Show that the dimension of vector space V of all 2x2 matrices with the real field R is 4 by giving a basis with 4 elements for V?

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Let `alpha=[[1,0],[0,0]]` , `beta=[[0,1],[0,0]]` , `gamma=[[0,0],[1,0]]` and `delta=[[0,0],[0,1]]` be four elements of V.

The subset S={`alpha, beta, gamma, delta` } of V is linearly independent because:

`aalpha+b beta+cgamma+ddelta=0`

`rArr a[[1,0],[0,0]]+b[[0,1],[0,0]]+c[[0,0],[1,0]]+d[[0,0],[0,1]]=0`

`rArr [[a,b],[c,d]]=[[0,0],[0,0]]`

`rArr a=0, b=0, c=0, d=0.`

Also, L(S)=V because if `[[a, b], [c, d]]` is any v``ector in V, then we can write:

`[[a,b],[c,d]]=aalpha+b beta+cgamma+ddelta`

**Therefore, S is a basis of V. Since the number of elements in S is 4, therefore dim. V=4.**

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