Linear Algebra/Polynomials of Maps and Matrices

Recall that the set of square matrices is a vector space under entry-by-entry addition and scalar multiplication and that this space ${ℳ}_{n\times n}$ has dimension $n^2$. Thus, for any $n\times n$ matrix $T$ the $n^2+1$-member set $\{I,T,T^2,\dots ,T^{n^2}\}$ is linearly dependent and so there are scalars $c_0,\dots ,c_{n^2}$ such that $c_{n^2}{T}^{n^2}+\dots +c_{1}T+c_{0}I$ is the zero matrix.

Of course, if $T={\mathrm{R}\mathrm{e}\mathrm{p}}_{B,B}(t)$ then $f(T)={\mathrm{R}\mathrm{e}\mathrm{p}}_{B,B}(f(t))$, which follows from the relationships $T^j={\mathrm{R}\mathrm{e}\mathrm{p}}_{B,B}(t^j)$, and $cT={\mathrm{R}\mathrm{e}\mathrm{p}}_{B,B}(ct)$, and $T_1+T_2={\mathrm{R}\mathrm{e}\mathrm{p}}_{B,B}(t_1+t_2)$.

As Example 1.2 shows, there may be polynomials of degree smaller than $n^2$ that zero the map or matrix.

A minimal polynomial always exists by the observation opening this subsection. A minimal polynomial is unique by the "with leading coefficient $1$" clause. This is because if there are two polynomials $m(x)$ and $\hat{m}(x)$ that are both of the minimal degree to make the map or matrix zero (and thus are of equal degree), and both have leading $1$'s, then their difference $m(x)-\hat{m}(x)$ has a smaller degree than either and still sends the map or matrix to zero. Thus $m(x)-\hat{m}(x)$ is the zero polynomial and the two are equal. (The leading coefficient requirement also prevents a minimal polynomial from being the zero polynomial.)

Using the method of that example to find the minimal polynomial of a $3\times 3$ matrix would mean doing Gaussian reduction on a system with nine equations in ten unknowns. We shall develop an alternative. To begin, note that we can break a polynomial of a map or a matrix into its components.

In particular, if a minimial polynomial $m(x)$ for a transformation $t$ factors as $m(x)=(x-{\lambda }_1{)}^{q_1}\cdots (x-{\lambda }_{\ell }{)}^{q_{\ell }}$ then $m(t)=(t-{\lambda }_1{)}^{q_1}\circ \cdots \circ (t-{\lambda }_{\ell }{)}^{q_{\ell }}$ is the zero map. Since $m(t)$ sends every vector to zero, at least one of the maps $t-{\lambda }_i$ sends some nonzero vectors to zero. So, too, in the matrix case— if $m$ is minimal for $T$ then $m(T)=(T-{\lambda }_{1}I{)}^{q_1}\cdots (T-{\lambda }_{\ell }I{)}^{q_{\ell }}$ is the zero matrix and at least one of the matrices $T-{\lambda }_{i}I$ sends some nonzero vectors to zero. Rewording both cases: at least some of the ${\lambda }_i$ are eigenvalues. (See Problem 17.)

Recall how we have earlier found eigenvalues. We have looked for $\lambda$ such that $T\overrightarrow{v}=\lambda \overrightarrow{v}$ by considering the equation $\overrightarrow{0}=T\overrightarrow{v}-x\overrightarrow{v}=(T-xI)\overrightarrow{v}$ and computing the determinant of the matrix $T-xI$. That determinant is a polynomial in $x$, the characteristic polynomial, whose roots are the eigenvalues. The major result of this subsection, the next result, is that there is a connection between this characteristic polynomial and the minimal polynomial. This results expands on the prior paragraph's insight that some roots of the minimal polynomial are eigenvalues by asserting that every root of the minimal polynomial is an eigenvalue and further that every eigenvalue is a root of the minimal polynomial (this is because it says "$1\le q_i$" and not just "$0\le q_i$").

The proof takes up the next three lemmas. Although they are stated only in matrix terms, they apply equally well to maps. We give the matrix version only because it is convenient for the first proof.

The first result is the key— some authors call it the Cayley-Hamilton Theorem and call Theorem 1.8 above a corollary. For the proof, observe that a matrix of polynomials can be thought of as a polynomial with matrix coefficients.

$\left(\begin{array}{cc}2x^2+3x-1& x^2+2\\ 3x^2+4x+1& 4x^2+x+1\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 3& 4\end{array}\right)x^2+\left(\begin{array}{cc}3& 0\\ 4& 1\end{array}\right)x+\left(\begin{array}{cc}-1& 2\\ 1& 1\end{array}\right)$

We sometimes refer to that lemma by saying that a matrix or map satisfies its characteristic polynomial.

Combining the prior two lemmas gives that the minimal polynomial divides the characteristic polynomial. Thus, any root of the minimal polynomial is also a root of the characteristic polynomial. That is, so far we have that if $m(x)=(x-{\lambda }_1{)}^{q_1}\dots (x-{\lambda }_i{)}^{q_i}$ then $c(x)$ must has the form $(x-{\lambda }_1{)}^{p_1}\dots (x-{\lambda }_i{)}^{p_i}(x-{\lambda }_{i+1}{)}^{p_{i+1}}\dots (x-{\lambda }_{\ell }{)}^{p_{\ell }}$ where each $q_j$ is less than or equal to $p_j$. The proof of the Cayley-Hamilton Theorem is finished by showing that in fact the characteristic polynomial has no extra roots ${\lambda }_{i+1}$, etc.

Exercises

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References

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