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.

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.)

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