Which are Invertible

Algebra Level 5

Which of the following is/are invertible linear transformations ?

$T_1: \mathbb{R}^3 \to \mathbb{R}^3$ is the transformation that takes $(x, \, y, \, z)$ to $(x - y,\, y - z,\, z - x)$ .
$T_2: \mathbb{C}^2 \to \mathbb{C}^2$ is the transformation that takes $(w, \, z)$ to $\big(\text{Re}(w) + \text{Im}(z) i, \, \text{Re}(z) + \text{Im}(w) i\big)$ .
$V$ is the vector space of all sequences of real numbers (vector addition creates a new sequence from the component-wise sums of the previous two). $T_3: V \to V$ is the "right shift" transformation that takes a sequence $\{a_n\}_{n \ge 0}$ and returns a sequence $\{b_n\}_{n \ge 0}$ satisfying $b_0 = 0$ and $b_n = a_{n - 1}$ for all $n \ge 1$ .

Assume $\mathbb{C}^2$ is a vector space over the complex numbers.

1 and 3 2 2 and 3 All None

1 solution

Henry Maltby
Jul 7, 2016

None of them are invertible linear transformations.

$T_1$ is not invertible because the matrix $\begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ -1 & 0 & 1 \end{pmatrix}$ is not invertible.

$T_2$ is not linear because $i \cdot T(1, \, i) = i \cdot (1 + i, \, 0) = (-1 + i, \, 0)$ but $T(i \cdot 1, \, i \cdot i) = T(i, \, -1) = (0, \, -1 + i)$ .

$T_3$ is not invertible because it has no right inverse. There is no linear transformation $R$ such that $T_3(R(v)) = v$ for all vectors $v$ . Consider, for instance, $v = \{1, \, 0, \, 0, \, \dots\}$ as the sequence beginning with a $1$ and then followed by $0$ 's. The output of $T_3$ always starts with a $0$ , so no matter what $R$ is, $T_3(R(v)) \ne v$ .

Note, however, that $T_3$ does have a left inverse given by the linear transformation known as "left shift".

Helpful and interesting but confused totally

Abdullahi Yusuf - 1 year, 11 months ago

Which are Invertible

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