91Ó°ÊÓ

Firstly, let's establish that the eigenvalues of matrix \(A\) and its transpose \(A^{t}\) are the same, meaning that their characteristic polynomials are equal. To prove this, we need to show that: \[ |A-\lambda I|=|A^{t}-\lambda I| \] Since the determinant of a matrix is equal to the determinant of its transpose, we have: \[ |A-\lambda I|=|(A-\lambda I)^{t}| \] Now, using the properties of transposes, we can simplify: \[(A-\lambda I)^{t}=A^{t}-(\lambda I)^{t}\] \[(A-\lambda I)^{t}=A^{t}-\lambda I\] Thus we have: \[ |A-\lambda I|=|A^{t}-\lambda I| \] Which shows that \(A\) and \(A^{t}\) have the same eigenvalues.

Now that we know the eigenvalues of \(A\) and \(A^t\) are the same, let's consider their eigenspaces. For an eigenvalue \(\lambda\) of both \(A\) and \(A^t\), the eigenspace of \(A\) corresponding to eigenvalue \(\lambda\) can be represented as: \[ E_{A}(\lambda) = \{x \in \mathbb{R}^{n} | Ax = \lambda x\} \] Similarly, the eigenspace of \(A^t\) corresponding to eigenvalue \(\lambda\) can be represented as: \[ E_{A^t}(\lambda) = \{x \in \mathbb{R}^{n} | A^tx = \lambda x\} \]

Recall that we need to show: \(\operatorname{rank}\left((A-\lambda I)^{r}\right)=\operatorname{rank}\left(\left(A^{t}-\lambda I\right)^{r}\right)\) We can see that: \(\operatorname{rank}\left((A-\lambda I)^{r}\right) = \operatorname{rank}\left(((A-\lambda I)^{t})^{r}\right) = \operatorname{rank}\left(\left(A^{t}-\lambda I\right)^{r}\right)\) According to the Rank-Nullity Theorem, \[\operatorname{rank}\left((A-\lambda I)^{r}\right) + \operatorname{nullity}\left((A-\lambda I)^{r}\right) = n\] \[\operatorname{rank}\left(\left(A^{t}-\lambda I\right)^{r}\right) + \operatorname{nullity}\left(\left(A^{t}-\lambda I\right)^{r}\right) = n\] Since the rank and nullity of \((A-\lambda I)^{r}\) and \(\left(A^{t} - \lambda I\right)^{r}\) both add up to the same value, it must be the case that: \(\operatorname{rank}\left((A-\lambda I)^{r}\right)=\operatorname{rank}\left(\left(A^{t}-\lambda I\right)^{r}\right) \)

By showing that the rank of \((A-\lambda I)^{r}\) and \(\left(A^{t}-\lambda I\right)^{r}\) are equal for any eigenvalue \(\lambda\) and positive integer \(r\), we have proved that the Jordan canonical forms of \(A\) and \(A^t\) are the same. This implies that: Matrix \(A\) and its transpose \(A^t\) are similar.