Okay, this problem is another proof that you would find in a first semester course of linear
algebra. It says, let B equal the set of four matrices listed here. Prove that B is a basis
for M22(R). Obviously, M22(R) is the notation for any two-by-two matrix with real values on all the
entries. This could be (X Y; Z W) and those are just real values. To prove this problem,
you basically just have to know the definition of what a basis is and prove that the given set forms
a basis. By definition, a basis spans whatever you're trying to form a basis of. In this case,
it has to-- B has to span M22(R) and also a basis--the span, whatever's spanning it,
has to be linearly independent. And by definition, if the values are linearly independent and they
span a, in this case, a matrix, then it does form a basis. If we do those things,
we can prove this problem. The first thing I would do is to prove that B spans M22(R).
I would find the dimension first of M22(R). One way of thinking of dimension is, by definition,
it's the number of matrices in this case that you need to form a basis. By definition for matrices,
the dimension of any m-by-n matrix is m by n. In this case because we have a 2 by 2, it's going to
be 2 times 2 equal to 4. The dimension of M22(R) will be 4. You need a set of 4 matrices to span
M22(R) and to form a basis. Not coincidentally, we have 4 matrices here. The thing with dimension is
that they do have to be linearly independent. We haven't proven yet that the dimension of B is 4,
but we can say, because the dimension of M22(R) is 4, it suffices to show that B
is linearly independent. Because if B is linearly independent, then the dimension of B is 4. Then B
will span M22(R) and be linearly independent and by definition will be a basis. Now we just have
to--we've basically proven that it spans M22(R), more or less. Now we just have to show that the
set B is linearly independent. The way I would do that would just be putting coefficients and,
you know, using the definition of what is linear independence if you have C1 times a matrix plus
C2 plus C3 plus C4 equal to 0 and then you have to show that if you have a linear combination,
the coefficients must be 0. Suppose C1, C2, C3, C4 exists in R, so these are scalars. Then C1 times
the first matrix plus C2 times the second matrix plus C3 times the third matrix plus C4 times the
fourth matrix. And this is a zero matrix. And just by matrix properties, we can rewrite that as (C11
C11; C11 C11) plus--we're basically trying to show that if we have this linear combination,
then the only solution will be the coefficients are 0 and that's a way
of proving linear independence. I'm just going to do that step by step. So (C2 0; 0 C2) plus (0 -C3;
C3 0) plus (C4 C4; C4 -C4) is equal to the zero matrix. If we combine across matrices,
we're going to get C11 plus C2 plus C4; C1 minus C3 plus C4; C1 plus C3 plus C4 and C1 plus C2
minus C4 is equal to 0. Next I would rewrite this as a linear system of equations. So C1 plus C2
plus C4 is equal to 0; C1 minus C3 plus C4 is equal to 0; C1 plus C3 plus C4 is equal to 0;
C1 plus C2 minus C4 is equal to 0. We can rewrite this again in matrix form as (1 1 0
1) equal to 0; (1 0 -1 1) is equal to 0; (1 0 1 1) is equal to 0; (1 1 0 -1) is equal to 0. If we let
A be the coefficient matrix, so A is equal to (1 1 0 1; 1 0 -1 1; 1 0 1 1; 1 1 0 -1), then we can do
row reduction. Let's do R3 minus R1, R4 minus R1, and that gives us (1 1 0 1; 0 -1 -1 0; 0 -1 1 0;
0 0 0 -2). And we're basically just trying to see if we get any zero rows. Then we're going to do R3
minus R2 and that's going to give us (1 1 0 1; 0 -1 -1 0; 0 0 2 0; 0 0 0 -2). We can stop here
because this is already in row echelon form and we see that we're not going to get any zero rows,
so we can see that this is linearly independent and we can go further and say A is invertible
from the above. Thus we can say, because we know that A was invertible, we had the equation A times
the coefficient (C1; C2; C3; C4) was equal to the zero vector. We can say that A inverse times (0;
0; 0; 0) is equal to (C1; C2; C3; C4) because we know A is invertible. Thus we know that any matrix
times zeros will give you 0, so we know that (C1; C2; C3; C4) is equal to 0. We've proven
that B is linearly independent because B spans M22(R) and is linearly--let me write--l.i.,
linearly independent. Thus by definition, B forms a basis for M22(R) and we're done with our proof.