2
$\begingroup$

In the following question I am trying to express the null space of $A$ as the span of certain vectors.

Matrix $A$
$\begin{bmatrix} 2&4&6&4 \\2&5&7&6 \\2&3&5&2 \end{bmatrix}$

My solution:

To solve this problem my logic was to reduce the matrix to echelon form. Which produced

$\begin{bmatrix} 1&2&3&2 \\0&1&1&2 \\0&0&0&0 \end{bmatrix}$

For null space, the matrix is the matrix of coefficients of a homogeneous system:

$\begin{bmatrix} 1&2&3&2 \\0&1&1&2 \\0&0&0&0 \end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$

Rewrite the matrix as a system of equations:

$ x_1 + 2x_2 + 3x_3 +2x_4 = 0 $
$ x_2 + x_3 + 2x_4 = 0 $
$ 0 = 0 $

From here I think I need to solve for the dependent variables and then write the solution as a sum of column vectors but I am not sure if this is correct or how to do it.

$\endgroup$
  • $\begingroup$ Your approach is correct - write the solution as a sum of column vectors, and then the vectors you get will be the basis of your null space. However, you should reduce the matrix to rref, rather than stopping at ref $\endgroup$ – Rob Bland Sep 19 '16 at 19:48
  • $\begingroup$ why reduced row echelon form instead of just row echelon form $\endgroup$ – jh123 Sep 19 '16 at 19:50
  • $\begingroup$ and how do I go about solving for the dependent variables and writing the solution as a sum of column vectors $\endgroup$ – jh123 Sep 19 '16 at 19:51
  • $\begingroup$ Well, you see you have an equation which you can solve for $x_1$ and an equation you can solve for $x_2$ - those are your pivot columns, identified from ref - however, without rref, your equation for $x_1$ will have an $x_2$ term in it, when really you should write your pivot variables in terms of $only$ the free variables $\endgroup$ – Rob Bland Sep 19 '16 at 19:52
  • $\begingroup$ Set $x_4 = s$, $x_3 = t$, Find $x_2$ in terms of $s$ and $t$. Next, find $x_1$ in terms of $s$ and $t.$ Now rewrite it $v_1 s+ v_2 t$ $\endgroup$ – Doug M Sep 19 '16 at 19:53
1
$\begingroup$

Solve your system. You have: $$ x_2=-x_3-2x_4 $$ $$ x_1=-x_3+2x_4 $$ so, for $x_3=u$ and $x_4=v$ the solution has the form: $$ \begin{pmatrix} -u+2v\\ -u-2v\\ u\\ v \end{pmatrix}= u\begin{pmatrix} -1\\ -1\\ 1\\ 0 \end{pmatrix} +v\begin{pmatrix} 2\\ -2\\ 0\\ 1 \end{pmatrix} $$ that is the span of two vectors.

$\endgroup$
  • $\begingroup$ how did you solve for x2 and x1? $\endgroup$ – jh123 Sep 19 '16 at 20:03
  • $\begingroup$ Yes. Obviously you can chose also any other two vectors in the same subspace. $\endgroup$ – Emilio Novati Sep 19 '16 at 20:04
  • $\begingroup$ how did you solve for x2 and x1? $\endgroup$ – jh123 Sep 19 '16 at 20:04
  • $\begingroup$ Find $x_2$ from the second equation of your system and substitute in the first. $\endgroup$ – Emilio Novati Sep 19 '16 at 20:05
  • $\begingroup$ I made a type in my problem in the first row of equations supposed to be x1 x2 x3 x4 not x1 x2 x4 x4 $\endgroup$ – jh123 Sep 19 '16 at 20:07

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.