# Calculate the dimension of $U = \{(x_1,x_2,x_3,x_4,x_5) : x_1+x_3+x_5=x_2+x_4=0\}$

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?

• What does "size of $V$" mean? – DiegoMath Jun 30 '14 at 21:18
• Dimension of $Ker(V)$ – Gabriele Salvatori Jun 30 '14 at 21:21
• What is the kernel of a vector space? – mfl Jun 30 '14 at 21:27

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.