Diagonalizing Symmetric Forms by Congruence

Why Change Basis for a Symmetric Form?

A symmetric bilinear form can be represented by a symmetric matrix, but its entries may mix coordinates in a way that makes the form difficult to interpret. A suitable basis removes those cross terms: in the new coordinates, the form is a sum of separate squared-coordinate contributions. The key operation is congruence, the matrix transformation that represents a bilinear form after a change of basis.

Let \(V\) be a finite-dimensional real vector space and let \(B:V\times V\to\mathbb{R}\) be a symmetric bilinear form. Bilinearity means that \(B\) is linear in each argument, and symmetry means \(B(\mathbf{u},\mathbf{v})=B(\mathbf{v},\mathbf{u})\). Its associated quadratic form is \(q(\mathbf{v})=B(\mathbf{v},\mathbf{v})\). Given an ordered basis \(\mathcal{E}=(\mathbf{e}_1,\dots,\mathbf{e}_n)\), the matrix \(\mathbf{A}\) of \(B\) in that basis has entries \(a_{ij}=B(\mathbf{e}_i,\mathbf{e}_j)\). Symmetry of \(B\) implies \(\mathbf{A}^{\top}=\mathbf{A}\).

How a Basis Change Acts

Let \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) be another ordered basis. Write each \(\mathbf{b}_j\) in \(\mathcal{E}\)-coordinates, and place those coordinate columns in the invertible matrix \(\mathbf{P}\). Thus, for any \(\mathbf{v}\in V\), its coordinate columns satisfy \([\mathbf{v}]_{\mathcal{E}}=\mathbf{P}[\mathbf{v}]_{\mathcal{B}}\). The entries of \(\mathbf{P}\) express the new basis vectors in the old basis; they do not express the old basis vectors in the new one.

Proof. Write \(\mathbf{b}_i=\sum_k p_{ki}\mathbf{e}_k\) and \(\mathbf{b}_j=\sum_\ell p_{\ell j}\mathbf{e}_\ell\), where \(p_{ki}\) is the entry of \(\mathbf{P}\) in row \(k\), column \(i\). Bilinearity gives \(B(\mathbf{b}_i,\mathbf{b}_j)=\sum_k\sum_\ell p_{ki}B(\mathbf{e}_k,\mathbf{e}_\ell)p_{\ell j}\). This is the \(i,j\) entry of \(\mathbf{P}^{\top}\mathbf{A}_{\mathcal{E}}\mathbf{P}\), which proves the formula for every entry. \(\square\)

The same formula follows from the quadratic form: if \(\mathbf{x}=[\mathbf{v}]_{\mathcal{E}}\), then \(q(\mathbf{v})=\mathbf{x}^{\top}\mathbf{A}_{\mathcal{E}}\mathbf{x}\). Substituting \(\mathbf{x}=\mathbf{P}\mathbf{y}\), where \(\mathbf{y}=[\mathbf{v}]_{\mathcal{B}}\), gives \(q(\mathbf{v})=\mathbf{y}^{\top}(\mathbf{P}^{\top}\mathbf{A}_{\mathcal{E}}\mathbf{P})\mathbf{y}\). In particular, the \(i,j\) entry in the new matrix is \(B(\mathbf{b}_i,\mathbf{b}_j)\); a diagonal matrix means distinct basis vectors are orthogonal with respect to \(B\).

Diagonalization Using Eigenvectors

For a real symmetric matrix, the real Spectral Theorem states that there is an orthonormal basis of eigenvectors. In matrix form, if \(\mathbf{A}^{\top}=\mathbf{A}\), there is an orthogonal matrix \(\mathbf{Q}\), meaning \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\), such that \(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}\) is diagonal. This is the standard Spectral Theorem from the earlier material on normal operators. Its eigenvector columns give a particularly convenient basis for diagonalizing a symmetric form.

Proof. Because \(B\) is symmetric, its matrix \(\mathbf{A}_{\mathcal{E}}\) is a real symmetric matrix. By the real Spectral Theorem, choose an orthogonal matrix \(\mathbf{Q}\) whose columns are an orthonormal eigenvector basis and for which \(\mathbf{A}_{\mathcal{E}}\mathbf{Q}=\mathbf{Q}\mathbf{D}\), where \(\mathbf{D}=\operatorname{diag}(\lambda_1,\dots,\lambda_n)\). Use the columns of \(\mathbf{Q}\) as the coordinate columns of a new basis \(\mathcal{B}\) relative to \(\mathcal{E}\). The change-of-basis formula gives \(\mathbf{A}_{\mathcal{B}}=\mathbf{Q}^{\top}\mathbf{A}_{\mathcal{E}}\mathbf{Q}\). Since \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\), this equals \(\mathbf{Q}^{\top}\mathbf{Q}\mathbf{D}=\mathbf{D}\). Therefore the form is diagonal in \(\mathcal{B}\), with the stated eigenvalues on the diagonal. \(\square\)

This application of the Spectral Theorem is stronger than merely finding some basis with no cross terms: it provides an orthonormal eigenvector basis relative to the original coordinate system. If \(\mathbf{y}\) denotes the coordinates in that basis, the quadratic form has the expression \(q(\mathbf{v})=\lambda_1y_1^2+\cdots+\lambda_ny_n^2\). Each coefficient tells how the form weights the square of one eigenvector coordinate.

Worked Diagonalizations

Diagonal by Congruence Does Not Require Eigenvectors

The Spectral Theorem gives an orthonormal eigenvector basis, but congruence allows any invertible basis change. The coefficients in a diagonal matrix obtained from an arbitrary basis need not be eigenvalues. The following example shows a form whose cross term disappears in a basis that is not an eigenvector basis.

Reading the Diagonal Coefficients

Whenever the matrix of a symmetric form in a basis is \(\operatorname{diag}(d_1,\dots,d_n)\), the associated quadratic form is \(q(y_1,\dots,y_n)=d_1y_1^2+\cdots+d_ny_n^2\). There are no mixed terms because each off-diagonal matrix entry is zero. A positive coefficient contributes a nonnegative term that increases with the squared coordinate; a negative coefficient contributes a nonpositive term; and a zero coefficient contributes nothing. These descriptions concern the displayed coordinates and should not be confused with a claim that every diagonal entry is invariant under a basis change.

For a diagonalization obtained from an orthonormal eigenvector basis, the diagonal coefficients are the eigenvalues of the original symmetric matrix. For a general basis, they are \(B(\mathbf{b}_i,\mathbf{b}_i)\), and they need not be eigenvalues. The third example makes this distinction concrete: its diagonal coefficients are \(1,1\), although the original matrix is not the identity. The next topic develops which aspects of the signs and zero coefficients are preserved under changes of basis; here the essential point is how to obtain and read a diagonal representation.

Common Pitfall: Using Similarity Instead

A frequent error is to use \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\) instead of \(\mathbf{P}^{\top}\mathbf{A}\mathbf{P}\). The former is the transformation used to represent a linear map in a new basis; a bilinear form has two arguments, so the change of coordinates acts on both sides, producing a transpose on the left. For the third example, \(\mathbf{P}^{-1}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\), and the incorrect calculation gives:

which is not diagonal. The correction is to use the form's congruence formula, \(\mathbf{P}^{\top}\mathbf{A}\mathbf{P}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\), as verified above. Similarity and congruence answer different change-of-basis questions.