Invariant Orthogonal Complements

The Orthogonal Complement as a Subspace

For a symmetric matrix, an eigenvector does more than identify one direction on which the matrix acts by scaling. The directions perpendicular to that eigenvector are preserved as a whole: applying the matrix to a vector perpendicular to the eigenvector produces another vector perpendicular to it. This fact is one of the key steps that lets us study a symmetric matrix on a smaller space while retaining symmetry.

Throughout, let \(\mathbf{A}\in\mathbb{R}^{n\times n}\) be symmetric, so \(\mathbf{A}^{\top}=\mathbf{A}\), and use the standard dot product \(\mathbf{x}\cdot\mathbf{y}=\mathbf{x}^{\top}\mathbf{y}\) on \(\mathbb{R}^n\). For a nonzero vector \(\mathbf{v}\), its orthogonal complement is the set of vectors perpendicular to it.

The condition concerns every vector in the complement, not just one chosen direction within it. If \(\mathbf{v}\) is an eigenvector with eigenvalue \(\lambda\), then \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). The key calculation compares \((\mathbf{A}\mathbf{x})\cdot\mathbf{v}\) with \(\mathbf{x}\cdot(\mathbf{A}\mathbf{v})\). Symmetry of the matrix is precisely what makes these quantities equal.

The Invariance Result

Proof. Let \(\mathbf{x}\in\mathbf{v}^{\perp}\). By definition, \(\mathbf{x}\cdot\mathbf{v}=0\). We must show that \(\mathbf{A}\mathbf{x}\in\mathbf{v}^{\perp}\), or equivalently that \((\mathbf{A}\mathbf{x})\cdot\mathbf{v}=0\). Using the dot product and the symmetry \(\mathbf{A}^{\top}=\mathbf{A}\), we obtain

This calculation also covers \(\mathbf{x}=\mathbf{0}\), since all its displayed dot products are zero. Thus every \(\mathbf{x}\in\mathbf{v}^{\perp}\) has \(\mathbf{A}\mathbf{x}\in\mathbf{v}^{\perp}\), which is the definition of invariance. \(\square\)

The result does not require \(\lambda\ne0\), and it does not require the eigenvalue to be distinct from any other eigenvalue. It uses only the eigenvector equation and the symmetry of \(\mathbf{A}\). The real setting matters for this dot-product calculation as written: the dot product is symmetric, and the scalar \(\lambda\) can be factored out of either slot.

Why the Restriction Is Still Symmetric

Invariance means that applying \(\mathbf{A}\) to a vector in \(\mathbf{v}^{\perp}\) does not leave that subspace. Consequently, the action of \(\mathbf{A}\) can be considered just on \(\mathbf{v}^{\perp}\): the restricted map takes inputs from that subspace and returns outputs there. The restriction is not a new arbitrary transformation; it inherits symmetry from the original matrix.

Proof. By invariance, both \(\mathbf{A}\mathbf{x}\) and \(\mathbf{A}\mathbf{y}\) belong to \(\mathbf{v}^{\perp}\), so the restriction is a map from that subspace to itself. For \(\mathbf{x},\mathbf{y}\in\mathbf{v}^{\perp}\), symmetry of the dot product and \(\mathbf{A}^{\top}=\mathbf{A}\) give

This is exactly the symmetry identity for the restricted map with the inherited dot product. \(\square\)

This conclusion is useful because it preserves the structure needed to study symmetric transformations after separating off one eigenvector direction. If an orthonormal basis of \(\mathbf{v}^{\perp}\) is chosen, the matrix of the restriction in that basis is symmetric. In finite-dimensional real inner-product spaces, an orthonormal basis can be obtained from a basis by the Gram-Schmidt process.

Coordinates Adapted to the Eigenvector

The invariance result has a concrete matrix consequence. Suppose \(\mathbf{q}_1=\mathbf{v}/\|\mathbf{v}\|\), and complete it to an orthonormal basis \((\mathbf{q}_1,\mathbf{q}_2,\ldots,\mathbf{q}_n)\) of \(\mathbb{R}^n\), with \(\mathbf{q}_2,\ldots,\mathbf{q}_n\in\mathbf{v}^{\perp}\). Let \(\mathbf{Q}\) be the matrix with these basis vectors as columns. The first basis direction is an eigenvector direction; the remaining basis directions lie in its orthogonal complement.

Proof. Since the columns of \(\mathbf{Q}\) are orthonormal, its \((i,j)\) entry in \(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}\) is \(\mathbf{q}_i\cdot(\mathbf{A}\mathbf{q}_j)\). The \((1,1)\) entry is \(\mathbf{q}_1\cdot(\lambda\mathbf{q}_1)=\lambda\), because \(\mathbf{q}_1\) is a unit eigenvector. For \(j\ge2\), invariance gives \(\mathbf{A}\mathbf{q}_j\in\mathbf{v}^{\perp}\), and \(\mathbf{q}_1\) is perpendicular to every vector in that subspace. Therefore \(\mathbf{q}_1\cdot(\mathbf{A}\mathbf{q}_j)=0\). For \(i\ge2\), symmetry gives \(\mathbf{q}_i\cdot(\mathbf{A}\mathbf{q}_1)=\mathbf{q}_i\cdot(\lambda\mathbf{q}_1)=0\). These are all off-diagonal entries connecting the first row or column to the remaining block. Finally, for \(i,j\ge2\), the entries satisfy \(\mathbf{q}_i\cdot(\mathbf{A}\mathbf{q}_j)=\mathbf{q}_j\cdot(\mathbf{A}\mathbf{q}_i)\), by symmetry of \(\mathbf{A}\). Hence the lower-right block \(\mathbf{B}\) is symmetric. \(\square\)

What Can Fail Without Symmetry

It is important not to infer invariance from the eigenvector equation alone. The proof depended on replacing \(\mathbf{A}^{\top}\) with \(\mathbf{A}\). For a nonsymmetric matrix, that equality is unavailable, and the conclusion can fail.

Common pitfall. A tempting but incorrect argument is: “\(\mathbf{v}\) is an eigenvector, so every vector perpendicular to \(\mathbf{v}\) stays perpendicular after applying the matrix.” The nonsymmetric example disproves that claim: \((0,1)^{\top}\) is perpendicular to \(\mathbf{v}\), but its image has dot product \(1\) with \(\mathbf{v}\). The correction is to check the matrix hypothesis and use \(\mathbf{A}^{\top}=\mathbf{A}\) in the inner-product calculation. The eigenvector equation alone does not supply that step.

For a symmetric matrix, the orthogonal complement of an eigenvector is therefore a smaller space on which the matrix still acts symmetrically. The adapted block form records exactly this separation: one diagonal entry describes the eigenvector direction, and a symmetric block describes the action on the complement. The result is local to one eigenvector and its complement; it does not, by itself, assert that the whole matrix has already been diagonalized.