Finding bases for ##W_1\cap W_2## and ##W_1+W_2##

[Terrell]

Homework Statement

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 $W_1+W_2$.

Homework Equations

$\alpha(1,2,3,6)+\beta(4,-1,3,6)=\gamma(1,-1,1,1)+\delta(2,-1,4,5)$

The Attempt at a Solution

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.

 

[Mark44]

Homework Statement

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 $W_1+W_2$.

Homework Equations

$\alpha(1,2,3,6)+\beta(4,-1,3,6)=\gamma(1,-1,1,1)+\delta(2,-1,4,5)$

The Attempt at a Solution

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.

First off, let's look at the geometry here. $W_1$ can be seen by inspection to be a plane in $\mathbb R^4$ (the 3rd vector in the set is the sum of the first two). $W_2$ is also a plane, since neither vector is a multiple of the other.
In the first matrix of your work above, you are solving the equation $c_1\vec{v_1} + c_2\vec{v_2} + c_3\vec{v_3} + c_4\vec{v_4} = \vec 0$ for the constants $c_1, c_2, c_3, c_4$. Here $v_1$ and $v_2$ are in $W_1$ and the other two vectors are in $W_2$.

Assuming your work in the second matrix is correct (I didn't check), what is this matrix saying about the constants $c_1, c_2, c_3, c_4$?

Does this mean that $W_1\cap W_2$ is a point/vector in 4-D space?

Points and vectors are different things. A vector consists of an infinite number of points. If your matrix had row-reduced to an identity matrix, that would mean that the solution is a unique point.

 

[Terrell]

Assuming your work in the second matrix is correct (I didn't check), what is this matrix saying about the constants c1,c2,c3,c4c1,c2,c3,c4c_1, c_2, c_3, c_4?

I think it is saying that $0.\overline{7}\overrightarrow{v_1} -0.\overline{4}\overrightarrow{v_2}=-3\overrightarrow{v_3}-\overrightarrow{v_5}$. Which in english means that as long as this parametric relationship is maintained, then this particular combination of vectors from $W_1$ and $W_2$ meet at a point in space. This continuity of points should trace a line in space. Is this a correct interpretation?

If your matrix had row-reduced to an identity matrix, that would mean that the solution is a unique point.

Why?

 

[Mark44]

I then obtained
$\begin{bmatrix} 1 & 0 & 0 & 0.\overline{7} \\ 0 & 1 & 0 & -0.\overline{4}\\ 0 & 0 & 1 & 3\\ 0 & 0 & 0 & 0\\ \end{bmatrix}$

Assuming your work in the second matrix is correct (I didn't check), what is this matrix saying about the constants $c_1, c_2, c_3, c_4$?

I think it is saying that $0.\overline{7}\overrightarrow{v_1} -0.\overline{4}\overrightarrow{v_2}=-\overrightarrow{v_3}-\overrightarrow{v_5}$. Which in english means that as long as this parametric relationship is maintained, then this particular combination of vectors from $W_1$ and $W_2$ meet at a point in space. This continuity of points should trace a line in space. Is this a correct interpretation?

No, it isn't. The matrix I was talking about, copied above, says this:
$c_1 = -7/9~c_4\\ c_2 = 4/9 c_4\\ c_3 = -3c_4\\ c_4 = c_4$
In other words, if we let $c_4 = 1$, which we can do because it is arbitrary, $-\frac 7 9 \vec{v_1} + \frac 4 9 \vec{v_2} - 3 \vec{v_3} + \vec{v_4} = \vec 0$. Since not all of the constants are 0, then the four vectors are linearly dependent. These four vectors are the columns of the first matrix, taking into account a typo you made in the 2nd entry of $\vec{v_3}$.

You can also infer from the 2nd matrix that the vectors $\vec{v_1}, \vec{v_2}, \vec{v_3}$ are linearly independent, and would form a basis for $W_1 + W_2$.

If your matrix had row-reduced to an identity matrix, that would mean that the solution is a unique point.

Why?

That point would be (0, 0, 0 0), which would mean that $c_1 = c_2 = c_3 = c_4 = 0$, and this would be the unique solution to the equation $c_1\vec{v_1} + c_2\vec{v_2} + c_3\vec{v_3} + c_4\vec{v_4} = \vec 0$. This would imply that these four vectors are linearly independent. In that case, $W_1 + W_2$ would be all of $\mathbb R^4$.

 

[Terrell]

No, it isn't. The matrix I was talking about, copied above, says this:
c1=−7/9 c4c2=4/9c4c3=−3c4c4=c4c1=−7/9 c4c2=4/9c4c3=−3c4c4=c4c_1 = -7/9~c_4\\ c_2 = 4/9 c_4\\ c_3 = -3c_4\\ c_4 = c_4
In other words, if we let c4=1c4=1c_4 = 1, which we can do because it is arbitrary, −79→v1+49→v2−3→v3+→v4=⃗0−79v1→+49v2→−3v3→+v4→=0→-\frac 7 9 \vec{v_1} + \frac 4 9 \vec{v_2} - 3 \vec{v_3} + \vec{v_4} = \vec 0. Since not all of the constants are 0, then the four vectors are linearly dependent. These four vectors are the columns of the first matrix, taking into account a typo you made in the 2nd entry of →v3v3→\vec{v_3}.

I do know and understand this idea to begin with, but what I do like to know is how can we use this equation/relationship to determine the intersection of $W_1$ and $W_2$ which I believe is a line, since the intersection of two plane must be a line. Moreover, also determine the equation of that line of intersection.

