Changing Variables by Congruence

What Changes When the Variables Change?

A quadratic form assigns a scalar to each vector. Its matrix depends on the coordinates used to describe that vector: the same input can be written using one coordinate vector \(\mathbf{x}\), or using another coordinate vector \(\mathbf{y}\) related by a linear substitution. The value of the form does not change when we describe the input in new coordinates, but the matrix that computes the value generally does.

Throughout, the scalar field is \(\mathbb{R}\). Let \(\mathbf{A}\in\mathbb{R}^{n\times n}\) be symmetric, and write the quadratic form as \(q(\mathbf{x})=\mathbf{x}^{\top}\mathbf{A}\mathbf{x}\). A change of variables has the form \(\mathbf{x}=\mathbf{P}\mathbf{y}\), where \(\mathbf{P}\in\mathbb{R}^{n\times n}\). If \(\mathbf{P}\) is invertible, each old coordinate vector \(\mathbf{x}\) corresponds to exactly one new coordinate vector \(\mathbf{y}\), and conversely. This invertibility condition is what makes the substitution a genuine change of coordinates rather than a map that may lose information.

The order of the factors records how the substitution acts. The vector \(\mathbf{y}\) is first sent to \(\mathbf{x}=\mathbf{P}\mathbf{y}\); the original form then evaluates \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}\). Substituting this expression gives

Here we used \((\mathbf{P}\mathbf{y})^{\top}=\mathbf{y}^{\top}\mathbf{P}^{\top}\), followed by associativity of matrix multiplication. Thus the new matrix is \(\mathbf{B}=\mathbf{P}^{\top}\mathbf{A}\mathbf{P}\), not a matrix obtained by putting the transpose on the other side.

Why the Transformed Matrix Is Still Symmetric

A quadratic form is represented here by a symmetric matrix. The congruence formula respects that symmetry. Indeed, transposing a product reverses the factor order, and \(\mathbf{A}^{\top}=\mathbf{A}\), so

This also gives a useful computational check: if the starting matrix is symmetric, the transformed matrix must be symmetric. A nonsymmetric result signals an error in the multiplication or transpose.

Proof. For reflexivity, take \(\mathbf{P}=\mathbf{I}_n\). Then \(\mathbf{I}_n^{\top}\mathbf{A}\mathbf{I}_n=\mathbf{A}\). For symmetry, suppose \(\mathbf{B}=\mathbf{P}^{\top}\mathbf{A}\mathbf{P}\) with \(\mathbf{P}\) invertible. Multiplying by \((\mathbf{P}^{-1})^{\top}\) on the left and by \(\mathbf{P}^{-1}\) on the right gives \(\mathbf{A}=(\mathbf{P}^{-1})^{\top}\mathbf{B}\mathbf{P}^{-1}\); since \(\mathbf{P}^{-1}\) is invertible, this exhibits \(\mathbf{A}\) as congruent to \(\mathbf{B}\). For transitivity, suppose \(\mathbf{B}=\mathbf{P}^{\top}\mathbf{A}\mathbf{P}\) and \(\mathbf{C}=\mathbf{Q}^{\top}\mathbf{B}\mathbf{Q}\), with \(\mathbf{P}\) and \(\mathbf{Q}\) invertible. Substitution gives \(\mathbf{C}=\mathbf{Q}^{\top}\mathbf{P}^{\top}\mathbf{A}\mathbf{P}\mathbf{Q}=(\mathbf{P}\mathbf{Q})^{\top}\mathbf{A}(\mathbf{P}\mathbf{Q})\). The product \(\mathbf{P}\mathbf{Q}\) is invertible, with inverse \(\mathbf{Q}^{-1}\mathbf{P}^{-1}\), so \(\mathbf{C}\) is congruent to \(\mathbf{A}\). \(\square\)

Computing the Congruent Matrix

For a practical calculation, first write the substitution explicitly as \(\mathbf{x}=\mathbf{P}\mathbf{y}\). Then calculate \(\mathbf{A}\mathbf{P}\), followed by \(\mathbf{P}^{\top}(\mathbf{A}\mathbf{P})\). Keeping the intermediate product visible helps prevent a common order error. Finally, compare the resulting quadratic expression with the original form after directly substituting for \(\mathbf{x}\).

