Check (2 1,0 0),(0 0,2 0),(3 -1,0 0),(0 3, 0 1) is a bais for M22 or not?

Lewis Harvey 2020-12-21 Answered
Check (21,00),(00,20),(31,00),(03,01) is a bais for M22 or not?
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Expert Answer

liingliing8
Answered 2020-12-22 Author has 95 answers

Recall: Basis Let V be a vector space over a field F. A set of S of vectors in V is snid to be a basis of V if (i) S is linearly independent in V, and
(ii) S generates V ie. linear span of S,L(S)=V
Theorem: Let V be a vector space of dimension n over a field F. Then any linearly independent set of n vectors of V is a basis of V.
M22 is the vector space of all 2×2 real matrices.
Let S={(2100),(0021),(3100),(0031)}
We have to show that 5 is a basis of M22
Let α1={(2100),α2=(0021),α3=(3100),α4=(0031)}
Let us consider the relation
c1α1+c2α2+c3α3+c4α4=0
where c1,c2,c3,c4RR.
Then we have c1=(2100),+c2(0021),+c3(3100),+c4(0031)=(0000)(2c1+3c3c1c32c2+3c3c2+c4)=(0000)
Fron the above equation we get,
2c1+3c3=0
c1c3=0
2c2+3c4=0
c2+c4=0
Solving these equation we get, c1=0,c2=0,c3=0,c4=0
This proves that the set α1,α2,α3,α4 is lineary independent.
Now, we know that M22 is a vector space of dimension 4 and its standard
basis is {(1000),(0100),(0010),(0001)}
As the dimension of M22 is 4 and α1,α2,α2,α4 is a linearly independent
set containing 4 vectors of M22,
therefore the set α1,α2,α3,α4
is a basis of M22.
Result:
{(2100),(0021),(3100),(0031)}
is a basis for M22

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