Is it possible to have this linear map

Question

Is it possible to have a linear map $T : \mathbb{R^3} → \mathbb{R^3}$ such that its range is the subspace spanned by $(3, 2, 1)$ and $(4,5,6)$ and whose null space is the subspace spanned by $(3, 2, 1)$?

My attempt: I think it is not possible.

Consider this example:

Let $T(x,y,z)=(3x+4y,2x+5y,x+6y)$, then, $\mathrm{null}\, T=\{0\}$.

How to show this for all such maps?

Answer 1

First addressing your particular example, try to evaluate $T(0,0,1)$.

Now guide for the original problem:

We would required $T(3,2,1)=(0,0,0)$.

Try to find two vectors $v_1, v_2$ such that $\{ v_1, v_2, (3,2,1)\}$ is a basis of $\mathbb{R}^3$.

Define $T(v_1)=(3,2,1)$ and $T(v_2)=(4,5,6).$

Verify if this construction satisfy the requirement of the question.

Answer 2

You're wrong: it is perfectly possible.

Let $v_2=(3,2,1)$, $v_3=(4,5,6)$. Note $v_2$ and $v_3$ are lineatly independent, so we may complete this set of vectors by a third vector $v_1$, to make up a basis of $\mathbf R^3$, say $v_1=(1,0,0)$.

Now consider the linear map $T$ defined by $$\begin{cases}T(v_1)=0,\\T(v_2)=v_2,\\T(v_3)=v_1.\end{cases}$$ $T$ fulfills all your conditions, its matrix the and you can have is matrix in the basis $(v_1,v_2,v_3)$ is $$\begin{pmatrix}0&0&1\\0&1&0\\0&0&0\end{pmatrix}. $$ You can obtain its matrix in the canonical basis using the change of basis formula.