Chapter 7 - Eigenvalues and Eigenvectors - 7.3 Diagonalization - Problems - Page 461: 30

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Since $A$ is similar to $B$, we have $B=S^{-1}AS$ then $\rightarrow B^T=(S^{-1}AS)^T\\ \rightarrow B^T=S^TA^T(S^{-1})^T\\ \rightarrow B^T=S^TA^T(S^T)^{-1}$ Since $S$ is invertible, $S^T$ is also invertible Hence, $A^T$ is similar to $B^T$