We are given three subspaces of $\mathbb{R}^3$:

$W_1 = \{(a_1,a_2,a_3)\in\mathbb{R}^3\mid a_1=3a_2\space$and$\space a_3=-a_2\}$

$W_2 = \{(a_1,a_2,a_3)\in\mathbb{R}^3\mid 2a_1-7a_2+a_3=0\}$

$W_3 = \{(a_1,a_2,a_3)\in\mathbb{R}^3\mid a_1-4a_2-a_3=0\}$

How to find the subspace formed by $W_1\cap W_2$, $W_1\cap W_3$ and $W_2\cap W_3$?


  • $\begingroup$ Do you know what each subspace defines geometrically? $\endgroup$ – mattos May 2 '15 at 10:41
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    $\begingroup$ @Ritu Certainly the intersection of two planes and a line (all through the origin) in $\Bbb R^3$ is generically trivial. It is only a line if the line subspace is contained in both planes, and this clearly a very special arrangement. $\endgroup$ – Travis Willse May 2 '15 at 10:48
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    $\begingroup$ @Ritu Plugging $W_1$ into $W_2$, we get $2(3a_2)-7a_2-a_2=-2a_2=0 \implies a_2=0$. Not what you are claiming $\endgroup$ – zed111 May 2 '15 at 12:30
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    $\begingroup$ @Ritu I only saw your message now, but it looks like everything has been resolved favorably! I encourage you to post an answer to your question and move your solution there. $\endgroup$ – Travis Willse May 2 '15 at 14:10
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    $\begingroup$ @Ritu seems right now $\endgroup$ – zed111 May 2 '15 at 14:10

If we put $a_1=3a_2$, $a_2=a_2$ and $a_3=-a_2$ in $2a_1-7a_2+a_3=0$, then $a_2=0$. Therefore, $\{0\}$ is the subspace formed by $W_1 \cap W_2$.

If we put $a_1=3a_2$, $a_2=a_2$ and $a_3=-a_2$ in $a_1-4a_2-a_3=0$; the equation is satisfied. So, I conclude $W_1 \subset W_3$. That is, $W_1$ is the subspace formed by $W_1\cap W_3$.

For finding subspace formed by $W_2\cap W_3$, the matrix $\left[ {\begin{array}{cc} 1 & -4 & -1\\ 2 & -7 & 1 \\ \end{array} } \right]$ can be row reduced to $\left[ {\begin{array}{cc} 1 & -4 & -1\\ 0 & 1 & 3 \\ \end{array} } \right]$. Therefore, $W_2\cap W_3 = \{(-11a_3,-3a_3,a_3)\in\mathbb{R}^3\mid a_3 \in \mathbb{R}\}$ .


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