## Transcribed Text

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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