Prove that linear operator is invertible

[gruba]

Homework Statement

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}.

Homework Equations

-Linear transformations
-Jordan decomposition

The Attempt at a Solution

[/B]
Linear mapping can be written as
\mathcal{A}<br /> \begin{bmatrix}<br /> x_1 \\<br /> x_2 \\<br /> x_3 \\<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> x_1+x_2-x_3 \\<br /> x_2+7x_3 \\<br /> -x_3 \\<br /> \end{bmatrix}<br />

Matrix of linear operator \mathcal{A} is T=<br /> <br /> \begin{bmatrix}<br /> 1 &amp; 1 &amp; -1 \\<br /> 0 &amp; 1 &amp; 7 \\<br /> 0 &amp; 0 &amp; -1 \\<br /> \end{bmatrix}<br /> <br />

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)?

 

[Ray Vickson]

Homework Statement

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}.

Homework Equations

-Linear transformations
-Jordan decomposition

The Attempt at a Solution

[/B]
Linear mapping can be written as
\mathcal{A}<br /> \begin{bmatrix}<br /> x_1 \\<br /> x_2 \\<br /> x_3 \\<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> x_1+x_2-x_3 \\<br /> x_2+7x_3 \\<br /> -x_3 \\<br /> \end{bmatrix}<br />

Matrix of linear operator \mathcal{A} is T=<br /> <br /> \begin{bmatrix}<br /> 1 &amp; 1 &amp; -1 \\<br /> 0 &amp; 1 &amp; 7 \\<br /> 0 &amp; 0 &amp; -1 \\<br /> \end{bmatrix}<br /> <br />

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)?

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.