MARTINA BALAGOVIC: Welcome.
Today's problem actually appeared in a quiz.
It appeared in quiz one in fall of 1999 as question four.
The problem puts the usual solve the following system
upside down by saying we have some matrix and we know that
all the solutions to ax equals this vector here, 1, 4, 1, 1,
all the solutions to this problem are given by x equals
0, 1, 1 plus any number c times 1, 2, 1.
And we're asked to say everything that we can about
the columns of the matrix A.
So I'm going to let you pretend that you are on an
exam, try to solve it yourself, and then come back
and compare your solution with mine.
OK, welcome back.
So the first thing that you should think about in this
sort of situation is what is the size of a?
Well if we want to multiply a with an x, that has three
entries, so a should have three columns.
Let me call those columns c1, c2, and c3.
And when I take some linear combinations of c1, c2 and c3,
I'm going to get this vector here, 1, 4, 1, 1.
So all the ci's, c1, c2, and c3 are vectors in R4.
Now, if you know about, if you learned about particular
solutions and special solutions, then my notation
here shouldn't surprise you.
I'm going to call this vector here xp, and
this vector here xs.
And I'm going to use the fact that xp plus c times xs
satisfies A times this equals b.
I will call this vector b, for any number c.
In particular, what I'm going to conclude is that when c
equals 0 we get a times xp equals b.
But also that when c equals 1, we get a times xp plus a times
xs equals b.
Replacing this by b, we get that this implies that a times
Axs equals 0.
So in trying to find what are the columns, c1, c2, and c3 of
the matrix A, let's look at these two equations.
xp satisfies a times xp equals b, and xs satisfies A
times xs equals 0.
Again, if you know what a particular and special
solutions are this shouldn't surprise you.
But we also know what xp and xs are, so let's use them to
try to calculate c1, c2, and c3.
A times xp equals b means that the linear combination of c1,
c2, and c3 encoded in the vector xp, which is 0, 1, 1,
gives the vector b.
So c1, c2, c3 times 0, 1, 1 gives us 1, 4, 1, 1.
In other words, c2 plus c3 equal b.
Let's turn our attention to A times xs equals 0.
This says that c1, c2, c3 times xs was defined to be 0,
2, 1 equals 0.
In other words, 2 times c2 plus c3 equals 0.
Now solving this system where the unknowns are vectors but
it's still just a linear system, we can see, for
example, from the second equation that c3
equals minus 2c2.
And plugging it back into the original equation getting c2
minus 2c2 equals b from which it follows that c2 is equal to
minus b, and that c3 is equal to 2 times b.
So from this tiny amount of information, we just knew the
solutions to this one particular equation involving
A, we got the second column of A and the third column of A
completely explicitly calculated.
Now, what can we say about the first column?
I said before that all the solutions of ax equals b are
of the form a particular solution plus some number
times a special solution.
And the information that we have is that there's just one
number here.
So we said everything, once we remove this vector here,
everything that we get here will satisfy A
times x equals 0.
And the fact that everything that satisfies A times x
equals 0 is a multiple of this one vector that was given to
us means that the null space of A has dimension one.
There is just one special solution.
So dimension of the null space of A is 1.
So rank of A is the number of columns minus this dimension
of null space, and it's equal to 2.
As rank of A is equal to 2, the number of linearly
independent columns needs to be 2 as well.
So the only thing that we can say at this point is if the
first column was also a multiple of b, as the second
and the third are, then the rank would be smaller than 2.
So that is the only thing that cannot happen.
So c1 is not a multiple of b.
Not any multiple, including not a 0 multiple.
And that's pretty much everything we can say.
Yes, if it was some other multiple of it, then we would
be able to find some other vector here and we would have
two parameters.
But it's not, and this is everything that we
can say about it.