# Thread: [SOLVED] Show that the following subsets U and W are subspaces of V

1. ## [SOLVED] Show that the following subsets U and W are subspaces of V

Let V be a vector space of nxn Matrices over a field $K$. Show that the following subsets $U$ and $W$ are subspaces of $V$.

1. $U = {A = (a_ij) \in V | a_ij = a_ji, \forall 1 \leq i, j \leq n}$.

2. Let $T$ be a fixed matrix in $V$ and $W = {A \in V | AT = TA}$.

So, $A$ are symmetrical matrices in $\mathbb{R}^n$, is there anything special about symmetrical matrices relevant for this question (part1)?

If the matrix is symmetrical than (part2) $AT \equiv TA$?

2. A subspace is just a vector space that is contained. You just gotta show that for both they are closed under addition and scalar multiplication. Also, that they contain the additive identity, but this is trivial for both of them.

1) This is simply the set of symmetric matrices. $A^T=A$
Let A and B be symmetric matrices.
$a_{ij}=a_{ji}$ and $b_{ij}=b_{ji}$ by definition.
A+B you just add compnantwise, so the ijth entry of A + B is $(A+B)_{ij}=a_{ij} + b_{ij} = a_{ji} + b_{ji} = (A+B)_{ji}$ (check)
Scalar multiplication.
A is a symmetric matrix so $a_{ij}=a_{ji}$. Let k be a scalar from the field K.

$ka_{ij}=ka_{ji}$ so kA is a symmetric matrix.
(check)
Clearly the 0 matrix is symmetric every entry is 0.
(check)
1) is a subspace.

2)T is fixed and this is just the orbit of T under conjugation (This first statement is only true in the set of invertible matrices, but I never used invertibility in the proof, I meant to remove this statement actually, but forgot). You are looking at the set of all matrices A such that AT=TA call it W.

Let A and B be matrices in W. Then AT=TA and BT=TB
(A+B)T=AT +BT= TA + TB=T(A+B)
so A+B is in W
(Check)
Let k be an element in the field.
T(kA)=(Tk)A=k(TA)=k(AT)=(kA)T
so kA is in W.
(check)
T(0)=0=(0)T
so 0 is in W
It is a subspace.

Done and done.

3. Thanks for the reply, Gamma. So, I just use the vector space rules to show that they are valid for these subspaces. I thought, maybe there is something special about this type of matrix. Thanks again!

4. Yup, that is all you have to do. You will find this to be the case a lot if you continue on in math. You will have to find things are subgroups of groups, subrings of rings, subfields of fields, subspaces of vector spaces, submanifolds of manifolds, it is a fairly common theme in mathematics. All these cases are examples of when you just need to make sure they are both a subset and satisfy the same rules as the mother object.

But in a way, you should not dismiss the uniqueness of these particular types of matrices. Their structure is important for them to actually be subspaces. Not just any subset of matrices will give you a subspace.