Prove that there does not exist a linear map

Question

I have to prove that there does not exist a linear map from $\mathbb{R}^5$ to $\mathbb{R}^2$ whose kernel is $\left\{\left(\begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{matrix}\right) | x_1 = 3x_2 \space and \space x_3 = x_4 = x_5 \right\}$

but I do not know where to start

Answer 1

If $\phi:\mathbb{R}^{5}\rightarrow\mathbb{R}^{2}$ is any linear map, then the rank-nullity theorem tells you that $$5=\dim(\text{Ker}(\phi))+\dim(\text{Im}(\phi)).$$ Since $\text{Im}(\phi)\subset\mathbb{R}^{2}$, its dimension is at most $2$, so that $\dim(\text{Ker}(\phi))\geq 3$. The subspace you are given is $$\text{Span}\{(3,1,0,0,0),(0,0,1,1,1)\},$$ which is $2$-dimensional. So it cannot possibly be the kernel of a linear map $\phi:\mathbb{R}^{5}\rightarrow\mathbb{R}^{2}$.

Answer 2

Suppose there actually exists a linear map T. We know $ker(T)+rank(T)=5$ by rank-nullity theorem. Now $rank(T)\leq2$, so $ker(T) \geq 3$. Now what is the dimension of the kernel given? Is it greater or equal to 3? Do you see a contradiction?