Question taken from Strang's Introduction to Linear Algebra:
The floor $\boldsymbol{V}$ and the wall $\boldsymbol{W}$ are not orthogonal subspaces, because they share a nonzero vector (along the line where they meet). No planes V and W in $\mathbf{R}^{3}$ can be orthogonal. Find a vector in the column spaces of both matrices:
$$
A=\left[\begin{array}{ll}1 & 2 \\ 1 & 3 \\ 1 & 2\end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{ll}5 & 4 \\ 6 & 3 \\ 5 & 1\end{array}\right]
$$
This will be a vector $A \boldsymbol{x}$ and also $B \widehat{\boldsymbol{x}}$. Think 3 by 4 with matrix $\left[\begin{array}{ll}A & B\end{array}\right]$.
Text Solution: $A \boldsymbol{x}=B \widehat{\boldsymbol{x}}$ means that $\left[\begin{array}{ll}A & B\end{array}\right]\left[\begin{array}{r}\boldsymbol{x} \\ -\widehat{\boldsymbol{x}}\end{array}\right]=\mathbf{0}$. Three homogenous equations in four unknowns always have a nonzero solution. Here $\boldsymbol{x}=(3,1)$ and $\widehat{\boldsymbol{x}}=(1,0)$ and $A \boldsymbol{x}=B \widehat{\boldsymbol{x}}=(5,6,5)$ is in both column spaces. Two planes in $\mathbf{R}^{3}$ must share a line.
My Question: I'm not following the problem at all. I understand the point with regard to the wall and floor. I understand $ n + (n – r) $ has to span all of n, but I do not follow how these two matrices deal with that. Where does $A x=\widehat{B x}$ even come from and how does it lead to the expression $\left[\begin{array}{ll}A & B\end{array}\right]\left[\begin{array}{r}x \\ -x\end{array}\right]=0$?
Edit: If a vector is in the column space of both matrices, then I understand that the vector would have to exist as the right hand side of Ax and Bx? (It would be a different vector x I presume.)
Best Answer
When finding common vectors in the column spaces of the matrices $A$ and $B$ (since in this case you only have two vectors(columns) per matrix you know they are not linearly dependent because they are not multiples of one another)
you are looking for common vectors in the spans of the form $a_1 \begin{bmatrix} 1\\1\\1\end{bmatrix}+a_2\begin{bmatrix} 2\\3\\2\end{bmatrix} = A\begin{bmatrix} a_1\\a_2\end{bmatrix} = A\vec{x} = b_1\begin{bmatrix} 5\\6\\5\end{bmatrix} + b_2\begin{bmatrix} 4\\3\\1\end{bmatrix} = B\begin{bmatrix} b_1\\b_2\end{bmatrix}=B\vec{y}$
(can a linear combination of the column vectors of $A$ equal a linear combination of the column vectors of $B$, i.e. the spaces share a common vector)
and $[\: A \:\: B \:] \begin{bmatrix} \vec{x}\\ -\vec{y}\end{bmatrix} = \vec{0}$ means essentially the same thing :=
$\begin{bmatrix} 1&2&5&4\\1&3&6&3\\1&2&5&1\end{bmatrix}\begin{bmatrix} a_1\\a_2\\-b_1\\-b_2\end{bmatrix} = a_1 \begin{bmatrix} 1\\1\\1\end{bmatrix}+a_2\begin{bmatrix} 2\\3\\2\end{bmatrix} - b_1\begin{bmatrix} 5\\6\\5\end{bmatrix} - b_2\begin{bmatrix} 4\\3\\1\end{bmatrix} = A\vec{x} -B\vec{y}=\vec{0}$
Since you are in $3$-space and you have two linearly independent vectors for each column space of $A$ and $B$, the subspaces generated by the span of those column vectors will be planes (if you had just one vector, the subspace would be a line, three linearly independent vectors, all of $3$-space), and as mentioned by the textbook answer, there are exists at least one nonzero solution (every homogeneous set of equations is consistent, has the zero solution) to the homogeneous set of equations
$$[\: A \:\: B \:] \begin{bmatrix} \vec{x}\\ -\vec{y}\end{bmatrix} = \vec{0} $$
because there are more unknowns than equations. So the planes either intersect in a line, or completely overlap.
This means the column spaces are not orthogonal by the definition of orthogonal because since they intersect, you can find a nonzero vector $\vec{v}$ s.t. $\vec{v}\in \text{col}(A),\text{col}(B)$ and $\vec{v} \cdot \vec{v} = ||\vec{v}||^2 \not= 0$.