Show that arbitrary $A$ and $A^T$ have same eigenvalue, algebraic and geometric multiplicity

Question

Show that an arbitrary $n \times n$ matrix $A$ and its transpose $A^T$ have the same eigenvalues, algebraic multiplicity and geometric multiplicity.

I'm not sure if I did it correctly and especially how to show that they have same geometric multiplicity?

same eigen values

Assume $A$ and $A^T$ have same eigenvalues, then they have the same chracteristic polynomial. So we need to show that $p_A(\lambda)=\det(A-\lambda I)$ is same as $p_{A^T}(\lambda)=\det(A^T-\lambda I)$.

So we have $$p_{A^T}(\lambda)=\det(A^T-\lambda I) = \det(A^T-\lambda I^T) = \det\left((A-\lambda I)^T\right) = \det(A-\lambda I)=p_A(\lambda)$$

We see their characteristic polynomials are same so their eigenvalues are same as well.

same algebraic multiplicity

I'm not sure if this is a correct reason proof but: Because the characteristic polynomials are same, we have that the algebraic multiplicities of the eigenvalues of $A$ nd $A^T$ sre the same.

same geometric multiplicity

I don't know? :/

Answer 1

Geometric Multiplicity of $\lambda$ in $A=\text{rank} (A-\lambda I)=\text{rank }(A-\lambda I)^T=\text{rank } (A^T-\lambda I)=$Geometric multiplicity of $\lambda$ in $A^T$