Exercise 15
For any square matrix \(A\), prove that \(A\) and \(A^t\) have the same characteristic polynomial (and hence the same eigenvalues).


Proof:

Suppose \(A\) is a square matrix of size \(n \times n\). By theorem 4.8, we know that for any matrix \(B \in M_{n\times n}\), \(\det(B) = \det(B^t)\). Therefore,

$$ \begin{align*} \det(A - \lambda I_n) &= \det((A - \lambda I_n)^t) \\ &= \det((A - \lambda I_n)^t) \\ &= \det(A^t - (\lambda I_n)^t) \\ &= \det(A^t - \lambda I_n). \end{align*} $$

From this, we see that both \(A\) and \(A^t\) have the same characteristic polynomial and therefore, they must have the same eigenvalues. \(\ \blacksquare\)

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