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Show that $A$ and $A^T$ have the exact same eigenvalues and that for each eigenvalue we have $\dim (N(A-\lambda I)) = \dim (N(A^T-\lambda I)) $

So this proof has basically two parts.

The first part went well I showed that the determinant $|A-\lambda I|=|A^T-\lambda I|$ and because of that $A$ and $A^T$ will have the characteristic polynomial and so they will have the same eigenvalues (with the same algebraic multiplication).

The second part is driving me a bit confused.Proving that $\dim (N(A-\lambda I)) = \dim (N(A^T-\lambda I))$ is equivalent to proove that the eigenspace of $A$ has the same dimension of the eigenspace of $A^T$.

I thought about the property "the geometric multiplicity of an eigenvector gives us the dimension of the eigenspace" but how do I know the eigenvalues have the same geometric multiplicity (we only guarantee that they have the algebraic multiplicity).

Is there any other property I should consider?

Hope someone can help me finishing

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  • $\begingroup$ Also you can think in terms of Jordan form of matrices. Let $J$ is Jordan form of $A$. Then $J=PAQ $ for some P and Q. This implies that $J^T=Q^TA^TP^T.$ Further $J$ and $J^T$ are similar via following matrix: $$\begin{bmatrix} 0&0&\cdots&0&1\\0&0&\cdots&1&0\\ \vdots&\vdots&\cdots& \vdots & \vdots\\ 0&1&\cdots&0&0 \\ 1&0&\cdots&0&0\end{bmatrix}.$$ $\endgroup$
    – Sry
    Commented May 27, 2016 at 12:55

2 Answers 2

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This is because of the Rank nullity theorem: as a matrix and its transpose have the same rank, we have $$\DeclareMathOperator\rk{rank} \rk(A-\lambda I)=\rk{}{}^\mathrm t\mkern-1.5mu(A-\lambda I)\iff \dim \ker(A-\lambda I)=\dim\ker{}{}^\mathrm t\mkern-1.5mu(A-\lambda I).$$

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Let $B$ be an $n\times n$-matrix and let $k:=\dim N(B)$. Then the row-echelon form has all zeroes in its last $k$ rows, so $B^{\top}$ has all zeroes in its last $k$ columns, meaning that $B^{\top}e_i=0$ for the last $k$ basis vectors $e_{n-k+1},\ldots,e_n$. Hence $\dim N(B^{\top})\geq\dim N(B)$ holds for all square matrices $B$. Then $$\dim N(B)\leq\dim N(B^{\top})\leq\dim N((B^{\top})^{\top})=\dim N(B),$$ which shows that equality holds for all $B$. In particular $$\dim N(A-\lambda I)=\dim N((A-\lambda I)^{\top})=\dim N(A^{\top}-\lambda I),$$ holds for all $A$ and all eigenvalues $\lambda$.

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