Similarity Transformation¶

Definition¶

Let $A$ and $B$ be real $n\times n$ matrices. $B$ is similar to $A$, written $A\sim B$$B$, if there is an invertible $n\times n$ matrix $P$ with

The map $A\mapsto P^{-1}AP$ is a similarity transformation. Some texts write $PAP^{-1}$ instead; this only replaces $P$ by $P^{-1}$.

Interpretation¶

Similar matrices represent the same linear operator in different bases. If the columns of $P$ form a basis $\mathcal{B}$ and $A$ is the matrix in standard coordinates, then $P^{-1}AP$ is the matrix in $\mathcal{B}$-coordinates: convert to standard coordinates ($P$), apply $A$, convert back ($P^{-1}$). Here $P$ is the transformation matrix of $\mathcal{B}$, written $F$ on the change of basis page.

The square uses the diagonalization example below. Going up with $P$, across with $A$ and down with $P^{-1}$ gives the same result as going across the bottom with $B$; this is the identity $B=P^{-1}AP$$P^{-1}AP$. Both rows describe one operator, so its eigenvalues $5$ and $2$ appear in both: as the roots of the characteristic polynomial of $A$ and as the diagonal entries of $B$.

Everything that describes the operator itself, rather than its coordinates, is therefore unchanged by similarity: eigenvalues, determinant, trace, rank. Eigenvectors are not invariant; they change coordinates along with everything else.

Two misconceptions. Equal invariants do not prove similarity (see the example below). And similarity is not row equivalence: $EA$ for an invertible $E$ generally has different eigenvalues from $A$.

Key Results¶

• Similarity is an equivalence relation on $n\times n$ matrices.
• Similar matrices have the same characteristic polynomial, hence the same eigenvalues with the same algebraic multiplicities; geometric multiplicities, determinant, trace and rank also agree.
• If $A\boldsymbol{v}=\lambda\boldsymbol{v}$ and $B=P^{-1}AP$$P^{-1}AP$, then $B(P^{-1}\boldsymbol{v})=\lambda\,P^{-1}\boldsymbol{v}$$\lambda P^{-1}\mathbf{v}$.
• $B^k=P^{-1}A^kP$ for all $k\ge0$$0$, and more generally $p(B)=P^{-1}p(A)P$ for polynomials $p$; likewise $\mathrm{e}^{Bt}=P^{-1}\mathrm{e}^{At}P$ for the matrix exponential.
• $A$ is diagonalizable if and only if it is similar to a diagonal matrix. Over $\mathbb{C}$, two matrices are similar if and only if they have the same Jordan normal form up to the order of the blocks.
• A real symmetric matrix is orthogonally similar to a diagonal matrix: $Q^{\top}AQ=\Lambda$ with $Q^{\top}=Q^{-1}$$Q^{-1}$.

Examples¶

Diagonalizing. $A=\begin{pmatrix}4&1\\2&3\end{pmatrix}$ has trace $7$ and determinant $10$, so eigenvalues $5$ and $2$, with eigenvectors $(1,1)^{\top}$ and $(1,-2)^{\top}$. With

$P^{-1}AP=\tfrac{1}{3}\begin{pmatrix}15&0\\0&6\end{pmatrix}=\operatorname{diag}(5,2)$$\mathrm{diag}(5,2)$. This is a diagonalization of $A$, and consequently $A^k=P\operatorname{diag}(5^k,2^k)P^{-1}$$P\mathrm{diag}(5^k,2^k)P^{-1}$.

Same invariants, not similar. $I_2$ and $J=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ share characteristic polynomial $(\lambda-1)^2$$1{)}^2$, trace $2$ and determinant $1$. But $P^{-1}I_2P=I_2\neq J$ for every invertible $P$, so they are not similar. An invariant that tells them apart is $\operatorname{rank}(M-I_2)$$I_2)$, which is $0$ for $M=I_2$ and $1$ for $M=J$$J$, because $P^{-1}MP-I_2=P^{-1}(M-I_2)P$$I_2)P$.

Prerequisites¶

• Inverse and Singular Matrices
• Eigenvalue and Eigenvector
• Characteristic Polynomial
• Change of Basis

Related Concepts¶

• Diagonalization
• Jordan Normal Form
• Matrix Exponential
• Singular Value Decomposition

Sources¶

• Math V: Advanced Linear Algebra · Week 6

Aliases: similar matrices, matrix similarity, conjugation by an invertible matrix, similarity transform Keywords: B = P^{-1} A P, similarity invariants, same characteristic polynomial, A^k = P D^k P^{-1}, orthogonal similarity Symbols: complex numbers, lambda

Last updated: 22 October 2026 19:04