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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? :/

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    $\begingroup$ Hint: Use the fact that $rank(A) = rank(A^T)$ and then use rank-nullity to deduce that $dim(ker(A-\lambda I)) = dim(ker(A^T-\lambda I))$ $\endgroup$ Commented May 28, 2018 at 8:08

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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 $A^T$

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    $\begingroup$ The geometric multiplicity of $\lambda$ is measured by the dimension of the kernel, not the rank. A line referencing rank-nullify to show that the dimensions of the kernels are equal would complete the proof. $\endgroup$ Commented May 28, 2018 at 8:09

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