# Thread: Matrices A subspaces of Mn(R)

1. ## Matrices A subspaces of Mn(R)

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?

IV) and the same for this one.

I look forward to hearing from you and really appreciate you reading this.

Thanks!

2. 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?
You are asked about the set of all n by n matrices, A, such that AB= BA for a specific B. You cannot assume that A is diagonal (which you do) and you cannot assume 2 by 2 matrices.

1) The set must be non-empty. Most often it is best to show that the 0 vector is in the set. Is it true that 0B= B0?

2) The set must be closed under addition. If AB= BA and CB= BC, that is, if A and B are in the set, is (A+ C)B= B(A+ C)?

3) The set must be closed under scalar multiplication. If AB= BA, that is, if A is in the set, and p is any scalar, is (pA)B= B(pA)?

IV) and the same for this one.

I look forward to hearing from you and really appreciate you reading this.

Thanks!
A counter-example will show that a general statement is NOT true, which is why your first examples were correct. But an example cannot prove that a statement is true.

### subspace of M_n (R) containing I

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