Prove that linear operator is invertible

[gruba]

Homework Statement

Let [itex]\mathcal{A}: \mathbb{R^3}\rightarrow \mathbb{R^3}[/itex] is a linear operator defined as [itex]\mathcal{A}(x_1,x_2,x_3)=(x_1+x_2-x_3, x_2+7x_3, -x_3)[/itex]
Prove that [itex]\mathcal{A}[/itex] is invertible and find matrix of [itex]\mathcal{A},A^{-1}[/itex] in terms of canonical basis of [itex]\mathbb{R^3}[/itex].

Homework Equations

-Linear transformations
-Jordan decomposition

The Attempt at a Solution

[/B]
Linear mapping can be written as
[tex]\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}
[/tex]

Matrix of linear operator [itex]\mathcal{A}[/itex] is [tex]T=

\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & 7 \\
0 & 0 & -1 \\
\end{bmatrix}

[/tex]

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

How to strictly prove that [itex]\mathcal{A}[/itex] is invertible (without matrix computation)?

 

[Ray Vickson]

Homework Statement

Let [itex]\mathcal{A}: \mathbb{R^3}\rightarrow \mathbb{R^3}[/itex] is a linear operator defined as [itex]\mathcal{A}(x_1,x_2,x_3)=(x_1+x_2-x_3, x_2+7x_3, -x_3)[/itex]
Prove that [itex]\mathcal{A}[/itex] is invertible and find matrix of [itex]\mathcal{A},A^{-1}[/itex] in terms of canonical basis of [itex]\mathbb{R^3}[/itex].

Homework Equations

-Linear transformations
-Jordan decomposition

The Attempt at a Solution

[/B]
Linear mapping can be written as
[tex]\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}
[/tex]

Matrix of linear operator [itex]\mathcal{A}[/itex] is [tex]T=

\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & 7 \\
0 & 0 & -1 \\
\end{bmatrix}

[/tex]

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

How to strictly prove that [itex]\mathcal{A}[/itex] 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
[tex] x_1+x_2-x_3 =y_1, \: x_2+7x_3 = y_2, \: -x_3 = y_3 [/tex]
to determine the ##x_i## in terms of the ##y_j##. That is a simple high-school algebra exercise.