# Prove that linear operator is invertible

1. Dec 7, 2015

### gruba

1. The problem statement, all variables and given/known data
Let $\mathcal{A}: \mathbb{R^3}\rightarrow \mathbb{R^3}$ is a linear operator defined as $\mathcal{A}(x_1,x_2,x_3)=(x_1+x_2-x_3, x_2+7x_3, -x_3)$
Prove that $\mathcal{A}$ is invertible and find matrix of $\mathcal{A},A^{-1}$ in terms of canonical basis of $\mathbb{R^3}$.

2. Relevant equations
-Linear transformations
-Jordan decomposition

3. The attempt at a solution

Linear mapping can be written as
$$\mathcal{A} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} = \begin{bmatrix} x_1+x_2-x_3 \\ x_2+7x_3 \\ -x_3 \\ \end{bmatrix}$$

Matrix of linear operator $\mathcal{A}$ is $$T= \begin{bmatrix} 1 & 1 & -1 \\ 0 & 1 & 7 \\ 0 & 0 & -1 \\ \end{bmatrix}$$

Using Jordan decomposition method, it is possible to find $T,T^{-1}$ in terms of canonical basis.

How to strictly prove that $\mathcal{A}$ is invertible (without matrix computation)?

2. Dec 7, 2015

### Ray Vickson

This belongs in the "Precalculus" math forum, not the Advanced Physics forum.

Anyway, you do not need to use matrices in this problem: you just want to show that ${\cal A}:\vec{x} \rightarrow \vec{y}$ can be inverted, which means there is a map ${\cal B}: \vec{y} \rightarrow \vec{x}$ with the property that $\vec{x} = {\cal B \, A} \vec{x}$. This amounts to simply solving the linear equations
$$x_1+x_2-x_3 =y_1, \: x_2+7x_3 = y_2, \: -x_3 = y_3$$
to determine the $x_i$ in terms of the $y_j$. That is a simple high-school algebra exercise.