# Thread: Show that the following matrices form a base for M22

1. ## Show that the following matrices form a base for M22

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.

2. Originally Posted by algie
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]

You can work with coordinates on the canonical basis $\displaystyle B$ of $\displaystyle M_{2\times 2}(\mathbb{R})$:

$\displaystyle \left[\begin{pmatrix}{a}&{b}\\{c}&{b}\end{pmatrix}\right]_B =(a,b,c,d)$

As $\displaystyle \dim M_{2\times 2}(\mathbb{R})=4$ you only need to prove that the rank of $\displaystyle A$ is $\displaystyle 4$ being $\displaystyle A$ the matrix whose rows are the corresponding coordinates of $\displaystyle M_i$ with respect to $\displaystyle B$ .

3. 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.

4. 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

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### Bases for M22 matrix

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