Computing Similarity Transforms by Hand

What the Similarity Transform Computes

A matrix for a linear operator depends on the coordinates used to describe its inputs and outputs. Changing coordinates changes the matrix, but not the operator itself. The product \(P^{-1}AP\) gives the matrix of the same operator in a basis whose vectors are the columns of \(P\), provided \(A\) is the matrix in the standard basis and \(P\) is invertible.

Throughout, work over \(\mathbb{R}\). Let \(T:\mathbb{R}^n\to\mathbb{R}^n\) be a linear map with standard-basis matrix \(\mathbf{A}\). Let \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) be an ordered basis and let \(\mathbf{P}=\begin{bmatrix}\mathbf{b}_1&\cdots&\mathbf{b}_n\end{bmatrix}\). A vector \(\mathbf{x}\) of \(\mathcal{B}\)-coordinates describes the standard-coordinate vector \(\mathbf{P}\mathbf{x}\). To apply \(T\), first form \(\mathbf{P}\mathbf{x}\), then apply \(\mathbf{A}\), then convert the result back to \(\mathcal{B}\)-coordinates using \(\mathbf{P}^{-1}\).

The order of multiplication records the coordinate journey. The factor on the right converts \(\mathcal{B}\)-coordinates to standard coordinates; the middle factor applies the operator; the factor on the left converts standard coordinates back to \(\mathcal{B}\)-coordinates. Matrix multiplication acts from right to left, so changing the order changes the calculation.

Why the Formula Gives the New Matrix

Proof. The vector with \(\mathcal{B}\)-coordinates \(\mathbf{x}\) has standard-coordinate vector \(\mathbf{P}\mathbf{x}\), because the columns of \(\mathbf{P}\) are the basis vectors. Applying \(T\) in standard coordinates gives \(\mathbf{A}(\mathbf{P}\mathbf{x})\). For any standard-coordinate vector \(\mathbf{y}\), its \(\mathcal{B}\)-coordinates are \(\mathbf{P}^{-1}\mathbf{y}\), since \(\mathbf{P}\mathbf{z}=\mathbf{y}\) has the unique solution \(\mathbf{z}=\mathbf{P}^{-1}\mathbf{y}\). Therefore the output coordinates are \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\mathbf{x}\), as required. \(\square\)

This also explains how to read the columns of the result: column \(j\) is the \(\mathcal{B}\)-coordinate vector of \(T(\mathbf{b}_j)\). Indeed, the \(j\)th standard coordinate vector \(\mathbf{e}_j\) represents \(\mathbf{b}_j\) in basis \(\mathcal{B}\), and multiplying the new matrix by \(\mathbf{e}_j\) selects its \(j\)th column.

Proof. If two matrices have the same action on every vector, they have the same action on each \(\mathbf{e}_j\). Multiplication by \(\mathbf{e}_j\) selects column \(j\), so their columns agree. Conversely, suppose their columns agree. Every \(\mathbf{x}\in\mathbb{R}^n\) can be written \(\mathbf{x}=\sum_{j=1}^n x_j\mathbf{e}_j\). By linearity, each matrix sends \(\mathbf{x}\) to the sum of its \(j\)th column multiplied by \(x_j\). Since the columns agree, these sums agree for every \(\mathbf{x}\). \(\square\)

A Reliable Hand Calculation

For a small explicit problem, compute the inverse first, then multiply \(\mathbf{A}\mathbf{P}\), and finally multiply on the left by \(\mathbf{P}^{-1}\). If the inverse is not immediately apparent, row-reduce \([\mathbf{P}\mid\mathbf{I}_n]\): each row operation applied to the whole augmented matrix preserves the equations being solved, and a left block reduced to \(\mathbf{I}_n\) leaves \(\mathbf{P}^{-1}\) in the right block. Check an inverse by multiplying it by \(\mathbf{P}\), and check the transformed matrix by applying both descriptions of the operator to the coordinate vectors.

Using Row Reduction and Checking Every Column

A Change of Basis Without Triangular Matrices

Common Mistakes and a Practical Check

Reversing the order is not a harmless change. In the first example, the correct result is \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}=\begin{bmatrix}1&0\\0&3\end{bmatrix}\). A wrong calculation using \(\mathbf{P}\mathbf{A}\mathbf{P}^{-1}\) gives \(\begin{bmatrix}1&4\\0&3\end{bmatrix}\): indeed, \(\mathbf{P}\mathbf{A}=\begin{bmatrix}1&5\\0&3\end{bmatrix}\), and \(\begin{bmatrix}1&5\\0&3\end{bmatrix}\begin{bmatrix}1&-1\\0&1\end{bmatrix}=\begin{bmatrix}1&4\\0&3\end{bmatrix}\). This is not the matrix for the stated coordinate conversion. Check the sequence: convert new coordinates to standard coordinates on the right, apply \(\mathbf{A}\), then convert back on the left.

Also distinguish the basis vectors from their coordinates: the columns of \(\mathbf{P}\) are vectors in standard coordinates, while an input \(\mathbf{x}\) to \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\) is a coordinate vector relative to those columns. The columns of the transformed matrix describe the images of the basis vectors, expressed in that same basis. Checking \(\mathbf{P}^{-1}\mathbf{P}=\mathbf{I}\) and checking each transformed column against the coordinate-conversion sequence catch many arithmetic and interpretation errors.