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Let $W_1=\{(1,1,2,1), (3,1,0,0)\}$ and $W_2=\{(-1,-2,0,1), (-4,-2,-2,-1)\}$

Apparently $\dim W_1=\dim W_2=2$.

For $\dim W_1\cap W_2$, since $(-4,-2,-2,-1)$ can be expressed as $-(1,1,2,1)-(3,1,0,0)$, $\dim W_1\cap W_2=1$. But what if the spans are complicated, how do you find $\dim W_1\cap W_2$. Do you try matrix on each vector and see which ones has a nontrivial solution?

For $\dim W_1+W_2$, how do you know it without calculating $\dim W_1\cap W_2$?

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  • $\begingroup$ To be pedantic, I think it's important to note the difference between the set of two vectors, and the span of that set. Here $W_1$ isn't a vector space, but taking the span makes it a vector space. So it only makes sense to talk about the dimension of $\text{span} W_1$. $\endgroup$
    – Alexander Heyes
    Feb 9, 2016 at 21:14

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Put the given vectors into the columns of matrices \begin{align*} w_1 &= \left[\begin{array}{rr} 1 & 3 \\ 1 & 1 \\ 2 & 0 \\ 1 & 0 \end{array}\right] & w_2 &= \left[\begin{array}{rr} -1 & -4 \\ -2 & -2 \\ 0 & -2 \\ 1 & -1 \end{array}\right] \end{align*} The dimensions of $W_1$ and $W_2$ are the ranks of the matrices $w_1$ and $w_2$ respectively. Row reducing gives $\DeclareMathOperator{rref}{rref}$ \begin{align*} \rref(w_1) &= \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0 \end{array}\right] & \rref(w_2) &= \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0 \end{array}\right] \end{align*} These reduced row-echelon forms show that both $w_1$ and $w_2$ have rank two. This proves that $\dim W_1=\dim W_2=2$, as you have observed.

Now, the dimension of $W_1+W_2$ is the rank of the matrix $$ [w_1,w_2] = \left[\begin{array}{rrrr} 1 & 3 & -1 & -4 \\ 1 & 1 & -2 & -2 \\ 2 & 0 & 0 & -2 \\ 1 & 0 & 1 & -1 \end{array}\right] $$ Row reducing gives $$ \rref[w_1,w_2]= \left[\begin{array}{rrrr} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ This proves that $\dim (W_1+W_2)=3$. The dimension formula then implies that $\dim(W_1\cap W_2)=1$.

This method also provides bases for the spaces in question. The pivot columns of $\rref[w_1,w_2]$ correspond to the columns in $[w_1,w_2]$ that form a basis for $W_1+W_2$. The "free" columns of $\rref[w_1,w_2]$ correspond to the columns of $[w_1,w_2]$ that span $W_1\cap W_2$. Here, we have $\DeclareMathOperator{Span}{Span}$ \begin{align*} W_1+W_2 &= \Span \left\{ \left[\begin{array}{r} 1 \\ 1 \\ 2 \\ 1 \end{array}\right], \left[\begin{array}{r} 3 \\ 1 \\ 0 \\ 0 \end{array}\right], \left[\begin{array}{r} -1 \\ -2 \\ 0 \\ 1 \end{array}\right] \right\} & W_1\cap W_1 &= \Span\left\{ \left[\begin{array}{r} -4 \\ -2 \\ -2 \\ -1 \end{array}\right] \right\} \end{align*}

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    $\begingroup$ The row reduction also shows the fourth vector is in the intersection of the two subspaces, so it makes a basis for the intersection. And that a basis for the sum consists of the first three vectors. $\endgroup$
    – egreg
    Feb 9, 2016 at 21:40
  • $\begingroup$ Correct. Updated! $\endgroup$
    – Brian Fitzpatrick
    Feb 9, 2016 at 23:08
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    $\begingroup$ Of course, you should mention that this method works only because $$(−4,−2,−2,−1) = \alpha (1,1,2,1) + \beta (3,1,0,0)$$ for some scalars $\alpha$ and $\beta$. If we had: $$(−4,−2,−2,−1) = \alpha (1,1,2,1) + \beta (3,1,0,0) + \gamma(−1,−2,0,1) $$ for some $\gamma \ne 0$ then the desired vector in $W_1 \cap W_2$ would be $\alpha (1,1,2,1) + \beta (3,1,0,0)$, and not $(−4,−2,−2,−1)$. It seems to be confusing people. $\endgroup$
    – mechanodroid
    Dec 26, 2017 at 23:07
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    $\begingroup$ @AkashPatalwanshi That is probably because the above method doesn't work in general. Consider $W_1, W_2 \le \mathbb{R}^4$ subspaces with bases $\left\{\begin{bmatrix}1 \\ 0 \\ 0 \\ 0 \end{bmatrix},\begin{bmatrix}0 \\ 1 \\ 0 \\ 0 \end{bmatrix}\right\}$ and $\left\{\begin{bmatrix}0 \\ 0 \\ 1 \\ 0 \end{bmatrix},\begin{bmatrix}1 \\ 1 \\ 1 \\ 0 \end{bmatrix}\right\}$ respectively. Then the corresponding matrix is $$\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0\end{bmatrix}$$ which is already in row echelon form. $\endgroup$
    – mechanodroid
    Mar 21, 2019 at 6:38
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    $\begingroup$ @AkashPatalwanshi The above answer claims that the basis for $W_1 \cap W_2$ is given by $$\left\{\begin{bmatrix}1 \\ 1 \\ 1 \\ 0 \end{bmatrix}\right\}$$ which is false since this vector is not in $W_1$. The correct basis should be for example $$\left\{\begin{bmatrix}1 \\ 1 \\ 0 \\ 0 \end{bmatrix}\right\}$$ $\endgroup$
    – mechanodroid
    Mar 21, 2019 at 6:40

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