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## Dimension of intersection of subspaces

Linear Algebra
Riemann
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### Dimension of intersection of subspaces

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 $\dim (W_1 \cap W_2 ).$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Papapetros Vaggelis
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### Re: Dimension of intersection of subspaces

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

$\displaystyle{\dim_{\mathbb{R}}(W_1\cap W_2)=2}$.