Transposes Preserve Eigenvalues

What Transposition Preserves

Transposing a matrix exchanges its rows and columns, so it changes the equations used to find eigenvectors. Yet it does not change the characteristic polynomial or the eigenvalues. Understanding both parts of this statement matters: the eigenvalues are preserved, but the vectors that produce them need not be.

Let \(\mathbf{A}\in\mathbb{F}^{n\times n}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\). Its characteristic polynomial is \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_n)\). A scalar \(\lambda\in\mathbb{F}\) is an eigenvalue if there is a nonzero vector \(\mathbf{v}\) such that \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\); the eigenspace \(E_\lambda\) consists of the zero vector and all vectors satisfying this equation. When working with a real matrix but allowing complex eigenvalues, we regard the matrix and its vectors as complex.

The key determinant fact is that transposing a square matrix leaves its determinant unchanged. We will prove that fact first, then apply it to \(\mathbf{A}-\lambda\mathbf{I}_n\). The characteristic polynomial identity will give equality of the eigenvalues, while direct calculations will show what can happen to the eigenspaces.

Why the Characteristic Polynomials Agree

Proof. For a square matrix \(\mathbf{M}=(m_{ij})\), the determinant formula sums over all permutations \(\sigma\) of \(\{1,\ldots,n\}\): \(\det\mathbf{M}=\sum_{\sigma}\operatorname{sgn}(\sigma)\prod_{i=1}^n m_{i,\sigma(i)}\). Applying this formula to the transpose gives \(\det(\mathbf{M}^{\top})=\sum_{\sigma}\operatorname{sgn}(\sigma)\prod_{i=1}^n m_{\sigma(i),i}\). Set \(\tau=\sigma^{-1}\). As \(\sigma\) runs over all permutations, so does \(\tau\); also \(\operatorname{sgn}(\sigma)=\operatorname{sgn}(\tau)\). Reindexing the product by \(j=\sigma(i)\) yields \(\prod_{i=1}^n m_{\sigma(i),i}=\prod_{j=1}^n m_{j,\tau(j)}\). Thus the sum for \(\det(\mathbf{M}^{\top})\) is exactly the determinant sum for \(\det\mathbf{M}\), and \(\det(\mathbf{M}^{\top})=\det\mathbf{M}\). Now take \(\mathbf{M}=\mathbf{A}-\lambda\mathbf{I}_n\). Since \(\mathbf{I}_n^{\top}=\mathbf{I}_n\), we have \((\mathbf{A}-\lambda\mathbf{I}_n)^{\top}=\mathbf{A}^{\top}-\lambda\mathbf{I}_n\). Therefore \(p_{\mathbf{A}^{\top}}(\lambda)=\det(\mathbf{A}^{\top}-\lambda\mathbf{I}_n)=\det((\mathbf{A}-\lambda\mathbf{I}_n)^{\top})=\det(\mathbf{A}-\lambda\mathbf{I}_n)=p_{\mathbf{A}}(\lambda)\). \(\square\)

Proof. The eigenvalue equation is equivalent to \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{v}=\mathbf{0}\) for some \(\mathbf{v}\ne\mathbf{0}\). A square matrix has a nonzero vector in its null space exactly when its determinant is zero. Hence \(\lambda\) is an eigenvalue of \(\mathbf{A}\) exactly when \(p_{\mathbf{A}}(\lambda)=0\), and the corresponding statement holds for \(\mathbf{A}^{\top}\). The theorem gives identical characteristic polynomials, so the roots, and their multiplicities as roots of the polynomial, agree. The algebraic multiplicity of an eigenvalue is its multiplicity as a root of the characteristic polynomial; identical polynomials therefore give identical algebraic multiplicities. \(\square\)

This result concerns the polynomial and its roots. It does not say that the eigenvector equations for the two matrices are identical. To compare eigenspaces, one must solve each matrix’s own equation.

One Eigenvalue, Different Eigenspaces

The equations also show why the change in eigenspaces is natural. An eigenvector \(\mathbf{v}\) of \(\mathbf{A}\) satisfies a column-vector equation, \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). An eigenvector of \(\mathbf{A}^{\top}\) satisfies a different column-vector equation, \(\mathbf{A}^{\top}\mathbf{w}=\lambda\mathbf{w}\). For a real matrix, transposing the latter equation gives \(\mathbf{w}^{\top}\mathbf{A}=\lambda\mathbf{w}^{\top}\): the transpose’s eigenvectors can be viewed as left eigenvectors of \(\mathbf{A}\), written as rows. This viewpoint does not make them right eigenvectors of \(\mathbf{A}\).

A Repeated Eigenvalue Does Not Force Matching Eigenspaces

This example separates two kinds of information. The characteristic polynomial records eigenvalues and their algebraic multiplicities. An eigenspace records all vectors solving a particular eigenvector equation. Equality of the first piece of information does not by itself identify the vectors in the second. In these examples the eigenspaces happen to have the same dimension for a matrix and its transpose, but the calculation here establishes their actual vectors directly rather than relying on a dimension claim.

When the Eigenspaces Do Agree

A matrix equal to its transpose is called symmetric in the real case. In that situation, the eigenspaces agree because the matrices are identical. For a matrix that is not symmetric, the first worked example warns against assuming the same conclusion: matching characteristic polynomials is not a reason to reuse one matrix’s eigenvectors for the other.

Common Pitfall: Reusing an Eigenvector

A tempting but incorrect shortcut is: “The eigenvalues of \(\mathbf{A}\) and \(\mathbf{A}^{\top}\) are the same, so their eigenvectors are the same too.” For the first example, \((1,0)^{\top}\) is an eigenvector of \(\mathbf{A}\) for eigenvalue \(1\). Reusing it for the transpose gives the wrong calculation: \(\mathbf{A}^{\top}(1,0)^{\top}=(1,1)^{\top}\ne(1,0)^{\top}\). The correct eigenvector for \(\mathbf{A}^{\top}\) at eigenvalue \(1\) is \((1,-1)^{\top}\), since \(\mathbf{A}^{\top}(1,-1)^{\top}=(1,-1)^{\top}\). The correction is to use the shared characteristic polynomial to identify the eigenvalues, then solve \((\mathbf{A}^{\top}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\) separately for the transpose’s eigenspace.

Keep the distinction clear: transposition guarantees equality of characteristic polynomials, eigenvalues, and algebraic multiplicities. It does not guarantee equality of eigenvectors or eigenspaces. For any particular matrix, calculate the eigenspaces from the appropriate matrix rather than transferring them without verification.