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March 8th, 2012, 03:30 PM   #1
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Proving matrices form a vector space

This problem asks us to show that matrices e i,j and Root(2)e i,j (1</= i, j</=2) form a basis for M2(Q[Root(2)]) considered as a Q-vector space.
It also asks to show the basis for M2(C) considered as a vector space over R.

I'm pretty lost when it comes to showing that something is a basis, the topic has not been covered well and the book doesn't do it justice. The only examples I can find online are concrete which makes it harder work out.

Thanks so much for the help.
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March 13th, 2012, 09:50 PM   #2
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Re: Proving matrices form a vector space

i assume you mean Eij is the matrix with 1 in the ij-th entry and 0's elsewhere.

to prove these 8 matrices form a basis for V = Mat(2,Q(?2)), we need to show 2 things:

they span V, and they are linearly independent.

suppose we have an arbitrary element A of Mat(2,Q(?2)): A =

[a b]
[c d].

since a,b,c,d are in Q(?2), we can write this as: A =

[s+t?2 u+v?2]
[w+x?2 y+z?2], where s,t,u,v,w,x,y and z are all rational.

thus A = sE11 + uE12 + wE21 + yE22 + t(?2E11) + v(?2E12) + x(?2E21) + z(?2E22), which shows the 8 matrices span V.


to prove linear independence, we need to show that if:

sE11 + uE12 + wE21 + yE22 + t(?2E11) + v(?2E12) + x(?2E21) + z(?2E22) is the 0-matrix, we MUST have:

s = t = u = v = w = x = y = z = 0.

start by "collecting terms", if such a linear combination is 0, it must be that:

(s+t?2)E11 + (u+v?2)E12 + (w+x?2)E21 + (y+z?2)E22 = 0 (*).

look at the 1,1 entry, the only matrix to contribute non-zero terms is E11. since the 1,1 entry of E11 is 1,

if the 1,1 entry of the sum (*) is 0, it must be the case that s+t?2 = 0. now show this means s = t = 0.

proceed similarly for the remaining 3 coordinates in the 2x2 matrix.

for M2(C), it might help to think of C as R(?(-1)).
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