Trace and Determinant from Eigenvalues

Why Eigenvalues Summarize Trace and Determinant

A real symmetric matrix can be expressed in orthogonal coordinates as a diagonal matrix. The diagonal entries in that representation are its eigenvalues, and the diagonal form makes two basic quantities especially transparent: the trace adds diagonal entries, while the determinant multiplies them. The important detail is that the eigenvalues must be counted with repetition. A repeated eigenvalue contributes once for each occurrence in the diagonal matrix.

Throughout, \(\mathbf{A}\) is an \(n\times n\) real symmetric matrix. The Spectral Theorem, established in the earlier material on orthogonal diagonalization, says that there are an orthogonal matrix \(\mathbf{Q}\) and a real diagonal matrix \(\boldsymbol{\Lambda}\) such that \(\mathbf{A}=\mathbf{Q}\boldsymbol{\Lambda}\mathbf{Q}^{\top}\). Orthogonal means \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\). We write the diagonal entries as \(\lambda_1,\dots,\lambda_n\), including repeated entries, so \(\boldsymbol{\Lambda}=\operatorname{diag}(\lambda_1,\dots,\lambda_n)\).

An eigenvalue is counted according to how many times it appears in this list. For symmetric matrices, that count also equals the dimension of its eigenspace. We will establish this fact below, along with the corresponding statement about multiplicity as a root of the characteristic polynomial. This distinction matters: a list with entries \(2,2,5\) contains three eigenvalues counted with repetition, even though it has only two distinct eigenvalue values.

The Trace and Determinant Formulas

Proof. Write \(\mathbf{A}=\mathbf{Q}\boldsymbol{\Lambda}\mathbf{Q}^{\top}\), where \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\) and \(\boldsymbol{\Lambda}\) has diagonal entries \(\lambda_1,\dots,\lambda_n\). First, for any two \(n\times n\) matrices \(\mathbf{X}\) and \(\mathbf{Y}\), the definition of trace gives \(\operatorname{tr}(\mathbf{X}\mathbf{Y})=\sum_i\sum_j x_{ij}y_{ji}=\sum_j\sum_i y_{ji}x_{ij}=\operatorname{tr}(\mathbf{Y}\mathbf{X})\). Applying this identity with \(\mathbf{X}=\mathbf{Q}\) and \(\mathbf{Y}=\boldsymbol{\Lambda}\mathbf{Q}^{\top}\) gives \(\operatorname{tr}\mathbf{A}=\operatorname{tr}(\boldsymbol{\Lambda}\mathbf{Q}^{\top}\mathbf{Q})=\operatorname{tr}\boldsymbol{\Lambda}=\sum_i\lambda_i\).

For the determinant, use the standard determinant product rule \(\det(\mathbf{X}\mathbf{Y})=\det\mathbf{X}\det\mathbf{Y}\), together with \(\det(\mathbf{Q}^{\top})=\det\mathbf{Q}\). Since \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\), these identities give \((\det\mathbf{Q})^2=1\). Therefore \(\det\mathbf{A}=\det\mathbf{Q}\det\boldsymbol{\Lambda}\det\mathbf{Q}^{\top}=(\det\mathbf{Q})^2\det\boldsymbol{\Lambda}=\prod_i\lambda_i\).

To check the multiplicities, let \(k\) be the number of entries of \(\boldsymbol{\Lambda}\) equal to a fixed eigenvalue \(\mu\). For a vector \(\mathbf{x}\), the equation \(\mathbf{A}\mathbf{x}=\mu\mathbf{x}\) is equivalent, after multiplying by \(\mathbf{Q}^{\top}\), to \(\boldsymbol{\Lambda}(\mathbf{Q}^{\top}\mathbf{x})=\mu(\mathbf{Q}^{\top}\mathbf{x})\). In the diagonal equation, a coordinate can be nonzero only where the corresponding diagonal entry is \(\mu\). There are exactly \(k\) such coordinates, so the solution space has dimension \(k\). The invertible coordinate change \(\mathbf{x}\mapsto\mathbf{Q}^{\top}\mathbf{x}\) preserves dimension, giving \(\dim E_\mu=k\).

Finally, \(\mathbf{A}-t\mathbf{I}_n=\mathbf{Q}(\boldsymbol{\Lambda}-t\mathbf{I}_n)\mathbf{Q}^{\top}\), so the determinant calculation already proved yields \(p_{\mathbf{A}}(t)=\prod_{i=1}^{n}(\lambda_i-t)\). If \(\mu\) occurs \(k\) times, this product has the factor \((\mu-t)^k\). Every factor from an entry different from \(\mu\) is nonzero at \(t=\mu\), so the root has multiplicity exactly \(k\). This proves all the claims. \(\square\)

The determinant product rule and the determinant identity for a transpose are standard determinant properties. The proof above uses them only after stating their role. The trace calculation, in contrast, shows directly why orthogonal change of coordinates preserves the sum of diagonal entries.

Worked Examples

What the Multiplicity Statement Adds

The trace and determinant formulas use the whole eigenvalue list, not merely its distinct values. If the distinct eigenvalues are \(\mu_1,\dots,\mu_r\) and their occurrence counts are \(k_1,\dots,k_r\), then \(k_1+\cdots+k_r=n\), and the same formulas can be grouped by distinct values:

This form is useful when an eigenspace has dimension greater than one: its eigenvalue is added that many times and multiplied by itself that many times. It also identifies immediate consequences. If any eigenvalue is zero, the determinant is zero. If all eigenvalues are positive, then the trace and determinant are positive; if one eigenvalue is zero, the determinant vanishes regardless of the other eigenvalues. These conclusions follow from the sum and product formulas, not from a new diagonalization argument.

Orthogonal diagonalization also provides a reliable way to check computations. The trace and determinant calculated from the entries must agree with the sum and product from the eigenvalue list. Agreement is a useful consistency check, although matching trace and determinant alone does not prove that a proposed eigenvalue list is correct: different lists can have the same sum and product.

Common Pitfall: Forgetting Repetitions

For \(\mathbf{B}=\operatorname{diag}(4,4,-2)\), a common incorrect calculation is to list only the distinct eigenvalues \(4\) and \(-2\), then claim \(\operatorname{tr}\mathbf{B}=4+(-2)=2\) and \(\det\mathbf{B}=4(-2)=-8\). Both values are wrong because the eigenvalue \(4\) occurs twice. The correction is to count all three diagonal entries: \(4+4-2=6\) and \(4\cdot4\cdot(-2)=-32\). In general, use an \(n\)-entry list for an \(n\times n\) matrix, retaining every repetition.