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Dimension of intersection of subspaces

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Dimension of intersection of subspaces


Post by Riemann » Thu Dec 07, 2017 8:14 am


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\}

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


Post by Papapetros Vaggelis » Wed Dec 27, 2017 4:08 pm

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 \(\displaystyle{(x,x,2\,x)\in W_1\cap W_2\cap W_3}\)


\(\displaystyle{W_1\cap W_2\cap W_3=\langle{(1,1,2)\rangle}\implies \dim_{\mathbb{R}}(W_1\cap W_2\cap W_3)=1}\)

Similarly, \(\displaystyle{W_1\cap W_2=\langle{(1,0,2)\,,(0,1,0)\rangle}}\) and

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