Answer to Question #87244 in Linear Algebra for Abhishek

Question

Define T : R
3 → R
3 by
T(x, y, x) = (−x, x−y,3x+2y+z).
Check whether T satisfies the polynomial (x−1)(x+1)
2
. Find the minimal
polynomial of T.

Answer 1

Denoting by $I$  the identity operation, we have 

$$(T + I) (x, y, z) = (0, x, 3x + 2y + 2z) \, , \\ (T + I)^2 (x, y, z) = (0, 0, 8x + 4y + 4z) \, , \\ (T - I) (x, y, z) = (- 2 x, x - 2 y, 3x + 2y) \, .$$

From the last two lines, we then have

$$(T - I) (T + I)^2 (x , y , z) = (0 , 0, 0) \, .$$

Therefore, $T$ satisfies the polynomial $p (x) = ( x - 1 ) ( x + 1 )^2$. Considering $v = (1 , 0 , 0) \in \R^3$, we have 

$$T (v) = (-1 , 1, 3) \, , \quad T^2 (v) = ( 1 , -2 , 2) \, .$$

The vectors $v$, $T (v)$ and $T^2 (v)$ are linearly independent; hence, because the space $\R^3$ is three-dimensional, the minimal polynomial of $T$ has degree three. Since $p (x)$ has degree three and is a monic polynomial (the nonzero coefficient of the highest degree is equal to 1), and since $p (T) = 0$, it is the minimal polynomial.