Finding the basis of intersection of two subspaces generated by given set of vectors

Question

I am trying to solve following question

Let $W_1 = (x, y, 0)$ where $x, y\in \mathbb{R}$ be a subspace of $\mathbb{R^3}$ over field $\mathbb{R}$ and $W_2$ be a subspace of $\mathbb{R^3}$ generated by $\{(1, 2, 3), (1, -1, 1) \}$. Then find out the basis of $W_1 \cap W_2$.

My efforts: $W_1$ is generated by $\{(1, 0, 0), (0, 1, 0) \}$ and hence $\dim W_1 = 2$. Given that $W_2$ is generated by $\{(1, 2, 3), (1, -1, 1) \}$ which are independent and so $\dim W_2 = 2$. Further I computed dimension of $W_1+W_2$ by writing generating vectors of $W_1$ and $W_2$ into a single matrix and then row-reducing it which come out to be $3$. So it follows the dimension of $\dim(W_1 \cap W_2)=1$. But how do I find a basis for it?

Please help me in clearing my doubt. Thank you so much

Answer 1

This intersection is the set of linear combinations of your basis of $W_2$ such that the 3rd component is $0.$ I.e. $a(1, 2, 3)+b(1, -1, 1)$ s.t. $3a+b=0.$ I.e. the line generated by $(1, 2, 3)-3(1, -1, 1)=(-2,5,0).$

Answer 2

By definition, the intersection of two subspaces $W_1$ and $W_2$ of a vector space $V$ is $W_1\cap W_2:=\{v\in V: v\in W_1\, ,\; v\in W_2\}$. This provides a systematic way to find the intersection: we find the conditions (implicit equations) that describe the subspace $W_1$ and the conditions (implicit equations) that describe the subspace $W_2$; all this in terms of an arbitrary vector $v\in V$. Finally, to find a basis for the intersection, we proceed exactly as you did in the first lines of your work.

If we start by knowing $W_1={\rm span}((1,0,0),(0,1,0))$ and it contains two linearly independent vectors in $R^{3}$, then there's no problem (if there were linearly dependent vectors, they could be removed within the ${\rm span}$). We take $v=(x,y,z)\in R^3$ and then seek to describe the subspace $W_1$. This can be done by setting $(x,y,z)=\alpha_1(1,0,0)+\alpha_2(0,1,0)$, which constructs a system of matrix equations $\begin{bmatrix} 1&0\\0&1\\0&0\end{bmatrix}\begin{bmatrix}\alpha_1\\\alpha_2\end{bmatrix}=\begin{bmatrix}x\\y\\z\end{bmatrix}\iff \begin{bmatrix}1&0&x\\0&1&y\\0&0&z\end{bmatrix}$, which indicates (by solving) that both $x$ and $y$ are free variables, i.e., $x,y\in R$, but $z$ is fixed at zero, i.e., $z=0$. Therefore, $W_1=\{(x,y,z)\in R^{3}: z=0,x,y\in R\}$. This makes it unnecessary work since we already knew this from the beginning, as $W_1=\{(x,y,0)\}=\{(x,y,z)\in R^{3}: z=0,x,y\in R\}$.

Similarly, if we start by knowing $W_2={\rm span}((1,2,3),(1,-1,1))$ and as you have already verified linear independence, we proceed as before. Let $v=(x,y,z)\in R^3$ and then seek to describe the subspace $W_2$, which can be done by $(x,y,z)=\alpha_1(1,2,3)+\alpha_2(1,-1,1)$, which can be seen in matrix form as $\begin{bmatrix} 1&1\\2&-1\\3&1\end{bmatrix}\begin{bmatrix}\alpha_1\\\alpha_2\end{bmatrix}=\begin{bmatrix}x\\y\\z\end{bmatrix}\iff \begin{bmatrix}1&1&x\\2&-1&y\\3&1&z\end{bmatrix}\sim \cdots \sim \begin{bmatrix}1&1&x\\0&3&2x-y\\0&0&5x+2y-3z\end{bmatrix}$, and this means that for consistency, we need $5x+2y-3z=0$, and this is the condition that applies to $W_2$. Thus, $W_2=\{(x,y,z)\in R^{3}: 5x+2y-3z=0\}$.

For the intersection, we then write $W_1\cap W_2=\{(x,y,z)\in R^{3}: \begin{cases}z=0,x,y,z\in R\\ 5x+2y-3z=0 \end{cases} \}=\{(x,y,z)\in R^{3}: 5x+2y=0, z=0\}$. Writing $5x+2y=0$ as $y=-\frac{5}{2}x$ with $x\in R$ and fixing $z=0$, we get $(x,y,z)=(x,-\frac{5}{2}x,0)=x(1,-\frac{5}{2},0)$.

Therefore, a basis for $W_1\cap W_2$ is given by $B=\{(1,-\frac{5}{2},0)\}$, or it can be scaled as $B=\{(-2,5,0)\}$, and furthermore $\dim(W_1\cap W_2)=1$ as you predicted would be the dimension.

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NB. In any case, I consider that Anne Bauval's answer (+1) has all the elements to be selected as a response to the question, as is customary. I am only writing it in the way that I would explain the answer to the question.