# Find a basis and the dimension of the subspace R^3 consisting of all vectors of x

Let $T_A : R^4 \rightarrow R^2$ be a multiplication by A. Find a basis and the dimension of the subspace $R^3$ consisting of all vectors of x for which $T_A(x)=0$ where A= $$\begin{bmatrix} 4 & 2 & 1 & -1 \\ 7 & -1 & 0 & 2 \\ \end{bmatrix}$$ I'm fairly new at this concept so I set up the problem but am confused on where to go. so I multiplied that matrix by $$\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\\ \end{bmatrix}$$ which equals $$\begin{bmatrix} 4x_1+2x_2+x_3-x_4\\ 7x_1-x_2+2x_4\\ \end{bmatrix}$$ so the equations are $4x_1+2x_2+x_3-x_4=0$ and $7x_1-x_2+2x_4=0$ but I'm not sure where to go from here.

• From there you have \begin{align} x_4 &= -\frac72 x_1+\frac12 x_2\\ x_3 &= 4x_1+2x_2 + \left(-\frac72 x_1+\frac12 x_2\right) = \frac12 x_1+\frac52 x_2. \end{align} – Math1000 May 8 '18 at 22:57

Continuing on from @Math1000's very helpful comment it is now apparent that $$\operatorname{null}T = \{(x_1,x_2,\frac{1}{2}x_1+\frac{5}{2}x_2,-\frac{7}{2} x_1+\frac{1}{2}x_2)\ |\ x_1,x_2\in\mathbf{R}\}$$ $\beta = \{(1,0,\frac{1}{2},-\frac{7}{2}),(0,1,\frac{5}{2},\frac{1}{2})\}\subseteq \operatorname{null}T$, and that that $\operatorname{span}(\beta) = \operatorname{null}T$ to prove linear independence argue to the contrary that for some $\alpha\in\mathbf{R}$ we have $$(1,0,\frac{1}{2},-\frac{7}{2}) = \alpha\cdot(0,1,\frac{5}{2},\frac{1}{2})$$ which implies that $\alpha = 0 = \frac{1}{5} = -7$ a contradiction consequently $\beta$ is linearily independent and is thus a basis and $\dim\operatorname{null}T=2$.