In my Linear Algebra class we have been asked to find the kernel and
the image of the following linear map $T : \mathbb{R}^4 \rightarrow \mathbb{R}^4$, then verify the rank-nullity theorem.
$T(x_1,x_2,x_3,x_4) \mapsto(x_1-x_2, x_2-x_3,x_3-x_4,x_4-x_1)$
$\ker(T)$ was trivial, it's obviously equal to the set $\{(v,v,v,v) \mid v \in \mathbb{R} \}$.
However, I am having difficulty finding $\mathrm{im}(T)$. Some observations:
- There is no ordered 4-tuple in the image such that all four elements have the same sign.
- $T$ is not surjective, since the vector $(1,0,0,0) \not \in \mathrm{im}(T)$.
- Since $\{(1,0,0,0), …, (0,0,0,1)\}$ spans $\mathbb{R}^4$, their images $\{(1,0,0,-1),…,(0,0,-1,1)\}$ must span the image.
Can anyone provide a hint to the solution?
Best Answer
Hints: Let's look at your third point in more detail. You concluded that the images of the four standard basis vectors $$\langle (1, 0, 0, -1)^T, (-1, 1, 0, 0)^T, (0, -1, 1, 0)^T, (0, 0, -1, 1)^T \rangle$$ spans the image. But is this spanning set a basis for the image? In other words, is this set linearly independent dependent? If it's independent, we're in trouble with rank-nullity because you found the 1-dimensional kernel. But if it is dependent, how do you modify this set to get a basis?