Finding a basis for the intersection of two subspaces in $\Bbb{R}^4$

Question

I am working through the questions of the Mathematics for Machine Learning book (Page 73 Q2.12, Here) and came across this question on finding a basis for the intersection of two subspaces, given by the following vectors which span $U_1$ and $U_2$, respectively.

$$ U_1 = \operatorname{span} \big[ \begin{pmatrix} 1 \\\\ 1 \\\\ -3 \\\\ 1\\\\ \end{pmatrix}, \begin{pmatrix} 2 \\\\ -1 \\\\ 0 \\\\ -1\\\\ \end{pmatrix}, \begin{pmatrix} -1 \\\\ 1 \\\\ -1 \\\\ 1\\\\ \end{pmatrix} \big]$$

$$ U_2 = \operatorname{span} \big[ \begin{pmatrix} -1\\\\ -2\\\\ 2\\\\ 1\\\\ \end{pmatrix}, \begin{pmatrix} 2\\\\ -2\\\\ 0\\\\ 0\\\\ \end{pmatrix}, \begin{pmatrix} -3\\\\ 6\\\\ 2\\\\ -1\\\\ \end{pmatrix} \big] $$

I have read through some helpful posts, namely (How to find a basis for the intersection of two vector spaces in $\mathbb{R}^n$?) which I found helpful to get started. My understanding is that for any vector $x$ that lies within the intersection of $U_1 \cap U_2$, then it must satisfy the following $\{x^4 \in \Bbb{R}^4 | \ A\lambda = x, By = x\}$ where $A$ and $B$ are matrices whose columns are the column vectors of $U_1$ and $U_2$, respectively.

Having then set up $A\lambda = By$ and then rearranging to $A\lambda - By = 0$ the task then becomes solving for the null space of $L = [ A \ | \ - B]$ but having went through the process of gaussian elimination in an attempt to reduce the matrix to reduced row echelon form, I seem to be stuck with how to proceed. The matrix that I have ended up with is as follows and I would appreciate your thoughts on the below questions / thoughts.

$$ L = \begin{pmatrix} 1 & 2 & -1 & 1 & -2 & 3\\\\ 0 & 1 & \frac{-2}{3} & \frac{-1}{3} & \frac{-4}{3} & 3\\\\ 0 & 0 & 0 & 0 & 0 & 0\\\\ 0 & 0 & 0 & -3 & -2 & 7\\\\ \end{pmatrix} $$

1. Would you agree that the matrix is in the correct form? I am not confident this is the case, even though I am not sure where I have gone wrong
2. If so, my understanding is that this implies the pivot columns of this matrix (columns $1$ and $2$, $c_1$ and $c_2$), should be able to express each of the other redundant (linearly dependent) columns as a linear combination to get the zero vector and while this seems fine to do for $c_3$, it seems impossible for $c_{i ; \ i = 4, .., 6}$ because of the entry in the fourth positions. Any ideas where I have gone wrong here?

I would greatly appreciate the communities thoughts / guidance of how to proceed here. Many thanks!

Answer 1

Let $A_1,A_2$ be matrices whose columns span given subspaces $U_1,U_2$ of $\mathbb{R}^n$.

We can use the fact that $(U_1\cap U_2)^{\perp}=N(A_1^T)+ N(A_2^T)$ to establish a nice method for finding $U_1\cap U_2$. Here it is.

• Find bases for $N(A_1^T),N(A_2^T)$.

• Put these basis vectors in the columns of a matrix $B$. Note the columns of $B$ span $(U_1\cap U_2)^{\perp}$.

• Find a basis $\beta$ for $N(B^T)$.

Then $U_1\cap U_2=\text{span}(\beta)$. In your problem we have $$N(A_1^T)=\text{span}\Bigg\{\begin{pmatrix}1\\ 2\\ 1\\ 0\end{pmatrix},\begin{pmatrix}0\\ -1\\ 0\\ 1\end{pmatrix}\Bigg\}$$ $$N(A_2^T)=\text{span}\Bigg\{ \begin{pmatrix}1\\ 1\\ 0\\ 3\end{pmatrix} \Bigg\}$$

If we set $B=\begin{pmatrix}1&0&1\\ \:2&-1&1\\ \:1&0&0\\ \:0&1&3\end{pmatrix}$ then we get $\text{rref}(B^T)=\begin{pmatrix}1&0&0&4\\ 0&1&0&-1\\ 0&0&1&-2\end{pmatrix}$ whose null space is precisely the intersection you seek: $$U_1\cap U_2=N(B^T)=\text{span}\Bigg\{\begin{pmatrix}-4\\ 1\\ 2\\ 1\end{pmatrix}\Bigg\}$$

Answer 2

Note that the list $\begin{pmatrix}1 \\ 1 \\ -3 \\ 1\end{pmatrix},\begin{pmatrix}2 \\ -1 \\ 0 \\ -1\end{pmatrix},\begin{pmatrix}-1 \\ 1 \\ -1 \\ 1\end{pmatrix}$ is linearly dependent. However, the list $\begin{pmatrix}2 \\ -1 \\ 0 \\ -1\end{pmatrix},\begin{pmatrix}-1 \\ 1 \\ -1 \\ 1\end{pmatrix}$ is a basis for $U_1$. Thus,

$$U_1=\operatorname{span}\left[\begin{pmatrix}2 \\ -1 \\ 0 \\ -1\end{pmatrix},\begin{pmatrix}-1 \\ 1 \\ -1 \\ 1\end{pmatrix}\right].$$

Let $x\in U_1\cap U_2$ be arbitrary, then $x\in U_1$ and $x\in U_2$. Since $x\in U_1$, we have

$$\tag{*}\label{*}x=f\begin{pmatrix}2 \\ -1 \\ 0 \\ -1\end{pmatrix}+g\begin{pmatrix}-1 \\ 1 \\ -1 \\ 1\end{pmatrix},$$

where $f,g\in\Bbb R$. At the same time, since $x\in U_2$, we have

$$\tag{**}\label{**}x=a\begin{pmatrix}-1 \\ -2 \\ 2 \\ 1\end{pmatrix}+b\begin{pmatrix}2 \\ -2 \\ 0 \\ 0\end{pmatrix}+c\begin{pmatrix}-3 \\ 6 \\ 2 \\ -1\end{pmatrix},$$

where $a,b,c\in\Bbb R$.

Since $x=x$, it follows from ($\ref{*}$) and ($\ref{**}$) that

$$\begin{cases}-a+2b-3c=2f-g \\ -2a-2b+6c=-f+g \\ 2a+2c=-g \\ a-c=-f+g\end{cases}$$

The solution of this system of equations is given by

$$\tag{***} \label{***} b = -\frac{3}{2}a \ ; \ c = 0 \ ; \ f = -3 a \ ; \ g = -2 a.$$ Substituting ($\ref{***}$) into ($\ref{*}$), we obtain

$$x=a\begin{pmatrix}-4 \\ 1 \\ 2 \\ 1\end{pmatrix}.$$

Therefore, $$U_1\cap U_2=\operatorname{span}\left[\begin{pmatrix}-4 \\ 1 \\ 2 \\ 1\end{pmatrix}\right].$$