# Show that matrix $A$ is NOT diagonalizable.

Let $A$ be a square matrix $A^2=0$ and $A\neq0$ and show that it is not diagonalizable.

I decided to use the sample matrix of $$A = \begin{bmatrix}0 & 1\\0 & 0 \end{bmatrix}$$ which satisfies the conditions above.

So my question is: how would I prove this is not diagonalizable. The matrix leads to a eigenvalue of $\lambda=0$ with an algebraic multiplicity of $2$.

I know that if the algebraic multiplicity and geometric multiplicity are equal, then it is diagonalizable.

But I am kind of stuck from here since when I use $\det(A-0I)=0$, it just leads to get $x_2=0$ but then also that $x_2=t$ so I don't really know what to do. Any help would be appreciated.

• If the $2\times 2$ matrix with only the eigenvalue zero had two linearly independent eigenvectors (geometric multiplicity two), that matrix would be the zero matrix. Jan 12, 2015 at 22:55

It isn’t enough to prove that your particular sample matrix isn’t diagonalizable: you must show that every non-zero square matrix $A$ such that $A^2=0$ is non-diagonalizable.

HINT: Suppose that $A^2=0$ and $A$ is diagonalizable. Then there are an invertible matrix $P$ and a diagonal matrix $D$ such that $D=P^{-1}AP$.

• What is $D^2$?
• What does this tell you about $A$?
• How does this prove the desired result?

The only eigenvalue is 0. If it's diagonalizable, it is similar to the zero matrix, so it's equal to the zero matrix. But it's not.

first the eigenvalues of $N = \pmatrix{0&1\\0&0}$ are $0,0$ if $N$ were digonalizable, then the diagonal matrix must be the zero matrix. so $UDU^-1$ will be the zero matrix too. therefore it cannot be $N$ and $N$ is not diagonalizable.

the same argument holds for any matrix $A \neq 0$ with $A^2 = 0$ reason is $A^2 = 0$ implies $0$ is the only eigenvalue.

If $$A^k=0$$ (in your case with $$k=2$$), then any eigenvalue of $$A$$ satisfies $$\lambda^k=0$$ (check what happens to an eigenvector) and therefore $$\lambda=0$$. Then if $$A$$ is diagonalisable, the only possibility is that $$A=0$$.

You have $det(A^2)=0$ then $det(A)=0$. This implies A has a zero eigenvalues which gives you that A is not diagonalizable