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$w_1=\{(x,y,z)\mid x+y-z=0\},\\ w_2=\{(x,y,z)\mid 3x+y-2z=0\},\\ w_3=\{(x,y,z)\mid x-7y+3z=0\}\\ \text{Find }\quad \dim(w_1\cap w_2\cap w_3) \quad\text{ and }\quad \dim(w_1+w_2)$

My approach:

Since the intersection of $3$ sets is the solution of the given $3$ equations, this solution is the basis of vector space $(w_1\cap w_2\cap w_3)$. Therefore, I think $\ \dim(w_1\cap w_2\cap w_3)=1$

Adding $w_1$ and $w_2$, we get $\,w_1+w_2=\{(x,y,z)\mid 4x+2y-3z=0\}$. I think its basis is $\{(1,0,0),(0,1,0),(0,0,1)\}$. So the dimension would be $\dim(w_1+w_2)=3$

I am not sure whether my answers are correct or not. Can some one confirm or correct me if I am wrong?

Thanks for help.

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  • $\begingroup$ You’ve got the correct answers, but for the wrong reasons. $\endgroup$ – amd Dec 12 '17 at 19:34
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Since the intersection of $3$ sets is the solution of the given $3$ equations, this solution is the basis of vector space $(w_1\cap w_2\cap w_3)$. Therefore, I think $\ \dim(w_1\cap w_2\cap w_3)=1$

You're right that a vector will be in the union if it satisfies the three equations, but then the dimension will still depend on this system of equations. Note that they're all planes (through the origin) in $\mathbb{R^3}$ and in general, you can get a plane, a line or simply the origin for the intersection. The dimension will only be $1$ if they intersect in a line - you can (and should) check this.

Adding $w_1$ and $w_2$, we get $\,w_1+w_2=\{(x,y,z)\ |\ 4x+2y-3z=0\}$. I think its basis is $\{(1,0,0),(0,1,0),(0,0,1)\}$. So the dimension would be $\,\dim(w_1+w_2)=3$

No, this is not how the addition of subspaces works, you cannot simply add these equations. A vector will be in $w_1+w_2$ if it can be written as a sum of vectors from $w_1$ and $w_2$.

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