## SectionCPCharacteristic Polynomial

We defined the characteristic polynomial of a square matrix a ways back in Definition CP. Now that we have the determinant of a matrix, we can provide an alternate formulation. Our definition of the characteristic polynomial requires knowing the algebraic multiplicity of an eigenvalue, obtained as the dimension of a generalized eigenspace, a type of invariant subspace. Most introductory texts instead introduce the determinant and use that to give a wildly different definition of the characteristic polynomial, that at first seems to have nothing to do with eigenvalues. In this “late determinant” version of this textbook, the upcoming Theorem CPD establishes the equivalence of these different definitions.

### SubsectionCPDCharacteristic Polynomial and Determinant

We now prove that the characteristic polynomial can be computed with the determinant.

By Theorem UTEC, $A$ is similar to an upper triangular matrix, $T\text{,}$ whose diagonal has each eigenvalue of $A$ repeated as many times as its algebraic multiplicity. Let $S$ be the nonsingular matrix achieving this similarity transformation, that is $\inverse{S}AS = T\text{.}$ Then we have

\begin{align*} \detname{xI_n - A} &= \detname{xI_n - ST\inverse{S}}\\ &= \detname{xSI_n\inverse{S} - ST\inverse{S}}&&\knowl{./knowl/definition-MI.html}{\text{Definition MI}}\\ &= \detname{SxI_n\inverse{S} - ST\inverse{S}}&&\knowl{./knowl/theorem-MMSMM.html}{\text{Theorem MMSMM}}\\ &= \detname{S\left(xI_n - T\right)\inverse{S}}&&\knowl{./knowl/theorem-MMDAA.html}{\text{Theorem MMDAA}}\\ &= \detname{S}\detname{xI_n - T}\detname{\inverse{S}}&&\knowl{./knowl/theorem-DRMM.html}{\text{Theorem DRMM}}\\ &= \detname{S}\detname{\inverse{S}}\detname{xI_n - T}&&\knowl{./knowl/property-CMCN.html}{\text{Property CMCN}}\\ &= \detname{S\inverse{S}}\detname{xI_n - T}&&\knowl{./knowl/theorem-DRMM.html}{\text{Theorem DRMM}}\\ &= \detname{I_n}\detname{xI_n - T}&&\knowl{./knowl/definition-MI.html}{\text{Definition MI}}\\ &= \detname{xI_n - T}&&\knowl{./knowl/theorem-DIM.html}{\text{Theorem DIM}} \end{align*}

The matrix $xI_n$ is a diagonal matrix, so $xI_n - T$ is upper triangular, and the entries on the diagonal are $x-\lambda_i\text{,}$ where $\lambda_i$ is an eigenvalue of $A\text{.}$ By expanding about the first row repeatedly, the determinant will equal the product of the diagonal entries (see Example DUTM. Since the eigenvalues of $A$ are repeated on the diagonal of $T$ according to the algebraic multiplicity, we have

\begin{equation*} \detname{xI_n - A} = \prod_{i=1}^{n}\left(x-\lambda_i\right) = \prod_{i=1}^{t}\left(x-\lambda_i\right)^{\algmult{A}{\lambda_i}} \end{equation*}

where the first product is over all the eigenvalues (with repeats) and the second product is over distinct eigenvalues. Since the latter expression equals the definition of $\charpoly{A}{x}$ in Definition CP, we have the desired result.

The eigenvalues of a matrix and its transpose are identical. The proof of this fact is considerably easier with a determinant. A proof that does not use the determinant can be formed by showing that a matrix and its transpose are similar, and then appealing to Theorem SMEE. But the similarity is well beyond the scope of this course. Now that we have a determinant, here is a proof.

Suppose $A$ has size $n\text{.}$ Then

\begin{align*} \charpoly{A}{x} &=\detname{xI_n-A}&& \knowl{./knowl/theorem-CPD.html}{\text{Theorem CPD}}\\ &=\detname{\transpose{\left(xI_n-A\right)}}&& \knowl{./knowl/theorem-DT.html}{\text{Theorem DT}}\\ &=\detname{\transpose{\left(xI_n\right)}-\transpose{A}}&& \knowl{./knowl/theorem-TMA.html}{\text{Theorem TMA}}\\ &=\detname{x\transpose{I_n}-\transpose{A}}&& \knowl{./knowl/theorem-TMSM.html}{\text{Theorem TMSM}}\\ &=\detname{xI_n-\transpose{A}}&& \knowl{./knowl/definition-IM.html}{\text{Definition IM}}\\ &=\charpoly{\transpose{A}}{x}&& \knowl{./knowl/theorem-CPD.html}{\text{Theorem CPD}}\text{.} \end{align*}

So $A$ and $\transpose{A}$ have the same characteristic polynomial, and by Theorem EMRCP, their eigenvalues are identical and have equal algebraic multiplicities. Notice that what we have proved here is a bit stronger than the stated conclusion in the theorem.

### ExercisesCPExercises

###### T20.

Give a new proof of Theorem EMRCP (eigenvalues are roots of the characteristic polynomial) using the version of the characteristic polynomial given in Theorem CPD.

Solution

Suppose $A$ has size $n\text{.}$ Then

\begin{align*} \lambda\text{ is an eigenvalue of }A &\iff A-\lambda I_n\text{ is singular}&&\knowl{./knowl/theorem-ESM.html}{\text{Theorem ESM}}\\ &\iff \lambda I_n-A\text{ is singular}\\ &\iff\detname{\lambda I_n-A}=0&&\knowl{./knowl/theorem-SMZD.html}{\text{Theorem SMZD}}\\ &\iff\charpoly{A}{\lambda}=0&&\knowl{./knowl/definition-CP.html}{\text{Definition CP}}\text{.} \end{align*}

So eigenvalues of $A$ are precisely the roots of the characteristic polynomial.