Similar Matrices

Diagonalization rewrote $A$ as $PDP^{-1}$. The same move with any invertible $P$ gives another matrix that describes the same map, just in different coordinates.

$A$ and $B$ are similar if $B = P^{-1}AP$ for some invertible $P$. As in change of basis, if the columns of $P$ form a basis $\mathcal{B}$, then $B$ is the matrix of $\mathbf{x} \mapsto A\mathbf{x}$ in $\mathcal{B}$-coordinates. Diagonalizable means similar to a diagonal matrix.

Key idea

Similar matrices describe one linear map in two bases, so they share everything that does not depend on coordinates: the characteristic polynomial, the eigenvalues (with multiplicities), the determinant, the trace and the rank.

Why it works

$\det(B - \lambda I)$ $= \det(P^{-1}AP - \lambda P^{-1}P)$ $= \det\big(P^{-1}(A - \lambda I)P\big)$ $= \det P^{-1}\det(A - \lambda I)\det P$ $= \det(A - \lambda I)$, since $\det P^{-1}\det P = 1$. Same characteristic polynomial, so the same eigenvalues; its constant term and its $\lambda^{n-1}$ coefficient give the same determinant and trace.

Worked example: Computing a similar matrix

Compute $B = P^{-1}AP$ for $A = \begin{bmatrix} 1 & 2 \\ 3 & 0 \end{bmatrix}$ and $P = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$.

$P^{-1} = \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}$ ($\det P = 1$).
$AP = \begin{bmatrix} 3 & 5 \\ 3 & 3 \end{bmatrix}$.
$B = P^{-1}(AP) = \begin{bmatrix} 3 & 7 \\ 0 & -2 \end{bmatrix}$.
Check the invariants: $\operatorname{tr} A = 1 = \operatorname{tr} B$ and $\det A = -6 = \det B$. $B$ is triangular, so its eigenvalues are $3$ and $-2$, and so are $A$'s.

Answer: $\begin{bmatrix} 3 & 7 \\ 0 & -2 \end{bmatrix}$

To show two matrices are not similar, it is enough to find one invariant that differs: a different trace, determinant, rank or set of eigenvalues settles it at once. To show they are similar, you usually exhibit $P$, or show both are similar to the same diagonal matrix.

Worked example: Same eigenvalues, not similar

Are $A = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}$ and $2I$ similar?

Both have eigenvalues $2, 2$, trace $4$ and determinant $4$.
But $P^{-1}(2I)P = 2P^{-1}P = 2I$ for every $P$: the only matrix similar to $2I$ is $2I$ itself.
$A \neq 2I$, so they are not similar. (Indeed $A$ has only one eigenvector line, while every vector is an eigenvector of $2I$.)

Answer: No.

Common mistake

Concluding that two matrices are similar because their eigenvalues match. Matching invariants is necessary, not sufficient. It is enough when both matrices are diagonalizable, for example when the shared eigenvalues are all distinct: then both are similar to the same diagonal matrix, and so to each other.

The map $A$ stretches by $3$ and by $-2$ along its two eigenvector lines. Any matrix similar to $A$ does the same stretching along lines in other positions.

Similarity also explains why diagonalizing helps: if $B = P^{-1}AP$, then $B^{k} = P^{-1}A^{k}P$, so powers, and anything built from powers, can be computed in whichever basis is easiest and translated back. Choosing $P$ to make $B$ as simple as possible is the main strategy for understanding a linear map.

Similarity is what "the same up to a change of coordinates" means. Every square matrix is similar to a nearly diagonal one called its Jordan form; the shear above is already in that form.

Similar Matrices

Practice

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