Similarity Transformations

Definition. The $n \times n$ matrices $\mathsf{A}$ and $\mathsf{B}$ are similar/conjugate if for some invertible matrix $\mathsf{P}$

\[\mathsf{B} = \mathsf{P^{-1}AP}\]

Definition. A similarity transformation is a map from $\mathsf{A} \mapsto \mathsf{P^{-1}AP}$.

Proposition. Similarity is an equivalence relation.

Proposition. The identity matrix (or a multiple of it) is similar only with itself (or a multiple of it).

Property. Similar matrices have the same determinant.

\[\begin{align*} \det(\mathsf{P^{-1}AP}) &= \det \mathsf{P}^{-1} \det \mathsf{A} \det \mathsf{P} \\ &= \det \mathsf{A} \det(\mathsf{P^{-1}P}) \\ &= \det \mathsf{A} \end{align*}\]

Property. Similar matrices have the same trace.

\[\begin{align*} Tr(\mathsf{P^{-1}AP}) &= \mathsf{P}^{-1}_{ij}\mathsf{A}_{jk}\mathsf{P}_{ki} \\ &= \mathsf{A}_{jk}\mathsf{P}_{ki}\mathsf{P}^{-1}_{ij} \\ &= \mathsf{A}_{jk} \delta_{kj} \\ &= Tr(\mathsf{A}) \end{align*}\]

From the above, the determinants and traces of matrices representing a map $\mathcal{A}$ with respect to different bases are the same. Therefore, the determinant and trace of a map is well-defined.

Property. Similar matrices have the same characteristic polynomial and hence the same eigenvalues.

\[\begin{align*} p_{\mathsf{B}}(\lambda) &= \det(\mathsf{B} - \lambda \mathsf{I}) \\ &= \det(\mathsf{P^{-1}AP} - \lambda \mathsf{P^{-1}IP}) \\ &= \det(\mathsf{P}^{-1}) \det(\mathsf{A} - \lambda \mathsf{I}) \det(\mathsf{P}) \\ &= \det(\mathsf{A} - \lambda \mathsf{I}) \det(\mathsf{P}^{-1}P) \\ &= p_{\mathsf{A}}(\lambda) \end{align*}\]

Therefore, the characteristic polynomial of linear map is well-defined.

Definition. The characteristic polynomial of a linear map $\mathcal{A}$ is the polynomial

\[p_{\mathcal{A}}(\lambda) = \det(\mathsf{A} - \lambda\mathsf{I})\]

where $\mathsf{A}$ is any matrix representation of $\mathcal{A}$.

Change of Basis

Every linear transformation is represented by a matrix depending on the choice of basis. Similar matrices actually represent the same transformation with respect to different bases.

Definition. Let $\Set{\mathbf{e}_i}$ and $\Set{\mathbf{e}_j’}$ be two sets of basis vectors of $\mathbb{F}^n$. Since $\Set{\mathbf{e}_i}$ is a basis, we can represent individual $\mathbf{e}_j’$ as

\[\mathbf{e}_j' = \sum_{i=1}^n P_{ij} \mathbf{e}_i\]

for some numbers $P_{ij}$.

The transformation matrix $\mathsf{P}$ is a square $n \times n$ matrix containing the numbers $P_{ij}$, i.e.

\[\mathsf{P} = \begin{pmatrix} P_{11} & P_{12} & \cdots & P_{1n} \\ P_{21} & P_{22} & \cdots & P_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ P_{n1} & P_{n2} & \cdots & P_{nn} \\ \end{pmatrix} = \begin{pmatrix} \mathbf{e}_1' & \mathbf{e}_2' & \cdots & \mathbf{e}_n' \end{pmatrix}\]

Similarily, we can also represent individual $\mathbf{e}_i$ as

\[\mathbf{e}_i = \sum_{k=1}^n Q_{ki} \mathbf{e}_k'\]

and hence the transformation matrix $\mathsf{Q}$ is

\[\mathsf{Q} = \begin{pmatrix} Q_{11} & Q_{12} & \cdots & Q_{1n} \\ Q_{21} & Q_{22} & \cdots & Q_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ Q_{n1} & Q_{n2} & \cdots & Q_{nn} \\ \end{pmatrix} = \begin{pmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \cdots & \mathbf{e}_n \end{pmatrix}\]

Property. The transformation matrices are inverses of each other, i.e.

\[\mathsf{PQ} = \mathsf{QP} = \mathsf{I}\]

Proof.

By substitution, we have

\[\mathbf{e}_j' = \sum_{i=1}^n P_{ij} \left( \sum_{k=1}^n Q_{ki} \mathbf{e}_k' \right) = \sum_{k=1}^n \mathbf{e}_k' \left( \sum_{i=1}^n Q_{ki} P_{ij} \right)\]

As the set $\Set{\mathbf{e}_j’}$ is a basis and so linearly independent, we have

\[\mathbf{e}_j' = \sum_{k=1}^n \mathbf{e}_j' \delta_{kj}\]

Hence,

\[\sum_{i=1}^n Q_{ki} P_{ij} = \delta_{kj}\]

and $\mathsf{QP} = \mathsf{I}$. Similarily, we can derive $\mathsf{PQ} = \mathbf{I}$ by a similar substitution.

Proposition. Consider a vector $\mathbf{u}$, we can represent it in component form respect to bases $\Set{\mathbf{e}_i}$ and $\Set{\mathbf{e}_j’}$, i.e.

\[\mathbf{u} = \sum_{i=1}^n u_i \mathbf{e}_i = \sum_{j=1}^n u_j' \mathbf{e}_j' = \sum_{j=1}^n u_j' \sum_{i=1} P_{ij} \mathbf{e}_i = \sum_{i=1}^n \mathbf{e}_i \sum_{j=1}^n P_{ij} u_j'\]

Therefore, we have

\[\mathsf{u} = \mathsf{Pu'} \quad \text{or} \quad \mathsf{u'} = \mathsf{P^{-1}u}\]

From the above, we can see that if $\mathsf{B} = \mathsf{P^{-1}AP}$, the transformation $\mathsf{B}\mathbf{v} = \mathsf{P^{-1}AP}\mathbf{v}$ consists of changing the basis by $\mathsf{P}$, applying the same transformation $\mathsf{A}$, and then changing it back by $\mathsf{P}^{-1}$. Therefore, $\mathsf{A}$ and $\mathsf{B}$ are the same transformation with respect to different bases.

References