Span(S) is the set of all 2 x 2 Symmetric Matrices

Posted by: tyr 8 years ago

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Let W_2 be the set of all 2 x 2 real symmetric matrices such that for all A in M_(2 x 2)(R), A_ij = A_ji; for all 1 ≤ i,j ≤ 2.

W_2 ≤ M_(2 x 2)(R)

We define A in W_2 such that A = [[a, c], [c, b]]; for all a,b,c in R.
Note: A = A^t, det(A) = ab - c^2. 

Let S = {M_1, M_2, M_3} = {[[1, 0],[0, 0]] , [[0, 0],[0, 1]] , [[0, 1],[1, 0]]}

To show that Span(S) = W_2 it is necessary to show that Span(S) ⊇ W_2 and Span(S) ⊆ W_2.

i)For all x,y,z in R:

[[a, c],[c,b]] = (x,y,z) ∙ S = x[M_1] + y[M_2] + z[M_3] => [[a, c] , [c,b]] = x[[1, 0],[0, 0]] + y[[0, 0],[0, 1]] + z[[0, 1],[1, 0]] => a = x, b = y, c = z.

Hence Span(S) ⊇ W_2.

ii)For all a,b,c in A:

{S * (a,b,c)} = {{M_1, M_2, M_3} * (a,b,c)} = {[M_1]a, [M_2]b, [M_3]c} = {[[1, 0],[0, 0]]a, [[0, 0],[0, 1]]b, [[0, 1],[1, 0]]c} = {[[a, 0],[0, 0]], [[0, 0],[0, b]], [[0, c],[c, 0]]} => [[a, c], [c, b]] = A.

Hence Span(S) ⊆ W_2.

Therefore Span(S) = W_2. []

• Tags:
• Linear Algebra
• algebra
• matrix
• proof
• real
• theory
• symmetric matrices
• span
• basis
• 2 x 2
• subset
• vectors

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