Change of Basis by Congruence

Why a Bilinear Form Has a New Matrix in a New Basis

A bilinear form assigns a real number to each ordered pair of vectors, linearly in each input. The form itself does not change when we choose a different basis, but the matrix used to calculate its values does. The key is to transform both inputs: one change-of-coordinates matrix acts on the first vector, and another acts on the second. This produces a transformation on both sides of the form’s matrix.

Let \(V\) be a finite-dimensional vector space over \(\mathbb{R}\), and let \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) be an ordered basis. A bilinear form on \(V\) is a function \(B:V\times V\to\mathbb{R}\) that is linear in each input when the other input is fixed. Its matrix in \(\mathcal{B}\), denoted \(\mathbf{A}_{\mathcal{B}}\), has entry \(B(\mathbf{b}_i,\mathbf{b}_j)\) in row \(i\), column \(j\). Thus, if \([\mathbf{x}]_{\mathcal{B}}\) and \([\mathbf{y}]_{\mathcal{B}}\) are the coordinate columns of \(\mathbf{x}\) and \(\mathbf{y}\), then \(B(\mathbf{x},\mathbf{y})=[\mathbf{x}]_{\mathcal{B}}^{\top}\mathbf{A}_{\mathcal{B}}[\mathbf{y}]_{\mathcal{B}}\).

We will write \(\mathbf{P}\) for this matrix and \(\mathbf{A}_{\mathrm{old}}\), \(\mathbf{A}_{\mathrm{new}}\) for the matrices of the same bilinear form in the old and new bases. The matrix \(\mathbf{P}\) is invertible: its columns are the coordinates of a basis, so they are linearly independent and span the coordinate space. We use this fact when reversing the change of basis.

Deriving the Congruence Rule

Proof. Fix arbitrary \(\mathbf{x},\mathbf{y}\in V\), and set \(\mathbf{x}_{\mathrm{old}}=[\mathbf{x}]_{\mathcal{B}}\), \(\mathbf{y}_{\mathrm{old}}=[\mathbf{y}]_{\mathcal{B}}\), \(\mathbf{x}_{\mathrm{new}}=[\mathbf{x}]_{\mathcal{C}}\), and \(\mathbf{y}_{\mathrm{new}}=[\mathbf{y}]_{\mathcal{C}}\). By the definition of \(\mathbf{P}\), \(\mathbf{x}_{\mathrm{old}}=\mathbf{P}\mathbf{x}_{\mathrm{new}}\) and \(\mathbf{y}_{\mathrm{old}}=\mathbf{P}\mathbf{y}_{\mathrm{new}}\). The old-basis formula for the form gives \(B(\mathbf{x},\mathbf{y})=\mathbf{x}_{\mathrm{old}}^{\top}\mathbf{A}_{\mathrm{old}}\mathbf{y}_{\mathrm{old}}=(\mathbf{P}\mathbf{x}_{\mathrm{new}})^{\top}\mathbf{A}_{\mathrm{old}}(\mathbf{P}\mathbf{y}_{\mathrm{new}})=\mathbf{x}_{\mathrm{new}}^{\top}(\mathbf{P}^{\top}\mathbf{A}_{\mathrm{old}}\mathbf{P})\mathbf{y}_{\mathrm{new}}\). The new-basis formula also gives \(B(\mathbf{x},\mathbf{y})=\mathbf{x}_{\mathrm{new}}^{\top}\mathbf{A}_{\mathrm{new}}\mathbf{y}_{\mathrm{new}}\). Since these equalities hold for every pair of new coordinate columns, take \(\mathbf{x}_{\mathrm{new}}=\mathbf{e}_i\) and \(\mathbf{y}_{\mathrm{new}}=\mathbf{e}_j\) for each \(i,j\). The \((i,j)\) entries of the two matrices must agree. Hence \(\mathbf{A}_{\mathrm{new}}=\mathbf{P}^{\top}\mathbf{A}_{\mathrm{old}}\mathbf{P}\). \(\square\)

The transpose appears because the first coordinate column is written as a row when it forms the scalar \(\mathbf{x}^{\top}\mathbf{A}\mathbf{y}\). The two factors of \(\mathbf{P}\) are both needed because both arguments of \(B\) change coordinates. No symmetry assumption on \(B\) or its matrix is required.

Computing the New Matrix

Entry-by-entry checking is especially useful when the matrix is not symmetric. The \((i,j)\) entry records \(B(\mathbf{c}_i,\mathbf{c}_j)\), while the \((j,i)\) entry records \(B(\mathbf{c}_j,\mathbf{c}_i)\). Bilinearity alone does not make those two numbers equal.

