1
$\begingroup$

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!

$\endgroup$
6
  • $\begingroup$ what you have find as im(A) and im(B)? $\endgroup$
    – David
    Jan 4, 2015 at 17:24
  • $\begingroup$ They both contain the first column of $A$ because it can be written as a linear combination of the columns of $B$. $\endgroup$
    – Dan Fox
    Jan 4, 2015 at 17:25
  • $\begingroup$ im(A)=span${(1,2,1),(0,1,-1)}$ and for im(B)=span${(0,1,0),(1,0,1)}$ $\endgroup$
    – Nash
    Jan 4, 2015 at 17:26
  • $\begingroup$ Basically i have used the column vectors since the image of a transformation is the span of the column vectors. $\endgroup$
    – Nash
    Jan 4, 2015 at 17:27
  • $\begingroup$ @Dan is there a way to come to that conclusion using an equation? $\endgroup$
    – Nash
    Jan 4, 2015 at 17:34

3 Answers 3

2
$\begingroup$

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 $im(B).$

$\endgroup$
2
  • $\begingroup$ whops, just overlapped, sorry $\endgroup$
    – N. Ciccoli
    Jan 4, 2015 at 17:42
  • $\begingroup$ Thank you jflipp for the awesomely explained answer! :) $\endgroup$
    – Nash
    Jan 4, 2015 at 18:09
0
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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 $B$?

$\endgroup$
1
  • $\begingroup$ I don't think it would be still in the image of B $\endgroup$
    – Nash
    Jan 4, 2015 at 17:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .