Question

define T: R3→R3 by
. T(x, y, x)=(x+y, y, 2x-2y+2z)
Check that T satisfies the polynomial (x-1)square(x-2). Find the minimal polynomial of T.

Answer 1

Answer on Question #75930 – Math – Linear Algebra

Question

Given: $T\colon \mathbb{R}^3 \to \mathbb{R}^3$ is a linear transformation defined by

$$T(x, y, z) = (x + y, y, 2x - 2y + 2z)$$

To prove or disprove: $(x - 1)^2(x - 2)$ is the characteristic polynomial and find minimal polynomial.

Solution

Consider the map $T\colon \mathbb{R}^3 \to \mathbb{R}^3$ defined by

$$T(x, y, z) = (x + y, y, 2x - 2y + 2z)$$

∴ Matrix representation of $T$ with respect to standard basis $\{e_1, e_2, e_3\}$ is as follows

$$T(1, 0, 0) = (1, 0, 2)$$ $$T(0, 1, 0) = (1, 1, -2)$$ $$T(0, 0, 1) = (0, 0, 2)$$ $$\Rightarrow \quad \text{Matrix representation } [T] = A = \begin{bmatrix} 1 & 0 & 2 \\ 1 & 1 & -2 \\ 0 & 0 & 2 \end{bmatrix}$$

Now we will check whether $(x - 1)^2(x - 2)$ is a characteristic polynomial.

So, replace $x$ by $A$ and $1$ by $I$, we get

$$(A - I)^2(A - 2I)$$ $$\Rightarrow \quad \begin{bmatrix} 0 & 0 & 2 \\ 1 & 0 & -2 \\ 0 & 0 & 1 \end{bmatrix}^2 \begin{bmatrix} -1 & 0 & 2 \\ 1 & -1 & -2 \\ 0 & 0 & 0 \end{bmatrix}$$ $$\Rightarrow \quad \begin{bmatrix} 0 & 0 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 2 \\ 1 & -1 & -2 \\ 0 & 0 & 0 \end{bmatrix}$$ $$\Rightarrow \quad \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right]$$

Therefore, $T$ satisfies the polynomial.

To find the minimal polynomial, first we will find the polynomial with a linear factor of characteristic polynomial.

So, the product of linear factors is $(x - 1)(x - 2)$

Replace $x$ by $A$ and 1 by $I$, we get

$$\begin{array}{l} (A - I)(A - 2I) \\ \Rightarrow \quad \left[ \begin{array}{ccc} 0 & 0 & 2 \\ 1 & 0 & -2 \\ 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{ccc} -1 & 0 & 2 \\ 1 & -1 & -2 \\ 0 & 0 & 0 \end{array} \right] \\ \Rightarrow \quad \left[ \begin{array}{ccc} 0 & 0 & 0 \\ -1 & 0 & 2 \\ 0 & 0 & 0 \end{array} \right] \neq \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] \end{array}$$

Hence, the minimal polynomial is same as the characteristic polynomial.

$$\therefore \quad m(x) = (x - 1)^2(x - 2)$$

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