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6 a All invertible A First We note that the Zero matrix sould be inclouded byevery subspace of V Sin Ce O-matrix - is not invertibles therefore the Set of invertible mathicos is not vector space Examplein 111 an last The following two matrices A and B are invertible As 2 1 B= -1 0 -1 - subspace should A+B also invertible, but A+ B 11. L. 0 a a ] which is not invertible (not closed under a edition) => Therefor it is not a subspact b all non-inventible of A This is not a Subsprace + because if we take matrices A= D- a J. I B= o 0 i] two which is non-inventible because the determinant equal to Zero for this two mutrics and A+B = O the Identity matrix \ which is invertible T that's mean the Subset is not (closed under addition -> it is not Subspace ) all A such that AB-BA - B.fixedmatri inv and Suppose A and A cw , W S subspace of 2 Vectorspace V, sattisly AB 6- BA, and AB 2 = BA 2 Let ceF be any Constant, then, (CA+A)B= 11 CAB+ AB 2 = CBA' + B A2 don't 27 (B(CA+A) \ 2 CA,A satisly the can Theorem that say ( A nun-empty subset w of V is a subst 0 of V if and only if for each Pair of vecto X, B in W and CEF => Cat B is in W Dall A such that A==A Suppose Char(F) #2 1 1) 2 Let A= A2 7 A = A 1 1) but A+A = !i] 2 + 11 a 1 1 2 I (A+A)2 2 = [20] 2 (2)2 0 4 = 11 _0 (2) 0 4 (A+A} # (A+A) A+A doesn't satisfy the Condition forasubspaci , so the Subset Can not be Subspace (IP) if we suppose F=1/202 Let AantB both satisfy A2-A and B2-B Let CEF 2.2 be any Scalar then, (CA+A)ZCA+ 2CAB +B2 11 2=0 C'A++ B2 if C = - =) B2 which equal to B if C= 1 A &+B2 which equal to A+B in both cases(CA+B)==(CA+B) in this Case the Subset is a subspace 4.14 A 1110 Now Let Char(F) = 2 but E not then polynomiaL x2-x-0 has most two Solutions in F,so 7 CEF Such that C consider the Identity matnix I [ij] then, I 2 s =) I 2 = I (A7A) (O7A) if I' = I is a subspac then, must CI also in the subspace ) must be (CI) 2 = (CI) which equivalent into C 2-1 = C . which Contradicts out 7 2 assumption C'+ C

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