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[SOLVED] Intersection of two spans

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
Hey!! :eek:

Let \begin{equation*}v_1:=\begin{pmatrix}1 \\ 2\\ -1 \\ 3\end{pmatrix}, v_2:=\begin{pmatrix}1 \\ 1\\ 1 \\ 1\end{pmatrix}, v_3:=\begin{pmatrix}-1 \\ 1\\ -5 \\ 3\end{pmatrix} , w_1:=\begin{pmatrix}1 \\ 2\\ -3 \\ 3\end{pmatrix}, w_2:=\begin{pmatrix}1 \\ 0\\ 0 \\ 1\end{pmatrix}\in \mathbb{R}^4\end{equation*}

I want to calculate the intersection of the spans $\text{Lin}(v_1, v_2, v_3)\cap \text{Lin}(w_1, w_2)$.


We have \begin{align*}&\text{Lin}(v_1, v_2, v_3)=\left \{\lambda_1v_1+\lambda_2v_2+\lambda_3v_3 : \lambda_1, \lambda_2, \lambda_3\in \mathbb{R}\right \}=\left \{\lambda_1\begin{pmatrix}1 \\ 2\\ -1 \\ 3\end{pmatrix}+\lambda_2\begin{pmatrix}1 \\ 1\\ 1 \\ 1\end{pmatrix}+\lambda_3\begin{pmatrix}-1 \\ 1\\ -5 \\ 3\end{pmatrix}: \lambda_1, \lambda_2, \lambda_3\in \mathbb{R}\right \} \\ & \text{Lin}(w_1, w_2)=\left \{\tilde{\lambda}_1w_1+\tilde{\lambda}_2w_2 : \tilde{\lambda}_1, \tilde{\lambda}_2\in \mathbb{R}\right \}=\left \{\tilde{\lambda}_1\begin{pmatrix}1 \\ 2\\ -3 \\ 3\end{pmatrix}+\tilde{\lambda}_2\begin{pmatrix}1 \\ 0\\ 0 \\ 1\end{pmatrix} : \tilde{\lambda}_1, \tilde{\lambda}_2\in \mathbb{R}\right \}\end{align*}

How could we continue?

Do we have to solve a system? We take a vector $(a,b,c,d)^T$ and try to write it as a linear combination of the $v_i$'s and then as a linear combination of the $w_i$'s ?

(Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
Hey mathmari !!

The intersection is all vectors that can both be written as $\lambda_1v_1+\lambda_2v_2+\lambda_3v_3$ and as $\tilde{\lambda}_1w_1+\tilde{\lambda}_2w_2$.
So let's solve:
$$
\lambda_1v_1+\lambda_2v_2+\lambda_3v_3 = \tilde{\lambda}_1w_1+\tilde{\lambda}_2w_2 \\

\implies \lambda_1\begin{pmatrix}1 \\ 2\\ -1 \\ 3\end{pmatrix}+\lambda_2\begin{pmatrix}1 \\ 1\\ 1 \\ 1\end{pmatrix}+\lambda_3\begin{pmatrix}-1 \\ 1\\ -5 \\ 3\end{pmatrix} = \tilde{\lambda}_1\begin{pmatrix}1 \\ 2\\ -3 \\ 3\end{pmatrix}+\tilde{\lambda}_2\begin{pmatrix}1 \\ 0\\ 0 \\ 1\end{pmatrix}\\

\implies \lambda_1\begin{pmatrix}1 \\ 2\\ -1 \\ 3\end{pmatrix}+\lambda_2\begin{pmatrix}1 \\ 1\\ 1 \\ 1\end{pmatrix}+\lambda_3\begin{pmatrix}-1 \\ 1\\ -5 \\ 3\end{pmatrix} - \tilde{\lambda}_1\begin{pmatrix}1 \\ 2\\ -3 \\ 3\end{pmatrix}-\tilde{\lambda}_2\begin{pmatrix}1 \\ 0\\ 0 \\ 1\end{pmatrix} = 0 \\

