Originally Posted by

**katiedavies1990** I am trying to revise for my upcoming exam and I am stuck on a question-

Let Mn(R) be the voctor space of all nxn real matrices, n>/2 (greater than or equal to). Which of the following set of matrices A in Mn(R) are subspaces of Mn(R)?

I) all invertible A

II) All non-invertib;e (i.e. singular) A

III) all A such that AB=BA for some fixed matrix B in Mn(R)

IV) all A such that A^2=A

i have proves I) because to be a subspace it must contain the zero element and o is not invertible.

II) Let a= (0 0 b= (1 0

0 1) 0 0)

a+b = (1 0

0 1) which is the identity and invertible so A is not closed under addition so therefore not a subspace

III) Here i let A= (a 0 and B= (c 0

0 b) 0 d)

then AB = (ac 0 BA= (ca 0

0 bd) 0 db)

then I have said it contains the identity so is not empty

A+B = (a+c 0

0 b+d) which is still in Mn(R) as it is diagonal

and then x (a 0 = xA

0 b)

= (xa 0

0 xb) which is also in Mn(R) as still diagonal therefore a subspace.

However my teacher says I am wrong because I need to prove for arbitrary nxn matrices. What does she mean?