Similarity Transformation

Definition.

Matrices \(\,\boldsymbol{A},\,\boldsymbol{B}\in M_n(C)\,\) are similar if there exists a matrix \(\ \boldsymbol{P}\in M_n(C)\ \) such that

\[\boldsymbol{B}\ =\ \boldsymbol{P}^{-1} \boldsymbol{A}\,\boldsymbol{P}\,.\]

Then the matrix \(\,\boldsymbol{B}\,\) is related with \(\,\boldsymbol{A}\,\) by a similarity transformation.

Theorem 4. \(\ \) Characteristic polynomials of similar matrices are equal.

Namely, if \(\ \boldsymbol{B}\,=\, \boldsymbol{P}^{-1} \boldsymbol{A}\,\boldsymbol{P},\ \ \ \boldsymbol{A},\boldsymbol{B},\boldsymbol{P}\in M_n(C),\ \det\boldsymbol{P}\neq 0\,,\ \) then \(\ \ w_{\boldsymbol{B}}(\lambda) = w_{\boldsymbol{A}}(\lambda)\,,\ \) where \(\ w_{\boldsymbol{X}}(\lambda) = \det(\boldsymbol{X}-\lambda\,\boldsymbol{I}_n)\,\) with \(\ \boldsymbol{I}_n\ \) an identity matrix of size \(n\) denotes a characteristic polynomial of matrix \(\boldsymbol{X} = \boldsymbol{A},\,\boldsymbol{B}\,.\)

Proof bases on Cauchy’s theorem on determinant of a product of matrices:

\[\begin{split}\begin{array}{rl} w_{\boldsymbol{B}}(\lambda) \!\! & =\ \ \det{(\boldsymbol{B}-\lambda\,\boldsymbol{I}_n)}\ = \\ & =\ \ \det{(\boldsymbol{P}^{-1}\boldsymbol{A}\,\boldsymbol{P} - \boldsymbol{P}^{-1}\lambda\,\boldsymbol{I}_n\ \boldsymbol{P})}\ = \\ & =\ \ \det{[\,\boldsymbol{P}^{-1} (\boldsymbol{A}-\lambda\,\boldsymbol{I}_n)\, \boldsymbol{P}\,]}\ = \\ & =\ \ \det{\boldsymbol{P}^{-1}}\cdot\ \det{(\boldsymbol{A}-\lambda\,\boldsymbol{I}_n)}\ \cdot\ \det{\boldsymbol{P}}\ = \\ & =\ \ (\det{\boldsymbol{P}})^{-1}\cdot\ \det{(\boldsymbol{A}-\lambda\,\boldsymbol{I}_n)}\,\cdot\, \det{\boldsymbol{P}}\ = \\ & =\ \ \det{(\boldsymbol{A}-\lambda\,\boldsymbol{I}_n)}\ \ =\ \ w_{\boldsymbol{A}}(\lambda)\,. \end{array}\end{split}\]

Corollary. Similar matrices have the same set of eigenvalues. The corresponding eigenvalues have the same algebraic multiplicities.

Theorem 5. \(\\\) The corresponding eigenvalues of similar matrices have the same geometric multilplicity.

Proof. \(\\\) Let \(\ \lambda\in C\ \) be an eigenvalue of the matrix \(\ \boldsymbol{A}\in M_n(C)\ \) with geometric mutliplicity \(\,k\). Then this eigenvalue is associated with \(\,k\,\) linearly independent eigenvectors \(\ \boldsymbol{x}_i\,\) of the matrix \(\ \boldsymbol{A}:\)

(1)\[ \begin{align}\begin{aligned}\boldsymbol{A}\,\boldsymbol{x}_i\ =\ \lambda\,\boldsymbol{x}_i\,, \quad i=1,2,\ldots,k\,;\\\sum_{i=1}^k\,\alpha_i\,\boldsymbol{x}_i\ =\ \boldsymbol{0} \quad\Rightarrow\quad\alpha_i=0,\ \ i=1,2,\ldots,k\,.\end{aligned}\end{align} \]

Let \(\ \boldsymbol{B}\,=\, \boldsymbol{P}^{-1}\boldsymbol{A}\,\boldsymbol{P},\ \) \(\ \boldsymbol{y}_i\,=\,\boldsymbol{P}^{-1}\,\boldsymbol{x}_i,\ \ i=1,2,\ldots,k\,,\ \) \(\ \boldsymbol{P}\in M_n(C),\, \det{\boldsymbol{P}}\neq 0.\ \) Then