 

[Mark44]

I do know and understand this idea to begin with, but what I do like to know is how can we use this equation/relationship to determine the intersection of $W_1$ and $W_2$ which I believe is a line, since the intersection of two plane must be a line. Moreover, also determine the equation of that line of intersection.

Yes, $W_1$ and $W_2$ are planes in $\mathbb R^4$. For a couple of reasons, which I won't go into, these planes intersect in a line (not a point -- that can't happen).
$W_1 = \{\vec{w_1} \in \mathbb R^4~ |~ \vec{w_1} = a\vec{v_1} + b\vec{v_2}\}$, with a and b being real numbers.
$W_2 = \{\vec{w_2} \in \mathbb R^4~ |~ \vec{w_2} = c\vec{v_3} + d\vec{v_4}\}$, with c and d being real numbers.
Here $\vec{v_1}$ and $\vec{v_2}$ are the first two columns of your first matrix, and $\vec{v_3}$ and $\vec{v_4}$ are the last two columns.
Set $\vec{w_1} = \vec{w_2}$ and solve this matrix equation, and you should get your solution -- a line in parametric form.

 

[Terrell]

Here is what I obtained, $\alpha(1,2,3,6)+\beta(4,-1,3,6)=\gamma(1,-1,1,1)+\delta(2,-1,4,5)$
\begin{align}
\Longleftrightarrow
\left[\begin{matrix}
1 & 4 & -1 & -2\\
2 & -1 & 1 & 1\\
3 & 3 & -1 & -4\\
6 & 6 & -1 & -5\\
\end{matrix}\right]
\cdot
\left[\begin{matrix}
\alpha\\
\beta\\
\gamma\\
\delta\\
\end{matrix}\right]
&=
\left[\begin{matrix}
0\\
0\\
0\\
0\\
\end{matrix}\right]\\
\end{align}
\begin{align}
\alpha - \frac{9}{7}&=0\\
\beta + \frac{4}{9}&=0\\
\gamma + 3\delta&=0\\
\end{align}
Therefore, the parametric form of the line is $(\alpha,\beta,\gamma,\delta)=t(9/7, -4/9, -3, 1)$ such that $t$ is a free variable, and the cartesian form of the line is $36\delta=28\alpha-81\beta-12\gamma$.

 

[Terrell]

What I see intuitively is that $(\alpha,\beta,\gamma,\delta)=t(9/7, -4/9, -3, 1)$ is the subspace that gets mapped to $(0,0,0,0)$ by the transformation matrix \begin{bmatrix} 1&4&-1&-2\\ 2&-1&1&1\\ 3&3&-1&-4\\ 6&6&-1&-5\end{bmatrix}.
However, I don't intuitively understand how this "solution vector",$(\alpha,\beta,\gamma,\delta)=t(9/7, -4/9, -3, 1)$, is the intersection between the plane formed by vectors $(1,2,3,6)$ & $(4,-1,3,6)$ and the plane formed by vectors $(1,-1,3,6)$ & $(2,-1,4,5)$.

 

[Terrell]

After reading a book, I realized the post I was reading from a different site is incorrect and my initial understanding of how this must be solved was correct. So please ignore my previous two posts.
Correction: equating $\alpha(1,2,3,6)+\beta(4,-1,3,6)=\gamma(1,-1,1,1)+\delta(2,-1,4,5)$, I then need to solve
$\left[\begin{matrix}1&4&-1&-2\\2&-1&1&1\\3&3&-1&-4\\6&6&-1&-5\\\end{matrix}\right]\left[\begin{matrix}\alpha\\\beta\\\gamma\\\delta\end{matrix}\right]=\left[\begin{matrix}0\\0\\0\\0\\\end{matrix}\right]\Rightarrow \left[\begin{matrix}\alpha\\\beta\\\gamma\\\delta\end{matrix}\right]=\frac{7}{9}\left[\begin{matrix}1\\2\\3\\6\end{matrix}\right]-\frac{4}{9}\left[\begin{matrix}4\\-1\\3\\6\end{matrix}\right]=-3\left[\begin{matrix}1\\-1\\1\\1\end{matrix}\right]+\left[\begin{matrix}2\\-1\\4\\5\end{matrix}\right]$
This method is the correct one right? thank you!

 

[Mark44]

After reading a book, I realized the post I was reading from a different site is incorrect and my initial understanding of how this must be solved was correct. So please ignore my previous two posts.
Correction: equating $\alpha(1,2,3,6)+\beta(4,-1,3,6)=\gamma(1,-1,1,1)+\delta(2,-1,4,5)$, I then need to solve
$\left[\begin{matrix}1&4&-1&-2\\2&-1&1&1\\3&3&-1&-4\\6&6&-1&-5\\\end{matrix}\right]\left[\begin{matrix}\alpha\\\beta\\\gamma\\\delta\end{matrix}\right]=\left[\begin{matrix}0\\0\\0\\0\\\end{matrix}\right]\Rightarrow \left[\begin{matrix}\alpha\\\beta\\\gamma\\\delta\end{matrix}\right]=\frac{7}{9}\left[\begin{matrix}1\\2\\3\\6\end{matrix}\right]-\frac{4}{9}\left[\begin{matrix}4\\-1\\3\\6\end{matrix}\right]=-3\left[\begin{matrix}1\\-1\\1\\1\end{matrix}\right]+\left[\begin{matrix}2\\-1\\4\\5\end{matrix}\right]$
This method is the correct one right? thank you!

Looks OK to me. One way to check would be to find the equations of the two planes, which could be done by finding a normal to each plane using the dot product. Once you have a normal and a point on the plane (such as (0, 0, 0, 0)), you can write the equation of that plane. The vector you found should satisfy each of the two planes.

 

[Terrell]

will keep that in mind. thank you a lot!