1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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} $\endgroup$ – Math1000 May 8 '18 at 22:57
1
$\begingroup$

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 $\dim\operatorname{null}T=2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.