First year linear algebra level.

A is a matrix $$ \begin{bmatrix} -1&1&1&3\\ 2&-2&1&-3\\ 3&-3&-2&-8\\ \end{bmatrix} $$ Then describe the kernel of A as a span of linearly independent vectors. I reduced it to RREF: $$ \begin{bmatrix} 1&-1&0&-2\\ 0&0&1&1\\ 0&0&0&0\\ \end{bmatrix} $$ But I can't connect it to how it becomes the answer, which is: $$ span( \begin{bmatrix} 2\\0\\-1\\1 \end{bmatrix}, \begin{bmatrix} 1\\1\\0\\0\\ \end{bmatrix} ) $$ Also I'm not sure if the 2 is a typo on the answer sheet.

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    $\begingroup$ Welcome to Mathematics Stack Exchange. It looks correct $\endgroup$ – J. W. Tanner Oct 6 '19 at 20:41
  • $\begingroup$ Cf. this question $\endgroup$ – J. W. Tanner Oct 6 '19 at 20:48

Call $C_k$ the columns of matrix $R$ (the RREF transform)

You see that :

$$C_1+C_2=0 \ \iff \ (1)C_1+(1)C_2+(0)C_3+(0)C_4=0 \tag{1}$$

Collect the coefficients (between parentheses) of this null linear combination : you will get the second of your vectors (presented in a line instead of a column).

For the other one, it is the same :


Explanation : Relationship (1) can be written under the form :

$$\underbrace{\left(C_1|C_2|C_3|C_4\right)}_{R}\underbrace{\begin{pmatrix}1\\1\\0\\0\end{pmatrix}}_{V}=\underbrace{\begin{pmatrix}0\\0\\0\\0\end{pmatrix}}_0 \ \ \implies \ \ V \in \ker R$$

  • $\begingroup$ I have added an explanation. $\endgroup$ – Jean Marie Oct 6 '19 at 20:48
  • $\begingroup$ I still struggling to see it: starting with the RREF, how would I have gotten those coefficients? Is there a procedural way I could've known that the first vector would have the two zeros (I can see visually that C1-C2 is zero since they're 1,-1 respectively), but the second vector involved all three columns? ................ In class, all we did was show the kernel is Ax=0 so I think my base is a little weak. $\endgroup$ – Five9 Oct 6 '19 at 21:15
  • $\begingroup$ For the second vector, it is possible to notice at first that doing $(-1)C_3+(1)C_4$, its second and third coefficient become $0$, therefore it is a multiple of the first column $C_1$ : $(-1)C_3+(1)C_4 =aC_1$ for a certain $a$ (which is $-2$) ; rearranging gives $(-a)C_1+(-1)C_3+(1)C_4 =0$ $\endgroup$ – Jean Marie Oct 6 '19 at 21:24
  • $\begingroup$ Turns out I had a total lapse in thinking...I understand now, thanks for bearing with me. However just noting that this is very notationally different from what my book/class has :O $\endgroup$ – Five9 Oct 7 '19 at 1:41

$$ \begin{bmatrix} 1&-1&0&-2\\ 0&0&1&1\\ 0&0&0&0\\ \end{bmatrix} $$ To find kernel and its basis you just need the solution of $Ax=0$, so you get $x-y-2w=0,z+w=0$. Any two linearly independent solutions of these equations will form the spanning set of kernel. It is easy to verify that $x=2,y=0,z=-1,w=1$ and $x=1,y=1,z=0,w=0$ are such two LI solutions.


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