\implies \begin{pmatrix}
1 & 1 & -1 & 1 & 1 \\
2 & 1 & 1 & 2 & 0 \\
-1 & 1 & -5 & -3 & 0 \\
3 & 1 & 3 & 3 & 1\end{pmatrix}\begin{pmatrix}\lambda_1 \\ \lambda_2\\ \lambda_3 \\ - \tilde{\lambda}_1\\ -\tilde{\lambda}_2\end{pmatrix} = 0
$$
Can we solve that? (Wondering)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
We apply the Gauss algorithm and we get the following: \begin{align*}\begin{pmatrix}
1 & 1 & -1 & 1 & 1 \\
2 & 1 & 1 & 2 & 0 \\
-1 & 1 & -5 & -3 & 0 \\
3 & 1 & 3 & 3 & 1\end{pmatrix} &\rightarrow \begin{pmatrix}
1 & 1 & -1 & 1 & 1 \\
0 & -1 & 3 & 0 & -2 \\
0 & 2 & -6 & -2 & 1 \\
0 & -2 & 6 & 0 & -2\end{pmatrix}\rightarrow \begin{pmatrix}
1 & 1 & -1 & 1 & 1 \\
0 & -1 & 3 & 0 & -2 \\
0 & 2 & -6 & -2 & 1 \\
0 & 0 & 0 & -2 & -1\end{pmatrix} \\ & \rightarrow \begin{pmatrix}
1 & 1 & -1 & 1 & 1 \\
0 & -1 & 3 & 0 & -2 \\
0 &0 & 0 & -2 & 1 \\
0 & 0 & 0 & -2 & -1\end{pmatrix}\rightarrow \begin{pmatrix}
1 & 1 & -1 & 1 & 1 \\
0 & -1 & 3 & 0 & -2 \\
0 &0 & 0 & -2 & 1 \\
0 & 0 & 0 & 0 & -2\end{pmatrix}\end{align*}

From here we get the following:
\begin{align*}\begin{pmatrix}
1 & 1 & -1 & 1 & 1 \\
2 & 1 & 1 & 2 & 0 \\
-1 & 1 & -5 & -3 & 0 \\
3 & 1 & 3 & 3 & 1\end{pmatrix}\begin{pmatrix}\lambda_1 \\ \lambda_2\\ \lambda_3 \\ - \tilde{\lambda}_1\\ -\tilde{\lambda}_2\end{pmatrix} = 0&\Rightarrow \begin{pmatrix}
1 & 1 & -1 & 1 & 1 \\
0 & -1 & 3 & 0 & -2 \\
0 &0 & 0 & -2 & 1 \\
0 & 0 & 0 & 0 & -2\end{pmatrix}\begin{pmatrix}\lambda_1 \\ \lambda_2\\ \lambda_3 \\ - \tilde{\lambda}_1\\ -\tilde{\lambda}_2\end{pmatrix} = 0 \\ & \Rightarrow \left\{\begin{matrix}\lambda_1+\lambda_2-\lambda_3-\tilde{\lambda}_1-\tilde{\lambda}_2=0 \\ -\lambda_2+\lambda_3+2\tilde{\lambda}_2=0 \\ 2\tilde{\lambda}_1-\tilde{\lambda}_2=0 \\ 2\tilde{\lambda}_2=0\end{matrix}\right.\end{align*}

From the last equation we get $\tilde{\lambda}_2=0$. From the third one we get then $\tilde{\lambda}_1=0$. From the second equation we get then $\lambda_2=\lambda_3$. Form the first equation we get $\lambda_1=0$.

Does this mean that the intersection contain only the zero vector? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
Suppose we fill in your solution:
$$\begin{pmatrix}
1 & 1 & -1 & 1 & 1 \\
2 & 1 & 1 & 2 & 0 \\
-1 & 1 & -5 & -3 & 0 \\
3 & 1 & 3 & 3 & 1\end{pmatrix}\begin{pmatrix}0 \\ \lambda_2\\ \lambda_2 \\ 0 \\ 0 \end{pmatrix}
= \begin{pmatrix}0 \\ 2\lambda_2\\ -4\lambda_2 \\ 4\lambda_2 \end{pmatrix}
$$

That is not always zero is it? (Wondering)

What did you do in the second step of the Gaussian elimination? (Worried)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
Suppose we fill in your solution:
$$\begin{pmatrix}
1 & 1 & -1 & 1 & 1 \\
2 & 1 & 1 & 2 & 0 \\
-1 & 1 & -5 & -3 & 0 \\
3 & 1 & 3 & 3 & 1\end{pmatrix}\begin{pmatrix}0 \\ \lambda_2\\ \lambda_2 \\ 0 \\ 0 \end{pmatrix}
= \begin{pmatrix}0 \\ 2\lambda_2\\ -4\lambda_2 \\ 4\lambda_2 \end{pmatrix}
$$

That is not always zero is it? (Wondering)

What did you do in the second step of the Gaussian elimination? (Worried)
I found a typo at the last step.

