Find the intersection of the images of two matrices I am studying for a big test and have this problem:
I am given two matrices A and B
$$A = \left(\begin{array}{crc}
 1 & 0\\
 2 & 1\\
 1 & -1\\
\end{array}\right)$$
$$B = \left(\begin{array}{crc}
 0 & 1\\
 1 & 0\\
 0 & 1\\
\end{array}\right)$$
1) Decide if the im(A) and the im(B) intersect and if yes find all the vectors of the intersection.Explain why the intersection is a subspace of $R^3$ and find its dimension.
2) Find a vector in the im(A) that is not in the im(B)
I have tried to solve it and from my understanding they do not intersect which is prob wrong.
Any help appreciated. Thanks!
 A: Here are some hints.
First, you write "from my understanding they do not intersect". That's definitely wrong, because $im(A)$ and $im(B)$ are subspaces of $\mathbb R^3,$ and so both contain $0,$ the zero vector. So, we have at least $\{0\} \subseteq im(A) \cap im(B).$
Second, we can use dimesions. By "visual inspection", we see that the columns of the matrix $A$ are linearly independent. So, since $A$ has two columns, we must have $\dim\ im(A) = 2.$ In a completely similar manner, we find $\dim\ im(B) = 2.$ Now, there is a "well known" formula in linear algebra which states that for subspaces $U$ and $V,$ we have
$$
\dim U + \dim V = \dim(U+V) + \dim(U \cap V).
$$
We can apply this to the subspaces $im(A)$ and $im(B)$ of $\mathbb R^3.$ Note that $im(A) + im(B) \subseteq \mathbb R^3,$ and so $\dim(im(A) + im(B)) \leq \dim \mathbb R^3 = 3.$ Plugging everything we know about $im(A)$ and $im(B)$ in the formula above, we get
$$
\begin{align}
4 & = 2 + 2 \\
& = \dim\ im(A) + \dim\ im(B) \\
& = \dim(im(A) + im(B)) + \dim(im(A)\cap im(B)) \\
& \leq 3 + \dim(im(A)\cap im(B)),
\end{align}
$$
which leads us to
$$
\dim(im(A)\cap im(B)) \geq 1.
$$
So there must be a nonzero vector in $im(A)\cap im(B).$ In order to find it, you can use the fact that $im(A)$ and $im(B)$ are planes in $\mathbb R^3,$ and thus described by their normal vectors. The normal vector of $im(A)$ is the cross product of the columns of $A$, and similar for $im(B)$. Now, find a vector that is orthogonal to both these normal vectors. This is a homogeneous system $X$ of two linear equations in three variables. One solution to this system $X$ is
$$
v = \begin{pmatrix}1 \\ 2 \\ 1 \end{pmatrix}.
$$
We also can see that $X$ has rank 1, and thus we must have $im(A) \cap im(B) = \mathbb R v.$
Finally, note that every element of $im(B)$ is of the form
$$
w = \begin{pmatrix}\lambda \\ \mu \\ \lambda \end{pmatrix}
$$
for arbitrary $\lambda,\mu \in \mathbb R.$ From this, we see that for every element of $im(B),$ the first and last coordinate must be equal. From this in turn, we see that the second column of $A$ is contained in $im(A),$ but not in $im(B).$
A: 1) First of all both matrices hav rank 2, therefore Dim Im(A)=2=dim Im(B) and by Grassmann theorem this implies that $dim A\cap B\ge 1$, since they are subspaces of $\mathbb R^3$.
2) If you want equations then $Im A=\{a(1,2,1)+b(0,1,-2)\}=\{(a,2a+b,a-2b\}$ which means that its elements are triples $(x,y,z)$ such that: $x=a,y=2a+b,z=a-2b$ which, after solving w.r. to $a$ and $b$ implies $a=x, b=y-2x$ and therefore $z=x-2(y-2x)$. This is the equation defining $Im(A)$. You can similarly write down the (more easy) equations for $Im B$. Intersection is now just the corresponding linear system.
A: Notice that the first column of $A$ is in the image of $B$.  In fact the image of $B$ consists of vectors of the form $(a,b,a)$.  What would happen if you added a nonzero multiple of the second column of $A$ to a multiple of the first column?  Would that still be in the image of $B$?
