Cayley-Hamilton Theorem explanation. I'm having trouble interpreting the meaning of $f(T)$.

Question

Using the Theorem from "Linear Algebra" (-Friedberg, Insel, Spence 5th edition).

(Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $f(t)$ be the characteristic polynomial of $T$, then $f(T)=T_0$, the zero transformation. That is T "satisfies" its characteristic Equation

It looks like we're inserting some linear operator as input for the characteristic polynomial. This notation is used in some proofs in the text that I can find, but I haven't been able to find any explanation. The $T$ seems to be used interchangeably with the linear operator applied to a matrix fairly often, but I can't make sense of that either. The $f(T)$ has come up in one of my assigned problems and I can't move forward until I have some understanding of what this means.

Answer 1

If $f(x) = c_nx^n + c_{n-1}x^{n-1} + \cdots + c_1 x + c_0$ is a polynomial, then $$f(T) = c_nT^n + c_{n-1}T^{n-1} + \cdots + c_1 T + c_0I,$$ where $I$ is the identity operator. Here, $T^n$ is the linear operator $T$ composed with itself $n$-times. If you fix a basis, then $T$ has an associated matrix $A_T$. In this case $f(A_T)$ makes sense where $A_T^n$ is the $n$-th power of the matrix. Both of these perpectives agree.

Answer 2

If you've seen some rings before there's another perspective you could use. If $V$ is a vector space over a field $F$, and $T$ is a linear operator on $V$, consider the polynomial ring $F[x]$ (which is a ring containing polynomials with coefficients from $F$). Then there is an ring homomorphism $\Phi: F[x] \rightarrow Hom(V,V)$ defined by $$ f(x) \mapsto f(T). $$ So what this means is that given an arbitrary polynomial with coefficients from $F$, you can associate to it a linear operator on $V$ by "replacing $x$ with $T$" in the polynomial. So for example if you had a real vector space $V$, then the polynomial $f(x) = 5x^2+3x+1$ would map to $$ 5x^2+3x+1 \mapsto 5T^2+3T+id_V $$ where here $id_V$ denotes the identity transformation on $V$.

The use of $\Phi$ being a ring homomorphism is that any of the standard algebraic manipulations you can do on elements of $F[x]$ will also work for linear operators in $Hom(V,V)$. And as Zorngo mentionned the powers of $T$ correspond to repeated composition and $T^0 = id_V$.

Also, the notation $f(T)v$ might be clearer now because I've tried to illustrate that $f(T)$ is itself another linear operator on $V$, hence $f(T)v$ just means evaluate the linear operator at $v$. Concretely, if we have the linear operator associated to the polynomial $f(x) = 5x^2+3x+1$ again, then $$ f(T)v = 5T^2(v)+3T(v)+id_V(v). $$