It should be as follows:

\begin{align*}\begin{pmatrix}
1 & 1 & -1 & 1 & 1 \\
2 & 1 & 1 & 2 & 0 \\
-1 & 1 & -5 & -3 & 0 \\
3 & 1 & 3 & 3 & 1\end{pmatrix}\begin{pmatrix}\lambda_1 \\ \lambda_2\\ \lambda_3 \\ - \tilde{\lambda}_1\\ -\tilde{\lambda}_2\end{pmatrix} = 0&\Rightarrow \begin{pmatrix}
1 & 1 & -1 & 1 & 1 \\
0 & -1 & 3 & 0 & -2 \\
0 &0 & 0 & -2 & 1 \\
0 & 0 & 0 & 0 & -2\end{pmatrix}\begin{pmatrix}\lambda_1 \\ \lambda_2\\ \lambda_3 \\ - \tilde{\lambda}_1\\ -\tilde{\lambda}_2\end{pmatrix} = 0 \\ & \Rightarrow \left\{\begin{matrix}\lambda_1+\lambda_2-\lambda_3-\tilde{\lambda}_1-\tilde{\lambda}_2=0 \\ -\lambda_2+3\lambda_3+2\tilde{\lambda}_2=0 \\ 2\tilde{\lambda}_1-\tilde{\lambda}_2=0 \\ 2\tilde{\lambda}_2=0\end{matrix}\right.\end{align*}

From the last equation we get $\tilde{\lambda}_2=0$. From the third one we get then $\tilde{\lambda}_1=0$. From the second equation we get then $\lambda_2=3\lambda_3$. Form the first equation we get $\lambda_1=0$.

And so with both linear combinations we get the zero vector, correct? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
From the last equation we get $\tilde{\lambda}_2=0$. From the third one we get then $\tilde{\lambda}_1=0$. From the second equation we get then $\lambda_2=3\lambda_3$. Form the first equation we get $\lambda_1=0$.

And so with both linear combinations we get the zero vector, correct?
If I substitute the new solution, I still don't get a zero vector. (Worried)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
If I substitute the new solution, I still don't get a zero vector. (Worried)
Oh sorry, in my previous post I didn't corrected the value of $\lambda_1$.

It should be:

From the last equation we get $\tilde{\lambda}_2=0$. From the third one we get then $\tilde{\lambda}_1=0$. From the second equation we get then $\lambda_2=3\lambda_3$. Form the first equation we get $\lambda_1=-2\lambda_3$.

Therefore we have the following:

\begin{align*}-2\lambda_3\begin{pmatrix}1 \\ 2\\ -1 \\ 3\end{pmatrix}+3\lambda_3\begin{pmatrix}1 \\ 1\\ 1 \\ 1\end{pmatrix}+\lambda_3\begin{pmatrix}-1 \\ 1\\ -5 \\ 3\end{pmatrix} &=\lambda_3\begin{pmatrix}-2 \\ -4\\ 2 \\ -6\end{pmatrix}+\lambda_3\begin{pmatrix}3 \\ 3\\ 3 \\ 3\end{pmatrix}+\lambda_3\begin{pmatrix}-1 \\ 1\\ -5 \\ 3\end{pmatrix} \\ & =\lambda_3\left [\begin{pmatrix}-2 \\ -4\\ 2 \\ -6\end{pmatrix}+\begin{pmatrix}3 \\ 3\\ 3 \\ 3\end{pmatrix}+\begin{pmatrix}-1 \\ 1\\ -5 \\ 3\end{pmatrix}\right ]\\ & =\lambda_3\begin{pmatrix}0 \\ 0\\ 0 \\ 0\end{pmatrix} \\ & =\begin{pmatrix}0 \\ 0\\ 0 \\ 0\end{pmatrix}\end{align*}
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
Ah okay.
So yes, then we have found indeed that the intersection is only the zero vector. (Nod)

Moreover, the vectors $v_1,v_2,v_3$ are linearly dependent. (Nerd)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
Ah okay.
So yes, then we have found indeed that the intersection is only the zero vector. (Nod)

Moreover, the vectors $v_1,v_2,v_3$ are linearly dependent. (Nerd)
Ah ok! Thank you!! (Sun)