\[ \begin{align}\begin{aligned}\boldsymbol{B}\,\boldsymbol{y}_i\ =\ (\boldsymbol{P}^{-1}\boldsymbol{A}\,\boldsymbol{P})\, (\boldsymbol{P}^{-1}\boldsymbol{x}_i)\ =\ \boldsymbol{P}^{-1}\boldsymbol{A}\,(\boldsymbol{P}\, \boldsymbol{P}^{-1})\ \boldsymbol{x}_i\ =\\\quad =\ \boldsymbol{P}^{-1}(\boldsymbol{A}\,\boldsymbol{x}_i)\ =\ \boldsymbol{P}^{-1}(\lambda\,\boldsymbol{x}_i)\ =\ \lambda\,(\boldsymbol{P}^{-1}\boldsymbol{x}_i)\ =\ \lambda\,\boldsymbol{y}_i\,,\\\sum_{i=1}^k\,\alpha_i\,\boldsymbol{y}_i\ =\ \boldsymbol{0} \quad\Leftrightarrow\quad \sum_{i=1}^k\,\alpha_i\,(\boldsymbol{P}^{-1}\,\boldsymbol{x}_i)\ =\ \boldsymbol{0} \quad\Leftrightarrow\quad \boldsymbol{P}^{-1}\,\sum_{i=1}^k\,\alpha_i\,\boldsymbol{x}_i\ =\ \boldsymbol{0} \quad\Leftrightarrow\quad\\\Leftrightarrow\qquad\sum_{i=1}^k\,\alpha_i\,\boldsymbol{x}_i\ =\ \boldsymbol{0} \quad\Rightarrow\quad \alpha_i=0,\ \ i=1,2,\ldots,k.\end{aligned}\end{align} \]

Hence, the matrix \(\ \boldsymbol{B}\ \) satisfies the conditions analogous to (1):

\[ \begin{align}\begin{aligned}\boldsymbol{B}\,\boldsymbol{y}_i\ =\ \lambda\,\boldsymbol{y}_i\,, \quad i=1,2,\ldots,k\,;\\\sum_{i=1}^k\,\alpha_i\,\boldsymbol{y}_i\ =\ \boldsymbol{0} \quad\Rightarrow\quad\alpha_i=0,\ \ i=1,2,\ldots,k\,,\end{aligned}\end{align} \]

which means that \(\ \lambda\ \) is an eigenvalue of the matrix \(\ \boldsymbol{B}\ \) with the same (as the matrix \(\,\boldsymbol{A})\) geometric multiplicity \(\,k\).

Theorem 6. \(\\\) Similar matrices have the same determinant and trace, and are of the same rank, i.e., if \(\ \boldsymbol{B}\,=\, \boldsymbol{P}^{-1}\boldsymbol{A}\,\boldsymbol{P},\ \) \(\boldsymbol{A},\boldsymbol{B},\boldsymbol{P}\in M_n(C), \ \det{\boldsymbol{P}}\neq 0,\ \ \) then

\[\det{\boldsymbol{B}}\,=\det{\boldsymbol{A}},\quad \text{Tr}\,{\boldsymbol{B}}\,=\,\text{Tr}\,{\boldsymbol{A}},\quad \text{rk}\,{\boldsymbol{B}}\,=\,\text{rk}\,{\boldsymbol{A}}\,.\]

First two equalities follow from a previous theorem on equality of characteristic polynomials of similar matrices. Namely, \(\ \det{\boldsymbol{A}},\,\det{\boldsymbol{B}}\ \) are constant terms (coefficients of \(\ \lambda^0\)), \(\ \) and \(\ \text{Tr}\,\boldsymbol{A},\,\text{Tr}\,\boldsymbol{B}\ \) are coefficients of the term \(\ \lambda^{(n-1)}\ \) in characteristic polynomials of matrices \(\ \boldsymbol{A},\,\boldsymbol{B}\). Equality of polynomials implies equality of the coefficients of the terms having the same power \(\ \lambda.\)

These relations may be also proved directly from properties of determinant and trace of a matrix. Cauchy’s theorem on determinant of a product of matrices implies that

\[ \begin{align}\begin{aligned}\det{\boldsymbol{B}}\ =\ \det{(\boldsymbol{P}^{-1}\boldsymbol{A}\,\boldsymbol{P})}\ =\\=\ \det{\boldsymbol{P}^{-1}}\cdot\ \det{\boldsymbol{A}}\ \cdot\ \det{\boldsymbol{P}}\ =\\=\ (\det{\boldsymbol{P}})^{-1}\cdot\ \det{\boldsymbol{A}}\ \cdot\ \det{\boldsymbol{P}}\ =\ \det{\boldsymbol{A}}\,,\end{aligned}\end{align} \]

and reordering cyclicly the factors under the trace symbol, we obtain

\[\text{Tr}\,\boldsymbol{B}\ =\ \text{Tr}\,(\boldsymbol{P}^{-1}\boldsymbol{A}\,\boldsymbol{P})\ =\ \text{Tr}\,(\boldsymbol{A}\,\boldsymbol{P}\,\boldsymbol{P}^{-1})\ =\ \text{Tr}\,\boldsymbol{A}\,.\]

Equality of ranks of similar matrices follows from the fact that multiplication of a given matrix by a square non-degenerate matrix (on the left or on the right) does not change its rank:

\[\text{rk}\,\boldsymbol{A}\ =\ \text{rk}\,(\boldsymbol{A}\,\boldsymbol{P})\ =\ \text{rk}\,(\boldsymbol{P}^{-1}\boldsymbol{A}\,\boldsymbol{P})\ =\ \text{rk}\,{\boldsymbol{B}}\,.\]