Eigenvalues of a Linear Transformation Homework assistance

Solution 1:

I agree with (A) and (C), although I think you should be clearer about the reason for (A). (For example, your reasoning implies that $2^2 = 4$ must be an eigenvalue of $T^2$, but this is not a root of $p_T$.)

(B) The eigenvalues of $T$ are the roots of its characteristic polynomial, so:

The only eigenvalues of $T$ are $0$, $\pm 1$, $\pm 2$. But the eigenvalues of $T^2$ are the squares of eigenvalues of $T$, so there are three distinct, not five.

(D) The eigenvalues of $T^2$ are the squares of eigenvalues of $T$, hence:

They are $0$, $1$, and $4$. The eigenvalues of $T^2 - 4 I$ are just these decreased by $4$, i.e. $-4$, $-3$, and $0$. At any rate, there aren't four.

(E) The eigenvalues of $T^3$ are the cubes $\lambda^3$ of the eigenvalues $\lambda$ of $T$, and similarly the eigenvalues of $T^3 - T$ are given by $\lambda^3 - \lambda$, hence:

The eigenvalues of $T^3 - T$ are then $0^3 - 0 = 0$, $1^3 - 1 = 0$, $(-1)^3 - (-1) = 0$, $2^3 - 2 = 6$, and $(-2)^3 - (-2) = -6$. This matches up with the given characteristic polynomial.