Reversing and Composing Changes

Proof. Multiply \(\mathbf{A}_{\mathrm{new}}=\mathbf{P}^{\top}\mathbf{A}_{\mathrm{old}}\mathbf{P}\) on the left by \((\mathbf{P}^{-1})^{\top}\) and on the right by \(\mathbf{P}^{-1}\). Since \((\mathbf{P}^{-1})^{\top}\mathbf{P}^{\top}=(\mathbf{P}\mathbf{P}^{-1})^{\top}=\mathbf{I}\) and \(\mathbf{P}\mathbf{P}^{-1}=\mathbf{I}\), the result is \((\mathbf{P}^{-1})^{\top}\mathbf{A}_{\mathrm{new}}\mathbf{P}^{-1}=\mathbf{A}_{\mathrm{old}}\). \(\square\)

For example, in the first worked example, \(\mathbf{P}^{-1}=\begin{bmatrix}0&1\\1&-1\end{bmatrix}\). Reversing the computed matrix gives

To see the intermediate multiplication, the first two factors give \(\begin{bmatrix}3&2\\7&3\end{bmatrix}\), and multiplying that by \(\mathbf{P}^{-1}\) gives \(\begin{bmatrix}2&1\\3&4\end{bmatrix}\), as required.

Changes can also be carried out in stages. If \(\mathbf{P}\) takes coordinates in a second basis to an old basis and \(\mathbf{Q}\) takes coordinates in a third basis to that second basis, then the third-basis coordinates map to the old coordinates by \(\mathbf{P}\mathbf{Q}\). Applying the rule twice gives \(\mathbf{Q}^{\top}(\mathbf{P}^{\top}\mathbf{A}_{\mathrm{old}}\mathbf{P})\mathbf{Q}=(\mathbf{P}\mathbf{Q})^{\top}\mathbf{A}_{\mathrm{old}}(\mathbf{P}\mathbf{Q})\), since \((\mathbf{P}\mathbf{Q})^{\top}=\mathbf{Q}^{\top}\mathbf{P}^{\top}\). Thus the direct and staged computations agree.

Common Mistake: Using Only One Change-of-Basis Factor

A common incorrect calculation is \(\mathbf{A}_{\mathrm{new}}=\mathbf{A}_{\mathrm{old}}\mathbf{P}\), which transforms the second input but leaves the first input in old coordinates. For the two-dimensional example, that incorrect rule gives \(\begin{bmatrix}3&2\\7&3\end{bmatrix}\), not the matrix of the form in the new basis. For instance, its \((1,1)\) entry would be \(3\), but the correct value is \(B(\mathbf{c}_1,\mathbf{c}_1)=10\). The correction is to transform the first input as well, giving \(\mathbf{P}^{\top}\mathbf{A}_{\mathrm{old}}\mathbf{P}\).

The inverse matrix belongs in the reverse direction, when recovering the old-basis matrix from the new-basis matrix. It does not replace \(\mathbf{P}\) in the forward rule: \(\mathbf{P}\) was defined by the coordinate equation \([\mathbf{x}]_{\mathrm{old}}=\mathbf{P}[\mathbf{x}]_{\mathrm{new}}\). Before multiplying, check which coordinates the columns of the proposed change-of-basis matrix convert.

As a practical method, first write the new basis vectors in old coordinates, then assemble \(\mathbf{P}\) with those vectors as columns. Compute \(\mathbf{A}_{\mathrm{old}}\mathbf{P}\), then multiply by \(\mathbf{P}^{\top}\) on the left. Finally, check selected entries—or all entries when the matrix is small—using \(B(\mathbf{c}_i,\mathbf{c}_j)\). This last check catches both a transposition error and an incorrectly assembled change-of-basis matrix.

What the Rule Keeps Fixed

The numerical entries depend on the basis, but they continue to encode the same function \(B\). For a particular vector pair, the old coordinates and old matrix give one scalar; the new coordinates and new matrix give the same scalar by the theorem’s calculation. This is why the formula is a coordinate rule rather than a change to the bilinear form.

Keep the order of operations visible: the columns of \(\mathbf{P}\) are new basis vectors in old coordinates; the right factor accounts for the second input; the transpose on the left accounts for the first. The formula requires a change between bases of the same finite-dimensional real space and a bilinear form represented in those bases. It does not require the form to be symmetric.