Okay, this problem is another proof that you  would find in a first semester course of linear  
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  
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  
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,  
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  
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,  
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,  
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  
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,  
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).  
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,  
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,  
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  
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  
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  
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,  
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  
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  
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  
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  
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,  
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  
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,  
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 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  
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  
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,  
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  
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;  
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,  
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  
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  
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;  
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  
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  
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  
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;  
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  
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  
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,  
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  
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  
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;  
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  
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  
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.,  
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.
linearly independent. Thus by definition, B forms  a basis for M22(R) and we're done with our proof.