# SOLUTION: Let V = matrix M(2×2)(Complex). Let W1 = {A = (Ajk) 1 <=j,k<=2 exists in V | A11 + A22 = 0 }. a) Find subspaces W2 and W3 of V such that V = W1 + W2, V = W1 + W3, W2 not e

Algebra ->  Algebra  -> College  -> Linear Algebra -> SOLUTION: Let V = matrix M(2×2)(Complex). Let W1 = {A = (Ajk) 1 <=j,k<=2 exists in V | A11 + A22 = 0 }. a) Find subspaces W2 and W3 of V such that V = W1 + W2, V = W1 + W3, W2 not e      Log On

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 Question 27385: Let V = matrix M(2×2)(Complex). Let W1 = {A = (Ajk) 1 <=j,k<=2 exists in V | A11 + A22 = 0 }. a) Find subspaces W2 and W3 of V such that V = W1 + W2, V = W1 + W3, W2 not equal to W3. (Note: One way to describe a subspace is to give a basis for it.) b) Explain why W2 (intersect) W3 = {0} for any pair of subspaces W2 and W3 satisfying the conditions of part a).Answer by khwang(438)   (Show Source): You can put this solution on YOUR website!Let V = matrix M(2×2)(Complex). Let W1 = {A = (Ajk) 1 <= j,k<= 2 exists in V | A11 + A22 = 0 }. a) Find subspaces W2 and W3 of V such that V = W1 + W2, V = W1 + W3, W2 not equal to W3. (Note: One way to describe a subspace is to give a basis for it.) b) Explain why W2 (intersect) W3 = {0} for any pair of subspaces W2 and W3 satisfying the conditions of part a). [ Note : Here W1+W2 should mean the direct sum instead of the general sum i.e W1 ^ W2 = {0}) Sol: a) Note dim V = 2x2 = 4 and since A11+A22 =0, W1 is a 3 dim subspace of V. By choosing a nonzero vector u , say u = I2 in V not in W1, set W2 to be generated by u, then we have W1 + W2 = V since dim W1 + W2 = 4. Similarly, choose w = (1 0) (0 0), and let W3 be generated by w. We get W1 + W3 = V. Since u != cw for any scalar(complex), we see that W2, W3 are not equal. b) If W2 & W3 are two distinct subspaces of V such that V = W1 + W2 = W1 + W3 (direct sum). Then dim W2 = dim W3 = 1. Claim: W2 ^ W3 = {0}. For otherwise, W2^W3 as a nonzero subspace of the 1-dim space W2 ( or W3), then dim (W2 ^ W3) = 1 and so W2^W3 = W2 = W3. This contradicts to W2, W3 are not equal. Hence, W2 ^W3 must be {0}. Kenny