Constructing an Orthogonal Diagonalization

From Eigenvectors to an Orthogonal Diagonalization

For a real symmetric matrix, the Spectral Theorem guarantees that its eigenvectors can be chosen to form an orthonormal basis. Constructing an orthogonal diagonalization turns that guarantee into a concrete calculation: find the eigenvalues and eigenspaces, choose unit eigenvectors that are mutually orthogonal, and place those vectors into a matrix. The diagonal entries must then record the eigenvalues belonging to those columns, in the same order.

Throughout, \(\mathbf{A}\) is a real symmetric \(n\times n\) matrix, and vectors are columns in \(\mathbb{R}^n\). A list \(\mathbf{q}_1,\dots,\mathbf{q}_n\) is an orthonormal basis if it is a basis, every vector has norm \(1\), and distinct vectors in the list have dot product \(0\). A matrix \(\mathbf{Q}\) is orthogonal if \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\). For a square matrix, this also gives \(\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\), since its columns form an orthonormal basis and \(\mathbf{Q}^{-1}=\mathbf{Q}^{\top}\).

To construct this expression, use each eigenvector as a column of \(\mathbf{Q}\). If the \(i\)th column \(\mathbf{q}_i\) has eigenvalue \(\lambda_i\), put \(\lambda_i\) in the \(i\)th diagonal position of \(\mathbf{D}\). The vectors must be an orthonormal basis, not just a collection of eigenvectors. For repeated eigenvalues, there may be several choices of basis within the eigenspace; choose an orthonormal one.

Why the Construction Works

Proof. The columns of \(\mathbf{Q}\) are orthonormal, so the \((i,j)\) entry of \(\mathbf{Q}^{\top}\mathbf{Q}\) is \(\mathbf{q}_i^{\top}\mathbf{q}_j\). This is \(1\) when \(i=j\) and \(0\) when \(i\ne j\), giving \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\). Since the columns are a basis, \(\mathbf{Q}\) is invertible; therefore \(\mathbf{Q}^{-1}=\mathbf{Q}^{\top}\), and hence \(\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\). The \(i\)th column of \(\mathbf{A}\mathbf{Q}\) is \(\mathbf{A}\mathbf{q}_i=\lambda_i\mathbf{q}_i\), which is also the \(i\)th column of \(\mathbf{Q}\mathbf{D}\). Thus \(\mathbf{A}\mathbf{Q}=\mathbf{Q}\mathbf{D}\). Multiplying on the right by \(\mathbf{Q}^{\top}\) gives \(\mathbf{A}\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}\). Since \(\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\), this is \(\mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}\). \(\square\)

For real symmetric matrices, the Spectral Theorem supplies the required orthonormal eigenbasis. In particular, eigenspaces for distinct eigenvalues are orthogonal, and an orthonormal basis can be chosen within each eigenspace. Combining these bases gives an orthonormal eigenbasis of \(\mathbb{R}^n\). These are standard results established in the earlier material on the Spectral Theorem and orthogonality of eigenspaces.

A Two-Dimensional Construction

Repeated Eigenvalues Require a Basis Choice

When an eigenvalue is repeated, its eigenspace may contain more than one independent direction. The diagonalization needs enough eigenvectors to form a basis of the whole space. Any basis of a repeated eigenspace can be used if it is orthonormal; the eigenspace’s eigenvalue is repeated in \(\mathbf{D}\) for each column chosen from it.

Checking a Block-Structured Example

Common Pitfall: Eigenvectors Must Be Unit Vectors

Eigenvectors may be scaled without changing their eigenvalues, but the columns of \(\mathbf{Q}\) must be unit vectors. For the first worked example, using the unnormalized eigenvectors as columns gives \(\mathbf{Q}_0=\begin{bmatrix}1&1\\1&-1\end{bmatrix}\). The wrong orthogonality check is to treat these columns as though they were unit length. In fact,

Using \(\mathbf{Q}_0\) and \(\mathbf{D}=\begin{bmatrix}3&0\\0&1\end{bmatrix}\) in the product also gives the wrong matrix:

The correction is to divide each eigenvector by its norm before assembling \(\mathbf{Q}\). A separate check is necessary even when the vectors are normalized: if two columns are exchanged, their corresponding diagonal entries must also be exchanged. Each diagonal entry belongs to the eigenvector in the same column.

Why the Checks Matter

The two checks answer different questions. Verifying \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\) confirms that the chosen columns are orthonormal. Verifying \(\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}=\mathbf{A}\) confirms that the eigenvalues match those columns and that the resulting product is the given matrix. Neither check should be omitted in a hand construction. In particular, a matrix of eigenvectors can diagonalize by similarity without being orthogonal; the transpose in \(\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}\) is justified by orthogonality, not merely by having eigenvectors as columns.