Equivalent Forms of Orthogonal Diagonalizability

Two Equivalent Ways to Describe a Matrix

A real square matrix can be described by its action on coordinates, or by its action along particular directions. Diagonalization expresses the second description: in a suitable basis, the matrix acts by scaling each basis vector. Orthogonal diagonalization requires that basis to be orthonormal, so the change of coordinates preserves lengths and angles. For real matrices, this stronger requirement is not merely a convenient choice: it is exactly what symmetry characterizes.

Throughout, let \(\mathbf{A}\in\mathbb{R}^{n\times n}\), where \(n\in\mathbb{N}\). A real square matrix \(\mathbf{Q}\) is orthogonal if \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\); for a square matrix this also gives \(\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\). A real diagonal matrix has all its off-diagonal entries equal to zero.

The columns of an orthogonal matrix form an orthonormal basis of \(\mathbb{R}^n\). If \(\mathbf{Q}=[\mathbf{q}_1\ \cdots\ \mathbf{q}_n]\) and \(\mathbf{D}=\operatorname{diag}(\lambda_1,\dots,\lambda_n)\), then \(\mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}\) says that each column \(\mathbf{q}_j\) is an eigenvector with eigenvalue \(\lambda_j\). In particular, the representation is a matrix form of having an orthonormal basis of eigenvectors.

Changing Between the Two Forms

Proof. Suppose first that \(\mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}\). Multiplying on the left by \(\mathbf{Q}^{\top}\) and on the right by \(\mathbf{Q}\) gives \(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}=\mathbf{Q}^{\top}\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\mathbf{D}\mathbf{I}_n=\mathbf{D}\). Conversely, suppose \(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}=\mathbf{D}\). Multiply on the left by \(\mathbf{Q}\) and on the right by \(\mathbf{Q}^{\top}\). Then \(\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}=\mathbf{Q}\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\mathbf{A}\mathbf{I}_n=\mathbf{A}\). Both directions use orthogonality, which gives \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\). \(\square\)

The Characterization by Symmetry

Proof. First suppose that \(\mathbf{A}\) is orthogonally diagonalizable. Then \(\mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}\) for an orthogonal \(\mathbf{Q}\) and a real diagonal \(\mathbf{D}\). Transposing this equation and using the rule \((\mathbf{X}\mathbf{Y})^{\top}=\mathbf{Y}^{\top}\mathbf{X}^{\top}\) gives \((\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top})^{\top}=\mathbf{Q}\mathbf{D}^{\top}\mathbf{Q}^{\top}\). A real diagonal matrix satisfies \(\mathbf{D}^{\top}=\mathbf{D}\), so \(\mathbf{A}^{\top}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}=\mathbf{A}\). Hence \(\mathbf{A}\) is symmetric.

For the reverse direction, suppose that \(\mathbf{A}\) is real and symmetric. The Spectral Theorem, proved by induction on dimension in the earlier material, states that every real symmetric \(n\times n\) matrix has an orthonormal basis of real eigenvectors. Let \(\mathbf{q}_1,\dots,\mathbf{q}_n\) be such a basis, with corresponding real eigenvalues \(\lambda_1,\dots,\lambda_n\). Form \(\mathbf{Q}=[\mathbf{q}_1\ \cdots\ \mathbf{q}_n]\) and \(\mathbf{D}=\operatorname{diag}(\lambda_1,\dots,\lambda_n)\). Orthonormality gives \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\), and the eigenvector equations \(\mathbf{A}\mathbf{q}_j=\lambda_j\mathbf{q}_j\), one for each \(j\), give \(\mathbf{A}\mathbf{Q}=\mathbf{Q}\mathbf{D}\). Multiplying this equality on the right by \(\mathbf{Q}^{\top}\), and using \(\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\), yields \(\mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}\). Thus \(\mathbf{A}\) is orthogonally diagonalizable. The two implications prove the equivalence. \(\square\)

This proof uses the Spectral Theorem specifically for a real symmetric matrix; it does not claim that every real matrix has real eigenvalues or an orthonormal eigenbasis. The criterion is an equivalence over \(\mathbb{R}\), with a real diagonal matrix and a real orthogonal matrix. Its reverse implication depends on the existence of the orthonormal eigenbasis guaranteed by that theorem.

What Symmetry Rules Out

The forward direction of the theorem is a useful test: if a real matrix is not symmetric, it cannot be orthogonally diagonalizable. This does not mean it cannot be diagonalized at all. Ordinary diagonalization allows a change-of-basis matrix that need not be orthogonal, so its columns need not be perpendicular or unit length.

Common mistake: treating every diagonalization as orthogonal. A tempting but incorrect inference is “\(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), so \(\mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}\).” In the example, the actual change-of-basis matrix is \(\mathbf{P}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\). Its columns have dot product \(1\), not \(0\), so \(\mathbf{P}\) is not orthogonal and \(\mathbf{P}^{-1}\ne\mathbf{P}^{\top}\). The correction is to check the hypotheses: the symmetric-matrix criterion guarantees an orthogonal diagonalization, while an arbitrary diagonalization does not.

Using the Criterion Efficiently

The theorem turns two different questions into one another. To decide whether a real square matrix could be orthogonally diagonalized, first check whether it is symmetric. If it fails the transpose criterion, the answer is no. If it is symmetric, the Spectral Theorem guarantees an orthogonal diagonalization; actually finding the eigenvalues, eigenvectors, and matrices \(\mathbf{Q}\) and \(\mathbf{D}\) is a separate computational task.

The useful conceptual distinction is the role of the columns of the change-of-basis matrix. In orthogonal diagonalization they are not just eigenvectors: they must form an orthonormal basis. Symmetry is precisely the condition that ensures such a basis exists for a real matrix. The factor \(\mathbf{Q}^{\top}\) is therefore not an interchangeable substitute for any inverse; it records the special geometry of an orthonormal coordinate system.