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In the vector space $V \subset \Bbb R^5$, considering the vectors $v_1,v_2,v_3$

$v_1 = (0,1,1,0,0)$ $v_2 = (1,1,0,0,1)$ $v_3 = (1,0,1,0,1)$

We have $V = \mathrm{span}(v_1,v_2,v_3)$ and $U = \{(x_1,x_2,x_3,x_4,x_5) : x_1+x_3+x_5=x_2+x_4=0\}$

a) Calculate the dimension of $V$ and a base b) Calculate the dimension of $U$

I answered to the question a using Gaussian elimination in order to get the dimension and a base for $V$. So, I could use the same argument for the question b but I can't figure out how to find the dimension of $U$ written in that way. Any advices?

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  • $\begingroup$ What does "size of $V$" mean? $\endgroup$ – DiegoMath Jun 30 '14 at 21:18
  • $\begingroup$ Dimension of $Ker(V)$ $\endgroup$ – Gabriele Salvatori Jun 30 '14 at 21:21
  • $\begingroup$ What is the kernel of a vector space? $\endgroup$ – mfl Jun 30 '14 at 21:27
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Note that you can find 3 linear independent vectors in $U$, for example $[1,0,-1,0,0]$, $[0,1,0,-1,0]$ and $[1,0,1,0,-2]$ and it isn't posiible to find more, because $x_2$ is determined by $x_4$ and $x_5$ is determined by $x_1$ and $x_3$. So the dimension of $U$ is 3.

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You can consider the equation-system: (with (0,0,0,0,0) on the right side)

\begin{pmatrix} 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}

This is what we obtain from the term which describes U. Each vector which satisfies the term is element of the kern of the matrix. Now you can see that dim(U) is 3 (Why?) and you can compute a basis of the kern which is also a basis of U

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