$f:\mathbb{R}^3 ⇒ \mathbb{R}^3$, a linear map with $f(x,y,z) = (x-y+2z,z-y,x+z)$

Question

If $f\colon\mathbb{R}^3\to \mathbb{R}^3$, a linear map with

$f(x,y,z) = (x-y+2z,z-y,x+z)$ then:

1. the matrix $A_f$ on some basis is invertible
2. $f^{-1}$ exists
3. the matrix $A_f$ on some usual basis has $\operatorname{rank}A=2$
4. the matrix $A_f$ on some usual basis has $\operatorname{dim}N(A)=2$

which of those is correct (1,2,3,4)?

I have excludes the 4th because $\operatorname{dim}N(A)=1$, i calculated the 3rd and i got $\operatorname{rank} A=2$, but i am not sure about the other 2 (1 and 2).

Can someone help me ?

Answer 1

Yes, $\dim\ker f=1$. So, $f$ is not invertible, and therefore neither is the matrix of $f$ with respect to any basis. Besides, since $f$ is not invertible, $f^{-1}$ does not exist. On the other hand, $f(1,0,0)=(1,0,1)$, $f(0,1,0)=(-1,-1,0)$, and$$f(0,0,1)=(2,1,1)=f(1,0,0)-f(0,0,1).$$So,$$f\left(\Bbb R^3\right)=\operatorname{span}\bigl\{(1,0,1),(-1,-1,0)\bigr\}$$and, since $\bigl\{(1,0,1),(-1,-1,0)\bigr\}$ is linearly independent, $\operatorname{rank}f=2$. So, the only correct assertion is the third one.

Answer 2

You can note that 1 and 2 are equivalent and exclude both 3 and 4. Moreover 3 and 4 contradict each other, because of rank-nullity.

Note also that the rank is an invariant: no matter what basis you choose, the rank of the associated matrix is the same.

Now, if you have determined that the null space has dimension one, you know that the rank is two (rank-nullity). And you have your answer.

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A simple way to ascertain that 3 is correct (so the others aren't) is to compute the matrix with respect to the standard basis and perform row reduction $$ \begin{bmatrix} 1 & -1 & 2 \\ 0 & -1 & 1 \\ 1 & 0 & 1 \end{bmatrix} \to \begin{bmatrix} 1 & -1 & 2 \\ 0 & -1 & 1 \\ 0 & 1 & -1 \end{bmatrix} \to \begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix} $$ This shows that the rank is two. Also it shows that $f(0,0,1)=f(1,0,0)-f(0,1,0)$ and therefore that $(1,-1,-1)$ belongs to the null space.