Question

Let T: R4-R4 be defined by
T(x1,x2,x3,x4)=(-x2,x1,-x4,x3)
Check whether T is a linear operator and T4=I. Is T invertible?

Answer 1

Answer on Question #66802, Math, Linear Algebra

Let $T \colon R^4 \to R^4$ be defined by

$$T(x_1, x_2, x_3, x_4) = (-x_2, x_1, -x_4, x_3)$$

Check whether $T$ is a linear operator and $T^4 = I$. Is $T$ invertible?

Solution.

For linear operator $A$:

$$A(\alpha x + \alpha' x') = \alpha A(x) + \alpha' A(x')$$

where $x$ and $x'$ are vectors in $R^4$.

We have:

$$x = (x_1, x_2, x_3, x_4); \quad x' = (-x_2, x_1, -x_4, x_3)$$ $$T(\alpha x + \alpha' x') = T(\alpha x_1 - \alpha' x_2, \alpha x_2 + \alpha' x_1, \alpha x_3 - \alpha' x_4, \alpha x_4 + \alpha' x_3) = \\ = (-\alpha x_2 - \alpha' x_1, \alpha x_1 - \alpha' x_2, -\alpha x_4 - \alpha' x_3, \alpha x_3 - \alpha' x_4)$$ $$\alpha T(x) = \alpha(-x_2, x_1, -x_4, x_3)$$ $$\alpha' T(x') = \alpha'(-x_1, -x_2, -x_3, -x_4)$$ $$(-\alpha x_2 - \alpha' x_1, \alpha x_1 - \alpha' x_2, -\alpha x_4 - \alpha' x_3, \alpha x_3 - \alpha' x_4) = \alpha(-x_2, x_1, -x_4, x_3) + \alpha'(-x_1, -x_2, -x_3, -x_4)$$

So far, $T$ is a linear operator.

$$T^2(x_1, x_2, x_3, x_4) = (-x_1, -x_2, -x_3, -x_4)$$ $$T^3(x_1, x_2, x_3, x_4) = (x_2, -x_1, x_4, -x_3)$$ $$T^4(x_1, x_2, x_3, x_4) = (x_1, x_2, x_3, x_4)$$ $$T^4(x_1, x_2, x_3, x_4) = I(x_1, x_2, x_3, x_4)$$ $$T^4 = I$$

We can write:

$$T^{-1}(-x_2, x_1, -x_4, x_3) = (x_1, x_2, x_3, x_4)$$

Answer on Question #66802, Math, Linear Algebra

Then:

$$T T ^ {- 1} (x _ {1}, x _ {2}, x _ {3}, x _ {4}) = (x _ {1}, x _ {2}, x _ {3}, x _ {4})$$ $$T ^ {- 1} T (x _ {1}, x _ {2}, x _ {3}, x _ {4}) = T (x _ {2}, - x _ {1}, x _ {4}, - x _ {3}) = (x _ {1}, x _ {2}, x _ {3}, x _ {4})$$ $$T T ^ {- 1} = T ^ {- 1} T = I$$

$T$ is invertible.

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