagonalizable?

Answer 1

The geometric multiplicity of the eigenvalue $1$ is $1$. If you look for solutions to $Ax=Ix$, you will find only $\operatorname{span}[0,0,1,0]^T$, as $$ A\begin{bmatrix}a\\b\\c\\d\end{bmatrix}=\begin{bmatrix}0\\0\\c+d\\d\end{bmatrix} $$ yielding the equations $a=0, b=0$ and $c=c+d$ (as well as the "useless" $d=d$).

The matrix $A$ (or rather the associated linear transformation) does as shear transformation on the plane spanned by the 3rd and 4th basis vectors. And a shear transformation is not diagonalizable.

For the extra question, yes, if the matrix is block diagonal, then it is diagonalizable iff every block is. This is because the "blockization" is already a partial diagonalization. You are allowed to study the spaces affected by each block separately, as they do not interact.

Not for general block matrices, though. In particular, we know symmetric matrices are always diagonalizable, but it isn't difficult to construct a symmetric matrix with a non-diagonalizable block.