Similarity Transformation

Definition.

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

$$\mathit{B}={\mathit{P}}^{-1}\mathit{A}\mathit{P}.$$

Then the matrix $\mathit{B}$ is related with $\mathit{A}$ by a similarity transformation.

Theorem 4. $$ Characteristic polynomials of similar matrices are equal.

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

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

$$\begin{array}{r}\begin{array}{rl}{w}_{\mathit{B}}(\lambda )& =det(\mathit{B}-\lambda {\mathit{I}}_n)=\\ & =det({\mathit{P}}^{-1}\mathit{A}\mathit{P}-{\mathit{P}}^{-1}\lambda {\mathit{I}}_n\mathit{P})=\\ & =det[{\mathit{P}}^{-1}(\mathit{A}-\lambda {\mathit{I}}_n)\mathit{P}]=\\ & =det{\mathit{P}}^{-1}\cdot det(\mathit{A}-\lambda {\mathit{I}}_n)\cdot det\mathit{P}=\\ & =(det\mathit{P}{)}^{-1}\cdot det(\mathit{A}-\lambda {\mathit{I}}_n)\cdot det\mathit{P}=\\ & =det(\mathit{A}-\lambda {\mathit{I}}_n)=w_{\mathit{A}}(\lambda ).\end{array}\end{array}$$

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 $\mathit{A}\in M_n(C)$ with geometric mutliplicity $k$. Then this eigenvalue is associated with $k$ linearly independent eigenvectors ${\mathit{x}}_i$ of the matrix $\mathit{A}:$

(1)$$\begin{array}{r}\begin{array}{c}\mathit{A}{\mathit{x}}_i=\lambda {\mathit{x}}_i,i=1,2,\dots ,k;\\ \sum _{i=1}^k{\alpha }_i{\mathit{x}}_i=\mathbf{0}\Rightarrow {\alpha }_i=0,i=1,2,\dots ,k.\end{array}\end{array}$$

Let $\mathit{B}={\mathit{P}}^{-1}\mathit{A}\mathit{P},$ ${\mathit{y}}_i={\mathit{P}}^{-1}{\mathit{x}}_i,i=1,2,\dots ,k,$ $\mathit{P}\in M_n(C),det\mathit{P}\ne 0.$ Then

$$\begin{array}{r}\begin{array}{c}\mathit{B}{\mathit{y}}_i=({\mathit{P}}^{-1}\mathit{A}\mathit{P})({\mathit{P}}^{-1}{\mathit{x}}_i)={\mathit{P}}^{-1}\mathit{A}(\mathit{P}{\mathit{P}}^{-1}){\mathit{x}}_i=\\ ={\mathit{P}}^{-1}(\mathit{A}{\mathit{x}}_i)={\mathit{P}}^{-1}(\lambda {\mathit{x}}_i)=\lambda ({\mathit{P}}^{-1}{\mathit{x}}_i)=\lambda {\mathit{y}}_i,\\ \sum _{i=1}^k{\alpha }_i{\mathit{y}}_i=\mathbf{0}\iff \sum _{i=1}^k{\alpha }_i({\mathit{P}}^{-1}{\mathit{x}}_i)=\mathbf{0}\iff {\mathit{P}}^{-1}\sum _{i=1}^k{\alpha }_i{\mathit{x}}_i=\mathbf{0}\iff \\ \iff \sum _{i=1}^k{\alpha }_i{\mathit{x}}_i=\mathbf{0}\Rightarrow {\alpha }_i=0,i=1,2,\dots ,k.\end{array}\end{array}$$

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

$$\begin{array}{r}\begin{array}{c}\mathit{B}{\mathit{y}}_i=\lambda {\mathit{y}}_i,i=1,2,\dots ,k;\\ \sum _{i=1}^k{\alpha }_i{\mathit{y}}_i=\mathbf{0}\Rightarrow {\alpha }_i=0,i=1,2,\dots ,k,\end{array}\end{array}$$

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

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

$$det\mathit{B}=det\mathit{A},\text{Tr}\mathit{B}=\text{Tr}\mathit{A},\text{rk}\mathit{B}=\text{rk}\mathit{A}.$$

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

$$\begin{array}{r}\begin{array}{c}det\mathit{B}=det({\mathit{P}}^{-1}\mathit{A}\mathit{P})=\\ =det{\mathit{P}}^{-1}\cdot det\mathit{A}\cdot det\mathit{P}=\\ =(det\mathit{P}{)}^{-1}\cdot det\mathit{A}\cdot det\mathit{P}=det\mathit{A},\end{array}\end{array}$$

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

$$\text{Tr}\mathit{B}=\text{Tr}({\mathit{P}}^{-1}\mathit{A}\mathit{P})=\text{Tr}(\mathit{A}\mathit{P}{\mathit{P}}^{-1})=\text{Tr}\mathit{A}.$$

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:

$$\text{rk}\mathit{A}=\text{rk}(\mathit{A}\mathit{P})=\text{rk}({\mathit{P}}^{-1}\mathit{A}\mathit{P})=\text{rk}\mathit{B}.$$