Show that if $ \operatorname{rank}(A) = \operatorname{rank}(A^{2}) $, then the algebraic and geometric multiplicities of $ \lambda = 0 $ are equal.

Question

Let $A \in M_n$ and suppose $0 \in \delta(A)$.Prove that if $\text{Rank} (A)= \text{Rank}(A^2)$ then the algebraic and geometric multiplicity of $\lambda =0$ are equal.

I am not sure how to prove this.

In the Jordan form the algebraic multiplicity of $\lambda=0$ is the sum of zeros along the diagonal

In the Jordan form the geometric multiplicity is the number of Jordan block with eigenvalue 0.

Answer 1

One always have that the sum of the dimensions of the kernel and the image of (the linear map with matrix) $A$ equals the dimension $n$ of the space (the size of the matrix); that is the rank-nullity theorem. Now the hypothesis that the ranks of $A$ and $A^2$ are equal means that the images of $A$ and $A^2$ are equal: obviously the former contains the latter, but these subspaces also have the same dimension. Call this image subspace $W$. We can restate what we just said by saying that the restriction $W\to W$ of (the linear map with matrix) $A$ to$~W$ is surjective (the image of the restriction is the image of $A^2$, which is $W$). A surjective linear operator on a finite dimensional space is also injective, which here means that the kernel $\ker(A)\cap W$ of the restriction is$~\{0\}$. Together with the rank-nullity fact we conclude $V=\ker(A)\oplus W$.

We can understand $A$ according to this direct sum decomposition: on the summand $\ker(A)$ it acts as $0$, while on the summand $W$ it acts in an invertible manner, so $0$ is not an eigenvalue of the action on$~W$. The geometric multiplicity of $\lambda=0$ is $d=\dim(\ker(A))$ (since $\ker(A)$ is the eigenspace for $\lambda=0$). For the algebraic multiplicity one needs the characteristic polynomial of$~A$, which by the direct sum decomposition is the product of the characteristic polynomials of the restrictions to both summands. The one for $\ker(A)$ is $X^d$, and the one for $W$ is not divisible by $X$ since $0$ is no eigenvalues there. This makes $d$ the algebraic multiplicity of the eigenvalue $\lambda=0$ of$~A$ too.

Answer 2

The dimension of the solution set of $A\mathbf{x}=0$ is $n-\text{rank} A$, where $n$ is the size of the matrix. Therefore your assumption implies that $$A\mathbf{x}=0\Leftrightarrow A^2\mathbf{x}=0.$$

Now

$$\text{geometric multiplicity}=\dim \{\mathbf{x}|A\mathbf{x}=0 \ \}$$ $$\text{algebraic multiplicity}=\dim \{\mathbf{x}|A^n\mathbf{x}=0 \ \text{some} \ n\}$$

Thus we easily see that they are the same.