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If \(A\) and \(B\) are \(n \times n\) matrices, we say that \(A\) and \(B\) are similar, if \(B = P^{-1}AP\) for some invertible matrix \(P\). In this case we write \(A \sim B\).
Similar matrices share many properties.
If \(A\) and \(B\) are \(n\times n\) matrices and \(A\sim B\), then
1.
\(\det (A) = \det (B)\),
2.
\(\mbox {rank}(A) = \mbox {rank}(B)\),
3.
\(A\) and \(B\) have the same characteristic equations, and
4.
\(A\) and \(B\) have the same eigenvalues.
Let \(B = P^{-1}AP\) for some invertible matrix \(P\).
Similarly, for th:properties_similar_rank\(\mbox {rank} B = \mbox {rank}(P^{-1}AP) = \mbox {rank} A\), because multiplication by an invertible matrix cannot change the rank. To see this, note that any invertible
matrix is a product of elementary matrices. Multiplying by elementary matrices is equivalent to performing elementary row
(column) operations on \(A\), which does not change the row (column) space, nor the rank. It follows that similar matrices have the
same rank.
Sharing the four properties in Theorem th:properties_similar does not guarantee that two matrices are similar. The matrices \(A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) and \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) have the same
determinant, rank, characteristic polynomial, and eigenvalues, but they are not similar because \(P^{-1}IP = I\) for any invertible matrix \(P\).
The next theorem shows that similarity is preserved under inverses, transposes, and powers:
If \(A\) and \(B\) are \(n\times n\) matrices and \(A\sim B\), then
1.
\(A^{-1} \sim B^{-1}\) (provided that the inverses exist),
We will refer to the eigenvectors you listed above as \(\vec {x}_1\), \(\vec {x}_2\) and \(\vec {x}_3\).
Form matrix \(P\) whose columns are the eigenvectors \(\vec {x}_1\), \(\vec {x}_2\), and \(\vec {x}_3\)
you found in the previous part. Use technology to find \(P^{-1}\).
Find the product \(P^{-1}AP\). What did you get? Which of the following is true?
\(A\sim P\)\(A\sim D\) for ANY diagonal matrix \(D\)\(A\sim D\) for the diagonal
matrix \(D\) whose diagonal entries are the eigenvalues of \(A\)
Let \(\lambda \) be an eigenvalue of \(A\) with corresponding eigenvector \(\vec {x}\). If \(B = P^{-1}AP\) is similar to \(A\), show that \(P^{-1}\vec {x}\) is an eigenvector of \(B\) corresponding to \(\lambda \).