# How to Calculate and Simplify a Matrix Polynomial

## Problem 47

Let $T=\begin{bmatrix} 1 & 0 & 2 \\ 0 &1 &1 \\ 0 & 0 & 2 \end{bmatrix}$.
Calculate and simplify the expression
$-T^3+4T^2+5T-2I,$ where $I$ is the $3\times 3$ identity matrix.

(The Ohio State University Linear Algebra Exam)

Contents

## Hint.

Use the Cayley-Hamilton theorem.

If $p(t)$ is the characteristic polynomial for an $n\times n$ matrix $A$, then the matrix $p(A)$ is the $n \times n$ zero matrix.

## Solution.

We use the Cayley-Hamilton theorem.

The Cayley-Hamilton Theorem.
If $p(t)$ is the characteristic polynomial for an $n\times n$ matrix $A$, then the matrix $p(A)$ is the $n \times n$ zero matrix.

To obtain the characteristic polynomial for $T$, we note that the matrix $T$ is upper triangular. Thus $T-tI$ is also upper triangular and recall that the determinant of an upper triangular matrix is the product of the diagonal entries. Thus the characteristic polynomial $p_T(t)$ for $T$ is
$p_T(t)=\det(T-tI)=(1-t)(1-t)(2-t)=-t^3+4t^2-5t+2.$

By the Cayley-Hamilton theorem, we have
$p_T(T)=-T^3+4T^2-5T+2I=O.$ (Don’t forget $I$.)
Here $O$ is the $3 \times 3$ zero matrix.

Now we compute
\begin{align*}
-T^3+4T^2+5T-2I&=(-T^3+4T^2-5T+2I)+(10T-4I)\\
&=p_T(T)+10T-4I=10T-4I\6pt] &=\begin{bmatrix} 10 & 0 & 20 \\ 0 &10 &10 \\ 0 & 0 & 20 \end{bmatrix} \begin{bmatrix} 4 & 0 & 0 \\ 0 &4 &0 \\ 0 & 0 & 4 \end{bmatrix}\\[6pt] &=\begin{bmatrix} 6 & 0 & 20 \\ 0 &6 &10 \\ 0 & 0 & 16 \end{bmatrix}. \end{align*} Therefore the answer is \[-T^3+4T^2+5T-2I=\begin{bmatrix} 6 & 0 & 20 \\ 0 &6 &10 \\ 0 & 0 & 16 \end{bmatrix}.

## Comment.

This problem is actually one of the final exam problems in the linear algebra class I taught.
Of course, you may calculate them directly but most students who computed directly made calculation mistakes (unfortunately).

The second common mistake is that students jumped to conclusion that the expression $-T^3+4T^2+5T-2I$ is the zero matrix after finding the characteristic polynomial.
Those students knew that they could use the Cayley-Hamilton theorem but were a bit careless. The given expression and the characteristic polynomial are slightly different.

Anyway, I didn’t subtract many points for the second mistake.
(And you get little points if you made a mistake for a direct calculation.)

Let $A$ be an $n\times n$ matrix such that $A^k=I_n$, where $k\in \N$ and $I_n$ is the $n \times n$...