Proving/Disproving injectivity with $f(x,y,z) = (x+z, y-2z, -x-y+z)$

Question

I know this is really simple, but it's aggravating the hell out of me.

Say we have $(a,b,c),(x,y,z)\in\mathbb R^3$ such that $f(a,b,c) = f(x,y,z)$.

This means that: \begin{align}a+c &= x+z \\ b-2c &= y-2z\\ -a-b+c &= -x-y+z\end{align}

I can't find solutions to these, which tells me its probably injective. To make sure, I'd like to find a value to show that it's not, but I don't have the intuition. What should I be doing in these cases? Thanks.

Answer 1

The function is linear so $f(a,b,c)=f(x,y,z)$ iff $f(a-x,b-y,c-z)=0$.

So to prove that $f$ is injective it is sufficient to prove thta $f(x,y,z)=0 \Rightarrow (x,y,z)=(0,0,0)$ ( the same method can be used for any linear function).

So you only have to solve the system: \begin{align} &x+z=0\\ &y-2z=0\\ &-x-y+z=0 \end{align}

Answer 2

The linear map $f\colon \mathbb{R}^{3}\to \mathbb{R}^{3}$ is given by the matrix $$ \begin{bmatrix}1 & 0 & 1\\0 & 1 & -2 \\ -1 & -1 & 1 \end{bmatrix} $$ Its determinant is $0$. Hence the kernel is non trivial and the map is not injective.

Answer 3

To show it is not injective, you have to show ONE of the following:

• that there is a vector in $ \mathbb{R}^3 $ for which there is no vector that maps to it.

• that the dimension of the range is less than the dimension of the domain. You can do this by noting that the mapping is from $ \mathbb{R}^3 $ to $ \mathbb{R}^3 $ but the dimension of the null space is greater than zero.

and there are a bunch of equivalent ways of showing it. The bottom line is that the natural way to show all this is to formulate the problem as a set of linear equations, with a matrix. So, don't fight that. Embrace it.