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Two square matrices $A,B$ over the same field are similar if $B=S^{-1}AS$ for an invertible matrix $S$. They can describe one linear operator using two different ordered bases, with the same basis used for input and output in each description. Fix the convention $x=Sy$: $x$ is the old coordinate vector, $y$ the new coordinate vector, and the columns of $S$ are the new basis vectors in old coordinates. This is the convention of the change-of-basis matrix.^[ref]

For a concrete linear operator on the physical plane, let $T$ double horizontal components and multiply vertical components by five. In the old standard basis $\mathcal{E}=(e_1,e_2)$, its matrix of a linear map is $A=\mathrm{diag}(2,5)$. Choose the new ordered basis $\mathcal{F}=(b_1,b_2)$ with $b_1=e_1$, $b_2=e_1+e_2$. ThenThe second column of $B$ says $T(b_2)=-3b_1+5b_2=2e_1+5e_2$. Its off-diagonal entry records how the chosen basis vectors describe the output; the physical linear operator still stretches along the old horizontal and vertical directions.

Take the physical vector $p=2e_1+e_2=b_1+b_2$. Its two coordinate vectors, and the two descriptions of $T(p)$, areReconstruction gives $T(p)=4e_1+5e_2=-b_1+5b_2$. The output coordinate vectors differ, but the endpoint in the physical plane is the same.

This is why matrix similarity is useful: a convenient basis can simplify calculations without changing the linear operator. A diagonalizable matrix can be replaced by a diagonal matrix, making repeated application a calculation with powers of its eigenvalues. In the example, $A$ makes the two independent stretch factors visible, while $B$ expresses the same action in directions chosen for another purpose. Matrix similarity invariants separate properties of the linear operator from features of its address system.

Matrix similarity invariants are quantities unchanged by replacing $A$ with $S^{-1}AS$. Important examples are the characteristic polynomial, trace, determinant, matrix rank, and eigenvalues with their algebraic multiplicities. The characteristic polynomial convention here is ${\chi }_A(t)=\det (tI-A)$, where $I$ is the identity matrix.^[ref]

The short determinant proof works over any field: because $tI-B=S^{-1}(tI-A)S$, multiplicativity of the determinant givesThus its roots, the eigenvalues, retain their algebraic multiplicities in any field extension where the polynomial splits. Also $\det B=\det A$. The cyclic property of the trace gives $\mathrm{tr}(S^{-1}AS)=\mathrm{tr}(AS{S}^{-1})=\mathrm{tr}A$; equivalently, the trace is determined by the coefficient of $t^{n-1}$ in the characteristic polynomial for an $n$-by-$n$ matrix.^[ref]

For the two matrices in the opening example,Their entries differ, yet these quantities agree. Agreement is a useful necessary test for matrix similarity; it is not sufficient in general, as same eigenvalues do not imply similarity illustrates.

The relation also preserves repeated application. Adjacent factors $S{S}^{-1}$ cancel to giveFor the opening example, matrix powers therefore have the explicit formAt $k=2$, this is $\left(\begin{array}{cc}4& -21\\ 0& 25\end{array}\right)$. Acting on $y=(1,1{)}^T$ gives $(-17,25{)}^T$, which reconstructs as $S(-17,25{)}^T=(8,25{)}^T=A^{2}x$. More generally $q(B)=S^{-1}q(A)S$ for every polynomial $q$, so the minimal polynomial is also unchanged. Multiplication by invertible matrices preserves matrix rank, giving another invariant without finding eigenvalues.

A similarity transformation replaces a square matrix by a similar matrix. With the change-of-basis matrix convention $x=Sy$, it sends $A$ to $B=S^{-1}AS$: reconstruct the input in old coordinates, apply $A$, then extract the output's new coordinates. These three operations explain the order of the factors.

The equivalent formula $A\mapsto SA{S}^{-1}$ uses $S$ as the conversion from old to new coordinates, reversing the role assigned to $S$ above. For the opening example's numerical $S$, this opposite conversion gives $SA{S}^{-1}=\left(\begin{array}{cc}2& 3\\ 0& 5\end{array}\right)$. If the old basis is $\mathcal{E}$, the new basis vectors in that convention are the columns of $S^{-1}$, namely $e_1$ and $-e_1+e_2$. Both formulas are similarity transformations; the direction of the coordinate conversion decides which one describes a specified basis.

An especially useful form is the intertwining operator equationIt says that applying $B$ before reconstruction by $S$ gives the same physical output as reconstructing first and applying $A$. Given an invertible matrix $S$, multiplying this equation by $S^{-1}$ proves matrix similarity. Without invertibility, the equation alone does not prove matrix similarity.