- (a)
- \(A^{-1} \sim B^{-1}\) (provided that the inverses exist),
- (b)
- \(A^T \sim B^T\), and
- (c)
- \(A^k \sim B^k\) for all integers \(k \geq 1\).
(See Theorem 5)
Similar Matrices
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.
Let \(B = P^{-1}AP\) for some invertible matrix \(P\).
For th:properties_similar_det, \(\det B = \det (P^{-1}) \det A \det P = \det A\) because \(\det (P^{-1}) = 1/ \det P\) (Theorem th:detofinverse).
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.
As for th:properties_similar_char_poly,
\begin{align*} \det (B-\lambda I) &= \det \{P^{-1}AP-\lambda (P^{-1}P)\} \\ &= \det \{ P^{-1}(A-\lambda I)P\} \\ &= \det (A-\lambda I), \end{align*}
so \(A\) and \(B\) have the same characteristic equation. Finally, th:properties_similar_char_poly implies th:properties_similar_eig because the eigenvalues of a matrix are the roots of its characteristic polynomial.
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:
See Practice Problem prob:similarproperties.
Practice Problems
Prove that if \(A\) and \(B\) are \(n\times n\) matrices and \(A\sim B\), then
Let
\[A=\begin{bmatrix}1 & 2 & 0\\2 & -1 & -2\\0 & -2 & 1\end{bmatrix}\]
The eigenvalues of \(A\) are known to be \(\lambda _1=-3\), \(\lambda _2=3\) and \(\lambda _3=1\). Find the corresponding
eigenspaces.
\[\mathcal {S}_{-3}=\text {span}\left (\begin{bmatrix}\answer {-1}\\\answer {2}\\1\end{bmatrix}\right )\]
\[\mathcal {S}_3=\text {span}\left (\begin{bmatrix}\answer {-1}\\\answer {-1}\\1\end{bmatrix}\right )\]
\[\mathcal {S}_1=\text {span}\left (\begin{bmatrix}\answer {1}\\\answer {0}\\1\end{bmatrix}\right )\]
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 \(P^{-1}\).
\[P^{-1}=\frac {1}{6}\begin{bmatrix}\answer {-1} & \answer {2} & \answer {1}\\\answer {-2} & \answer {-2} & \answer {2}\\\answer {3} & \answer {0} & \answer {3}\end{bmatrix}\]
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 \(\lambda \).
Text Source
The text in this section is a compilation of material from Section 7.2.1 of Ken Kuttler’s A First Course in Linear Algebra (CC-BY) and Section 5.5 of Keith Nicholson’s Linear Algebra with Applications (CC-BY-NC-SA).
Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, p. 362-364.