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Let $W_1=\langle (1,2,3,6),(4,-1,3,6)(5,1,6,12))\rangle$ and $W_2=\langle (1,-1,1,1),(2,-1,4,5)\rangle$ be subspaces of $\Bbb{R}^4$. Find the bases for $W_1\cap W_2$ and $W_1+W_2$.

I began by determining if the vectors in $W_1$ are linearly independent, then found out that $dim(W_1)=2$, then likewise $dim(W_2)=2$ and $W_2$ is linearly independent. Now would it be correct to put the basis vectors of $W_1$ & $W_2$ in a matrix to find $W_1\cap W_2$? Since if I let $\{(1,2,3,6),(4,-1,3,6)\}$ and $\{(1,-1,1,1),(2,-1,4,5)\}$ be bases for $W_1$ & $W_2$, respectively, then $W_1\cap W_2$ can be expressed in the following equation $\alpha(1,2,3,6)+\beta(4,-1,3,6)=\gamma(1,-1,1,1)+\delta(2,-1,4,5)$ which can be solved by performing r.r.e.f. on the following matrix \begin{bmatrix} 1 & 4 & 1 & 2 \\ 2 & -1 & -11 & -1\\ 3 & 3 & 1 & 4\\ 6 & 6 & 1 & 5\\ \end{bmatrix}

I then obtained

\begin{bmatrix} 1 & 0 & 0 & 0.\overline{7} \\ 0 & 1 & 0 & -0.\overline{4}\\ 0 & 0 & 1 & 3\\ 0 & 0 & 0 & 0\\ \end{bmatrix}

Does this mean that $W_1\cap W_2$ is a point/vector in 4-D space? Furthermore, a basis for $W_1+W_2$ is the set $\{(1,2,3,6),(4,-1,3,6),(1,-1,1,1)\}$. I need help since my book didn't provide a solution.

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    $\begingroup$ Before we start, may I ask whether you're sure that the $(2,3)$ element of your matrix is $-11$ and not $-1$ ? And ask why you've changed the signs of the coefficients of $\gamma$ and $\delta$? $\endgroup$ – ancientmathematician Feb 4 '18 at 17:05
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$$W_1=\{(1,2,3,6),(4,-1,3,6),(5,1,6,12)\}$$ $$W_2=\{(1,-1,1,1),(2,-1,4,5)\}$$

$v\in \mathbb{R}^4$ is in the $v\in Span(W_1)$ $\iff$

$$ \left(\begin{array}{rrr|r} 1 & 4 & 5 & a\\ 2 & -1 & 1 & b \\ 3 & 3 & 6 & c\\ 6 &6 &2&d \end{array}\right) \Rightarrow \left(\begin{array}{rrr|r} 1 & 4 & 5 & a\\ 0 & -9 & -9 & b-2a \\ 0& 0 & 0 & c-b-a\\ 0 &0 &0&d-2a-2b \end{array}\right)$$

$v\in \mathbb{R}^4$ is in the $v\in Span(W_1)$ $\iff$

$$ \left(\begin{array}{rr|r} 1 & 2 & a\\ -1 & -1 & b \\ 1 & 4 & c\\ 1 &5 &d \end{array}\right) \Rightarrow \left(\begin{array}{rr|r} 1 & 2 & a\\ 0 & 1 & b+a \\ 0 & 0 & -3a-2b+c\\ 0 &0 &-4a-3b+d \end{array}\right)$$

So for $v$ to be in $W_1\cap W_2$ it need to follow:

$$\begin{pmatrix} -1 & -1& 1 &0 \\ -2 & -2 &0 &1\\ -3&-2& 1 &0\\ -4&-3&0&1 \end{pmatrix}\iff \begin{pmatrix} 1 & 0& 0 &\frac{1}{2}\\ 0 & 1 &0 &-1\\ 0&0& 1 &-\frac{1}{2}\\ 0&0&0&0 \end{pmatrix}$$

So basis for $W_1\cap W_2$ is $(\frac{1}{2},-1,-\frac{1}{2},0)$

For $W_1+W_2$ we know that $Span(W_1+W_2)=Span(W_1\cup W_2)$ so we can look at

$$\begin{pmatrix} 1 & 4& 5 &1&2\\ 2 & -1 &1 &-1&-1\\ 3 & 3& 6 &1&4\\ 6 & 6 & 12&1&5 \end{pmatrix}\Rightarrow \begin{pmatrix} 1 & 4& 5 &1&2\\ 0 & 9 &9 &3&5\\ 0 & 0& 0 &1&3\\ 0 & 0 & 0&0&0 \end{pmatrix}$$

So basis for $W_1+W_2$ is $\{(1,2,3,6),(4,-1,3,6),(1,-1,1,1)\}$

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  • $\begingroup$ where did this come from? How did you came up to this matrix? $$\begin{pmatrix} -1 & -1& 1 &0 \\ -2 & -2 &0 &1\\ -3&-2& 1 &0\\ -4&-3&0&1 \end{pmatrix}$$ $\endgroup$ – TheLast Cipher Feb 5 '18 at 5:09
  • $\begingroup$ @TheLastCipher from the equations for which a general vector is in the span of $w_1$ and $w_2$ $\endgroup$ – gbox Feb 5 '18 at 7:47
  • $\begingroup$ $\left(\begin{array}{rrr|r} 1 & 4 & 5 & a\\ 0 & -9 & -9 & b-2a \\ 0& 0 & 0 & c-b-a\\ 0 &0 &0&d-2a-2b \end{array}\right)$ and $\left(\begin{array}{rr|r} 1 & 2 & a\\ 0 & 1 & b+a \\ 0 & 0 & -3a-2b+c\\ 0 &0 &-4a-3b+d \end{array}\right)$ $\endgroup$ – gbox Feb 5 '18 at 7:48
  • $\begingroup$ i know that's the matrices you were referring to, but how did you combine these two to get the final matrix? thank you! $\endgroup$ – TheLast Cipher Feb 5 '18 at 8:07
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I don't think gbox's method is correct. I think employing the method shown here (from Ted Shifrin's text) is the correct one and it's more intuitive. enter image description here

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