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Show that H is a subspace of M sub2x2, by finding a subset of H such that span(S) = H

  • #1
H = ([a,b;c,d] : a+d =0}

Dim(M2x2)= 4, so a basis would have 4 components?

I got this far and am stuck.

[a, b ; c , -a] = a[1,0; 0,-1] + b[0, 1;0,0] +c[0,0; 1,0]
 

Answers and Replies

  • #2
33,179
4,859


H = ([a,b;c,d] : a+d =0}

Dim(M2x2)= 4, so a basis would have 4 components?
A basis for M2x2 would have 4 vectors/matrices, but how many would be in a basis for H?
I got this far and am stuck.

[a, b ; c , -a] = a[1,0; 0,-1] + b[0, 1;0,0] +c[0,0; 1,0]
Now, do these matrices span H? I.e., can every matrix in H be written as a linear combination of the three matrices above?

What's left to do is to show that H is a subspace of M2x2. To do this, you need to show three things:
That the zero matrix is in H.
That if A and B are in H, then A + B is in H.
That if A is in H and c is a scalar, then cA is in H.
 
  • #3


Thank you
 

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