The off-diagonal entries of a symmetric matrix contribute twice in its quadratic form: \(b_{12}y_1y_2+b_{21}y_2y_1=2b_{12}y_1y_2\). This is why the coefficient \(6\) on \(y_1y_2\) corresponds to entries \(3\) in positions \((1,2)\) and \((2,1)\) of \(\mathbf{B}\).

Invertibility and What the Change Preserves

The algebraic identity \(q(\mathbf{P}\mathbf{y})=\mathbf{y}^{\top}\mathbf{P}^{\top}\mathbf{A}\mathbf{P}\mathbf{y}\) holds for any matrix \(\mathbf{P}\) of the appropriate size. But if \(\mathbf{P}\) is not invertible, some \(\mathbf{x}\) may not be expressible as \(\mathbf{P}\mathbf{y}\), and different \(\mathbf{y}\) values may give the same \(\mathbf{x}\). In that case it is a substitution, but not a change of coordinates covering the whole original space.

For an invertible \(\mathbf{P}\), every vector \(\mathbf{x}\in\mathbb{R}^n\) has the unique corresponding vector \(\mathbf{y}=\mathbf{P}^{-1}\mathbf{x}\). Consequently the set of values taken by the quadratic form is unchanged: evaluating the original form over all \(\mathbf{x}\) gives the same possible real numbers as evaluating the transformed form over all \(\mathbf{y}\). In particular, any statement that concerns only whether the form is positive, negative, or zero at every vector transfers between the two coordinate systems.

Proof. For every \(\mathbf{y}\), the transformation identity gives \(\mathbf{y}^{\top}\mathbf{P}^{\top}\mathbf{A}\mathbf{P}\mathbf{y}=q(\mathbf{P}\mathbf{y})\), so every value of the transformed form is a value of the original form. Conversely, for any \(\mathbf{x}\), invertibility gives \(\mathbf{y}=\mathbf{P}^{-1}\mathbf{x}\), and then \(\mathbf{P}\mathbf{y}=\mathbf{x}\). Therefore \(q(\mathbf{x})\) is a value of the transformed form as well. The two sets of values are equal. If the original form is nonnegative for every \(\mathbf{x}\), the identity makes the transformed form nonnegative for every \(\mathbf{y}\). If the transformed form is nonnegative for every \(\mathbf{y}\), use \(\mathbf{y}=\mathbf{P}^{-1}\mathbf{x}\) for each \(\mathbf{x}\) to conclude \(q(\mathbf{x})\geq0\). \(\square\)

Common Pitfall: Reversing the Transpose

A frequent error is to use \(\mathbf{P}\mathbf{A}\mathbf{P}^{\top}\) in place of \(\mathbf{P}^{\top}\mathbf{A}\mathbf{P}\). The error can still produce a symmetric matrix, so symmetry alone does not catch it. In the first worked example, the incorrect order gives

This would claim that the new form is \(7y_1^2+8y_1y_2+3y_2^2\). But at \(\mathbf{y}=\begin{bmatrix}1\\0\end{bmatrix}\), the substitution gives \(\mathbf{x}=\begin{bmatrix}1\\0\end{bmatrix}\), and the original form equals \(2\), not \(7\). The correction is to transpose the left factor in \((\mathbf{P}\mathbf{y})^{\top}\): it is \(\mathbf{y}^{\top}\mathbf{P}^{\top}\), which forces the order \(\mathbf{P}^{\top}\mathbf{A}\mathbf{P}\).

Congruence should also not be confused with similarity. The expression for changing a quadratic form is \(\mathbf{P}^{\top}\mathbf{A}\mathbf{P}\); similarity uses \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\) and concerns a different transformation. For quadratic forms, the reliable procedure is to substitute \(\mathbf{x}=\mathbf{P}\mathbf{y}\) into \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}\), multiply in that order, and check the resulting expression by direct substitution.