Dimension of intersection of subspaces
Posted: Thu Dec 07, 2017 8:14 am
by Riemann
If
\begin{align*}
W_1 &=\{(x, y, z)| x + y- z =0\}\\
W_2 &= \{(x, y, z)| 3x +y- 2z =0\}\\
W_3 &=\{(x, y, z)| x -7y+3z =0\}
\end{align*}
then find $\dim (W_1 \cap W_2 \cap W_3 )$ and $\dim (W_1 \cap W_2 ).$
Re: Dimension of intersection of subspaces
Posted: Wed Dec 27, 2017 4:08 pm
by Papapetros Vaggelis
Hi Riemann.
Let \(\displaystyle{(x,y,z)\in W_1\cap W_2\cap W_3}\). Then,
\(\displaystyle{(x,y,z)\in W_1\implies x+y-z=0\,\,(I)}\)
\(\displaystyle{(x,y,z)\in W_2\implies 3\,x+y-2\,z=0\,\,(II)}\)
\(\displaystyle{(x,y,z)\in W_3\implies x-7\,y+3\,z=0\,\,(III)}\).
The relations \(\displaystyle{(I)\,,(II)}\) give us
\(\displaystyle{x+y-z=3\,x+y-2\,z\iff z=2\,x}\) and then the relation \(\displaystyle{(III)}\) becomes
\(\displaystyle{x-7\,y+6\,x=0\iff 7\,x-7\,y=0\iff x=y}\)
So, \(\displaystyle{(x,y,z)=(x,x,2\,x)=x\,(1,1,2)\in\langle{(1,1,2)\rangle}}\).
Conversely, if \(\displaystyle{(x,x,2\,x)\in\langle{(1,1,2)\rangle}}\), then,
\(\displaystyle{x+x-2\,x=0\implies (x,x,2\,x)\in W_1}\)
\(\displaystyle{3\,x+x-2\,2\,x=4\,x-4\,x=0\implies (x,x,2\,x)\in W_2}\)
\(\displaystyle{x-7\,x+3\,2\,x=7\,x-7\,x=0\implies (x,x,2\,x)\in W_3}\)
which means that \(\displaystyle{(x,x,2\,x)\in W_1\cap W_2\cap W_3}\)
So,
\(\displaystyle{W_1\cap W_2\cap W_3=\langle{(1,1,2)\rangle}\implies \dim_{\mathbb{R}}(W_1\cap W_2\cap W_3)=1}\)
Similarly, \(\displaystyle{W_1\cap W_2=\langle{(1,0,2)\,,(0,1,0)\rangle}}\) and
\(\displaystyle{\dim_{\mathbb{R}}(W_1\cap W_2)=2}\).