The question I'm given is this:

Let $S$ be the subspace of $R^{4}$ consisting of the solutions to the following system of equations:




I am asked to give a basis for $S$.

I have attempted to solve this by putting this into a matrix and solving for my $x_{1}$, $x_{2}$, $x_{3}$, and $x_{4}$, but I don't really know where to go from there. I am sure I'm missing something important.

Any help would be greatly appreciated!

  • $\begingroup$ What is the rank of associated matrix? $\endgroup$ – user261263 May 25 '16 at 8:07
  • $\begingroup$ I believe the rank of the matrix is 3. $\endgroup$ – greeneggs May 25 '16 at 8:13
  • $\begingroup$ And the principal minor? $\endgroup$ – user261263 May 25 '16 at 8:14
  • $\begingroup$ I don't quite understand how to find the principal minors. Is it just taking the determinant? $\endgroup$ – greeneggs May 25 '16 at 8:22

You can also do it without matrices, simply by solving the equations. Label them (1),(2),(3).

4(1)-(2): $3x_1+3x_2+3x_4=0$ (4). Then -(2)-4(3): $11x_1+11x_2+7x_4=0$ (5). Now 11(4)-3(5): $x_4=0$ and so $x_1+x_2=0$. (2) now gives $x_2+2x_3=0$. So putting $x_3=k$ we have $x_1=2k,x_2=-2k,x_3=k,x_4=0$.

Check: (1) $2k-4k+2k=0$, (2) $2k-10k+8k=0$, (3) $-6k+8k-2k=0$.

So a basis for $S$ is $(2,-2,1,0)$.


You are on the right way. Rewriting this question using matrices becomes: Find a basis of the kernel of $$ \begin{pmatrix}1&2&2&2\\1&5&8&5\\-3&-4&-2&-3\end{pmatrix}. $$ You can either solve this like @almagest suggested or you can use some more soffisticated techniques. In this example I suggest to perform a row reduction. How to do this is explained here wikibooks.

  • $\begingroup$ If you struggle you can add a comment and I will help you through the steps. $\endgroup$ – YannickSSE May 25 '16 at 8:33
  • $\begingroup$ So I have solved the matrix to find x1=2t, x2=-2t, x3=t, x4=0. What do I do with this after I find the solutions? $\endgroup$ – greeneggs May 25 '16 at 8:45
  • $\begingroup$ Solved the matrix is not the right term but you are done. You have written the solution as 'x=tv' with 'v=(2,-2,1,0)' hence the solution space is one dimensional with basis $\{v\}$. $\endgroup$ – YannickSSE May 25 '16 at 9:10
  • $\begingroup$ Thank you, the answer was there but I was thinking too complex for this question. $\endgroup$ – greeneggs May 25 '16 at 14:06

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