Decide whether or not $\phi:\mathbb{R}^2\rightarrow\mathbb{R}^2$ given by $\phi(x,y) = (x+y,2x-y)$ is an isomorphism.

Question

Decide whether or not $\phi:\mathbb{R}^2\rightarrow\mathbb{R}^2$ given by $\phi(x,y) = (x+y,2x-y)$ is an isomorphism.

I know that to be an isomorphism, it must be both injective and surjective.

Injective meaning ker$\phi = 0$

Surjective meaning $\phi(V)=W$

But, how does one go about proving those with the given values above?

Answer 1

Various way are possible. Let's do the pedantic way:

1.Injective ker$\phi = 0$

If $(x+y,2x-y)$ then from $x+y=0$ we obtain $x=-y$ and then substituing in the other variable we obtain $-3y=0$ and then $x=0$. So if $ \phi =0$ then $(x,y)=(0,0)$

2.Surjective $\phi(V)=W$

For every $(z,w)$ we can find $(x,y)$ such that $(x+y,2x-y)=(w,z)$. Suffices consider $x=(w-z)/3$ and $y=(2w+z)/3$

As I said this is the pedantic way. Other ways could be considering the Jacobian of the application or finding directly an inverse

Answer 2

In this case both the domain and codomain have the same dimension, so it suffices to show that $\phi$ is injective. This amounts to showing that $(x,y) = (0,0)$ is the only vector that satisfies $\phi(x,y) = (0,0)$, or $$\begin{align*}x + y &= 0 \\ 2x - y &= 0\end{align*}$$

This is just a linear system you can solve.

Answer 3

Judging by the tags, you wish to prove that $\phi$ is a linear isomorphism.

To do so, note that $\phi$ can be written as $\phi(\vec x)=A\vec x$ where $$ A= \begin{bmatrix} 1 & 1\\ 2 & -1 \end{bmatrix} $$ Note that $\phi$ is an isomorphism if and only if $A$ is invertible (do you see why?). Since $\det A=-3\neq 0$ we immediately see that $\phi$ is indeed an isomorphism.

Answer 4

Alternative: $\psi(x,y)=(\frac13(x+y),2x-y)$ can be verified to be (left and right) inverse of $\phi$.

Answer 5

For a linear mapping, it is an equivalence for a linear mapping to be either bijective or injective or surjective. So, once you have proven for example injectivity, you are done. And injectivity is equivalent with $det(M) \neq 0$ ($M$ being the associated matrix with respect to the canonical base, here with 1st line $1,1$ and second line, $2, -1$). The value of this determinant is $-3 \neq 0$ proving the result.

Answer 6

For a linear transformation $T$ between finite dimensional vector spaces of the same dimension, there is a theorem that states that if $T$ is injective then is it necessarily an isomorphism.

This comes from the fact that if $T$ is injective, then it maps linearly independent sets to linearly independent sets (you should verify this). Since the image of the basis of the domain will be linearly independent, it must span the whole space since the domain and codomain have the same dimension.

Knowing this, your problem reduces to showing that your map is an injection, i.e. that the only element that gets mapped to $0$ is $0$. So this problem is just asking you to solve the system of equations obtained by setting the first and second coordinate of the image equal to zero i.e.

$x+y=0$

$2x-y=0$