Orthogonal Similarity Preserves Symmetry

Symmetry Under an Orthogonal Change of Coordinates

A real symmetric matrix represents a linear transformation whose matrix has matching entries across its main diagonal. When coordinates are changed using an orthonormal basis, the representing matrix changes from \(\mathbf{A}\) to \(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}\), where \(\mathbf{Q}\) is orthogonal. A useful structural fact is that the new matrix is still symmetric. This matters because it lets us change orthonormal coordinates without losing symmetry; it does not, by itself, say anything about diagonalization or eigenvalues.

All matrices in this tutorial are real and square, and products are assumed to have compatible sizes. Recall that a matrix \(\mathbf{A}\) is symmetric when \(\mathbf{A}^{\top}=\mathbf{A}\). A square matrix \(\mathbf{Q}\) is orthogonal when \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}\); equivalently, \(\mathbf{Q}^{-1}=\mathbf{Q}^{\top}\). The expression \(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}\) is called an orthogonal similarity of \(\mathbf{A}\): orthogonality makes this expression the usual similarity \(\mathbf{Q}^{-1}\mathbf{A}\mathbf{Q}\).

Proof. Set \(\mathbf{B}=\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}\). The transpose of a product reverses the factor order, so \((\mathbf{X}\mathbf{Y})^{\top}=\mathbf{Y}^{\top}\mathbf{X}^{\top}\) for compatible matrices. Applying this rule twice, and using \((\mathbf{Q}^{\top})^{\top}=\mathbf{Q}\), gives \[ \mathbf{B}^{\top} =(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q})^{\top} =\mathbf{Q}^{\top}\mathbf{A}^{\top}\mathbf{Q}. \] Since \(\mathbf{A}\) is symmetric, \(\mathbf{A}^{\top}=\mathbf{A}\). Therefore \[ \mathbf{B}^{\top}=\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}=\mathbf{B}, \] which is the definition of symmetry. \(\square\)

Notice what the proof uses: symmetry of \(\mathbf{A}\) and the transpose rule for products. It does not use \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}\). Thus a stronger observation follows: for any real square \(\mathbf{Q}\), not necessarily orthogonal, \(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}\) is symmetric whenever \(\mathbf{A}\) is symmetric. Orthogonality matters for calling the operation a similarity and for interpreting it as a change between orthonormal coordinates. The broader operation \(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}\) is called a congruence.

Computing the Transformed Matrix

The theorem gives a quick structural check, but it is still important to compute the entries correctly. A reliable approach is to form \(\mathbf{A}\mathbf{Q}\) first and then multiply on the left by \(\mathbf{Q}^{\top}\). The columns of \(\mathbf{Q}\) determine which combinations of the columns of \(\mathbf{A}\) appear in \(\mathbf{A}\mathbf{Q}\); the left multiplication then combines rows. The examples below show how different orthogonal matrices produce different symmetric matrices.

What Orthogonality Adds

The symmetry conclusion is a consequence of the matching outer factors \(\mathbf{Q}^{\top}\) and \(\mathbf{Q}\). Orthogonality adds the identity \(\mathbf{Q}^{\top}=\mathbf{Q}^{-1}\), so the same expression can be read as a similarity. This is why orthogonal similarity is the relevant operation for changing coordinates by an orthonormal basis: it changes the matrix representing the transformation while preserving the symmetric structure.

It is useful to separate two claims that can otherwise be confused:

The first row follows by the same transpose computation as the theorem: \((\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q})^{\top} =\mathbf{Q}^{\top}\mathbf{A}^{\top}\mathbf{Q} =\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}\). It does not assert that an arbitrary \(\mathbf{Q}\) gives a similarity, because \(\mathbf{Q}^{\top}\) need not be the inverse of \(\mathbf{Q}\). The second row is the point where orthogonality is essential.

Common Mistake: Multiplying on Only One Side

A tempting but incorrect calculation is to argue that \(\mathbf{A}\mathbf{Q}\) must be symmetric because \(\mathbf{A}\) is symmetric and \(\mathbf{Q}\) is orthogonal. For a concrete check, use the matrices from the first example. We found \(\mathbf{A}\mathbf{Q}=\begin{bmatrix}1&-2\\3&-1\end{bmatrix}\). Its transpose is \(\begin{bmatrix}1&3\\-2&-1\end{bmatrix}\), which is not equal to it. So \(\mathbf{A}\mathbf{Q}\) is not symmetric. The correction is to compute the full expression, including the left factor: \(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}=\begin{bmatrix}3&-1\\-1&2\end{bmatrix}\), which is symmetric.

Another potential overstatement is that orthogonality is required for symmetry to survive. It is not. For instance, take \(\mathbf{A}=\begin{bmatrix}1&2\\2&3\end{bmatrix}\) and \(\mathbf{C}=\begin{bmatrix}2&0\\0&1\end{bmatrix}\), which is not orthogonal. Then

This result is symmetric, but this calculation is a congruence, not an orthogonal similarity. Keeping the two statements separate prevents the right conclusion from being supported by the wrong hypothesis.

Why the Result Is Useful

Symmetric matrices often arise from relationships that do not depend on the order in which two coordinates are paired: the \((i,j)\) and \((j,i)\) entries agree. When an orthonormal coordinate system is changed, the matrix entries may change—as the examples show—but the resulting matrix remains symmetric. Thus symmetry is a structural property of the represented transformation that survives orthogonal changes of coordinates, rather than a feature tied to one particular coordinate system.

For calculations, use this short checklist:

• Confirm that \(\mathbf{A}\) is square and \(\mathbf{A}^{\top}=\mathbf{A}\).
• Confirm that \(\mathbf{Q}\) is square of the same size and orthogonal if the transformation is being called an orthogonal similarity.
• Form \(\mathbf{A}\mathbf{Q}\), then multiply by \(\mathbf{Q}^{\top}\) on the left.
• Check the final result by comparing it with its transpose.