Recipe

Evaluate a polynomial at an operator or matrix

Use when: you're given a polynomial $p(x)=c_0+c_1x+\cdots+c_nx^n$ and an operator or matrix $A$ , and asked to compute $p(A)$

Goal: the single matrix (or operator) $p(A)$ , built entirely from powers of $A$ and the identity

Taught in The algebra L(V): composition, powers, polynomials in T

Quick recipe

List the powers of $A$ the polynomial actually needs
Compute each required power by matrix multiplication
Multiply each power by its coefficient, including the constant term
Add every term together, entry by entry

1
Read off the highest power $n$ appearing in $p$ , then plan to build $A^2,A^3,\dots,A^n$ in order, each one from the previous power.

Example

For $p(x)=x^2-2x+3$ and $A=\begin{pmatrix}2&1\\0&3\end{pmatrix}$ , only $A^2$ is needed beyond $A$ itself.

Why: building each power from the one before avoids recomputing $A$ multiplied by itself from scratch every time.
2
Multiply the previous power by $A$ one more time, entry by entry, and double-check the result before moving to the next power.

Example

$A^2=\begin{pmatrix}2&1\\0&3\end{pmatrix}\begin{pmatrix}2&1\\0&3\end{pmatrix}=\begin{pmatrix}4&5\\0&9\end{pmatrix}$

Why: every later term of the polynomial is built from these powers, so an arithmetic slip here carries into the final answer.
3
Multiply $A^k$ by $c_k$ for every power, and multiply the identity matrix $I$ , not just the number $c_0$ , by the constant coefficient.

Example

$c_2A^2=\begin{pmatrix}4&5\\0&9\end{pmatrix},\quad c_1A=-2\begin{pmatrix}2&1\\0&3\end{pmatrix}=\begin{pmatrix}-4&-2\\0&-6\end{pmatrix},\quad c_0I=3\begin{pmatrix}1&0\\0&1\end{pmatrix}=\begin{pmatrix}3&0\\0&3\end{pmatrix}$

Why: $p(A)$ must stay a matrix throughout, and only $c_0I$ , not the bare number $c_0$ , has matching dimensions to add to the other terms.
4
Sum all the coefficient-scaled powers found in Step 3, matching each matrix entry to the entry in the same position of every other term.

Example

$p(A)=\begin{pmatrix}4&5\\0&9\end{pmatrix}+\begin{pmatrix}-4&-2\\0&-6\end{pmatrix}+\begin{pmatrix}3&0\\0&3\end{pmatrix}=\begin{pmatrix}3&3\\0&6\end{pmatrix}$

Why: this final sum is the definition of $p(A)$ itself, once every power has already been computed and scaled correctly.

Common mistakes

Adding the constant term as a bare number instead of that number times the identity matrix.
Computing $A^k$ by multiplying $A$ by itself $k$ separate times instead of reusing the previous power.
Dropping a negative sign when a coefficient is negative, especially on the linear term.
Forgetting that $p(A)$ commutes with $A$ itself, a fact worth using to catch a computational slip.