Show that the following matrices form a base for M22.
[3 6] , [0 -1], [0 -8] ,,, [1 0]
[3 -6], [-1 0], [-12 -4], [-1 2]
c1M1+c2M2+c3M3+c4M4=B
c1M1+c2M2+c3M3+c4M4=
[0 0]
[0 0]
c1M1+c2M2+c3M3+c4M4=
[a b]
[c d]
After addition I get:
[1 0]
[0 1]
[c1 0] = [a b]
[0 c4] = [c d]
a=d
b=c
Not linearly independent, but it should be.
Thanks for your help.
Actually I do not know what rank is, I read about it just now and from my understanding the rank of a 2x2 matrix cannot be higher than 2.
I still do not understand, sorry.
In my book they have a really simple example that prove The standard basis for M_mn.
Example)
Show that
M1=e1=
[1 0]
[0 1]
M2=e2
M3=e3
M4=e4
form a basis for the vector space M_22.
Solution:
1. Prove linear independence. / Show that c1M1+c2M2+c3M3+c3M4=0 has only the trivial solution.
2. Show that every 2x2 matrix
B=
[a b]
[c d]
can be expresesd as c1M1+c2M2+c3M3+c4M4=B.
In the end they have
[c1 c2]=[0 0]
[c3 c4] [0 0]
and
[c1 c2]=[a b]
[c3 c4] [c d]
First eq. has c1=c2=c3=c4=0, has only solution 0, linear independence proven.
Second eq. c1=a, c2=b, c3=c, c4=d, conclusion; they span M_22, and form a basis of M_22.
My problem is, after addition and row operations on c1M1+c2M2+c3M3+c3M4 I get
[1 0]
[0 1]
and since I get zeroes, I do not understand if this is still a basis of M_22.
I mean, if this forms a base for M_22, then I might as well get anything in that matrix and say it form the basis of M_22.
I am sorry I cannot understand everything that is said, I failed this exam and really try to understand now!
I did it in another way but get another, "wrong?" answer again.
Create a matrix of c1 c2 c3 c4, they span M_22.
[3 0 0 1]
[6 -1 -8 0]
[3 -1 -12 -1]
[-6 0 -4 2]
After row operations I get
[1 0 0 0]
[0 0 1 0]
[0 0 0 1]
[0 1 0 0]
c1=c2=c3=c4=0
Is linear independent.
But still the same since
c1[1 0]+c2[0 0]+c3[0 1]+c4[0 0]
...[0 0].....[0 1]......[0 0].....[1 0]
equals
[c1 0]
[0 0]
I really do not know what I try to do, might as well skip this problem