When is this linear transformation an isomorphism?

[lep11]

Homework Statement

Let L: ℝ²→ℝ² such that L(x₁, x₂)^T=(1, 2 ; 3, α)(x₁, x₂)^T=Ax
Determine at what values of α is L an isomorphism. Obviously L is given in matrix form.

The Attempt at a Solution

First of all a quick check, dim (ℝ²)=dim(ℝ²)=2 Ok.

An isomorphism means linear transformation which is bijective. $Det (A)=α-6$ so for A to be invertible and therefore L to be bijective must hold $α≠6.$ On the other hand it suffices to determine when L is injective because L is bijective iff it's injective. If $α$ was 0, ker(L) would include infinitely many vectors and L would neither be injective nor bijective.
So $α≠6.$ and $α≠0.$
What's the best approach to this particular problem?

 

[pasmith]

A map from $\mathbb{R}^n$ to itself is invertible if and only if its determinant is non-zero. You have shown that the determinant of your map is $\alpha - 6$.

When $\alpha = 0$ the inverse of $\begin{pmatrix} 1 & 2 \\ 3 & 0 \end{pmatrix}$ is $\begin{pmatrix} \frac12 & -\frac16 \\ 0 & \frac13 \end{pmatrix}$.

 

[lep11]

A map from $\mathbb{R}^n$ to itself is invertible if and only if its determinant is non-zero. You have shown that the determinant of your map is $\alpha - 6$.

We haven't actually been told that in our class (yet?) so we haven't proven that either. Is it that simple? So is the final solution that L is an isomorphism when α∈ℝ\{6}?

When $\alpha = 0$ the inverse of $\begin{pmatrix} 1 & 2 \\ 3 & 0 \end{pmatrix}$ is $\begin{pmatrix} \frac12 & -\frac16 \\ 0 & \frac13 \end{pmatrix}$.

What about kernel of L and injectivity if α=0?

 

[pasmith]

We haven't actually been told that in our class (yet?) so we haven't proven that either. Is it that simple? So is the final solution that L is an isomorphism when α∈ℝ\{6}?What about kernel of L and injectivity if α=0?

The kernel is trivial: if $$\begin{pmatrix} 1 & 2 \\ 3 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = 0$$ then the bottom row gives $3x = 0$, so that $x = 0$. The top row then gives $x + 2y = 2y = 0$ so $y = 0$ also.

Similarly it is injective: if $$\begin{pmatrix} 1 & 2 \\ 3 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a \\ b\end{pmatrix}$$ then $x = \frac13 b$ and $y = \frac12a - \frac16b$.

(The inverse given in my previous post appears to be incorrect; as appears from the above it should be $\begin{pmatrix} 0 & \frac13 \\ \frac12 & - \frac16\end{pmatrix}$