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

$$A = \left(\begin{array}{crc}
 1 &amp; 0\\
 2 &amp; 1\\
 1 &amp; -1\\
\end{array}\right)$$

$$B = \left(\begin{array}{crc}
 0 &amp; 1\\
 1 &amp; 0\\
 0 &amp; 1\\
\end{array}\right)$$

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!
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
$$
\begin{align}
4 &amp; = 2 + 2 \\
&amp; = \dim\ im(A) + \dim\ im(B) \\
&amp; = \dim(im(A) + im(B)) + \dim(im(A)\cap im(B)) \\
&amp; \leq 3 + \dim(im(A)\cap im(B)),
\end{align}
$$
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 $im(B).$
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 $B$?