It is proved that all symmetric matrices of order n form (n ^ 2 + 2n) / 2-dimensional linear space, so anti symmetric matrices of order n form (n ^ 2-N) / 2-dimensional linear space;

The main control elements of n-order symmetric matrix are the elements above (including the main diagonal)
Let eij be an n-order matrix with row I, column J element 1, row J, column I element 1, and the rest are all 0
Then eij, I

Let u be a space composed of all real matrices of order n, where symmetric matrices form linear subspace V and antisymmetric matrices form linear subspace W. it is proved that u = V ⊕ W
Please, teacher!

Do all invertible matrices constitute linear space over real number field? What about all n-order matrices? If so, the dimension of the space and a set of bases are requested

Are all invertible matrices linear spaces over real number fields?
No, because the addition of inverse to matrix is not closed, the sum of inverse matrix is not necessarily invertible
All n-order matrices
It can form linear space over real number field
Let ε ij be the matrix of order n with the elements of row I and column J being 1 and the rest being 0
Then ε ij, I, j = 1,2,..., N forms a group of bases
So the dimension of space is n ^ 2

It is proved that any square matrix of order n can be expressed as the sum of a symmetric matrix and an antisymmetric matrix
I know I'm stupid, but I can't help it,

For any n-order square matrix A, let B = (a + a ') / 2, C = (A-A') / 2, then it is easy to verify
A = B + C
And B is symmetric (B '= b) and C is antisymmetric (C' = - C)
Here X 'denotes the transpose of X

A is a matrix of order n. for any n * 1 matrix, a has at * a * a = 0. It is proved that a is an antisymmetric matrix

Let a (I, J), I, j = 1,2,... N take: at = (0,0... 1., 0,... 0) (the i-th is 1, the rest is 0), then by at * a * a = 0, we can get: a (I, I) = 0, I = 1,2,... N. further take: at = (0,... 1,0,... 1,0,0) (the i-th and the j-th are 1, the rest are 0), then by at * a * a = 0, we can get: a (I, J) + a (J, I) = 0

Let a be a symmetric matrix of order n and B be an antisymmetric matrix of order n. It is proved that (a + b) (a-b) is a symmetric matrix

From the known a ^ t = a, B ^ t = - B
So [(a + b) (a-b)] ^ t
= (A-B)^T(A+B)^T
= (A^T-B^T)(A^T+B^T)
= (A+B)(A-B)
So (a + b) (a-b) is a symmetric matrix

A is a square matrix of order n. how to prove that a * a ^ t is a positive semidefinite matrix
The matrix obtained by multiplying a by the transpose of A

x^T(AA^T)x
= (A^Tx)^T(A^Tx)
This is the inner product of a ^ TX and a ^ TX
>=0
So AA ^ t is positive semidefinite
(symmetry omitted)

It is proved that for any m * n matrix A, a ^ t * A and a * a ^ t are symmetric matrices

Because (AA ^ t) ^ t = (a ^ t) ^ TA ^ t = AA ^ t
So AA ^ t is a symmetric matrix
Similarly, because (a ^ TA) ^ t = a ^ t (a ^ t) ^ t = a ^ TA
So a ^ TA is a symmetric matrix
Property: (AB) ^ t = B ^ TA ^ t
Any more questions
To prove d symmetry, we must prove d ^ t = D

It is proved that for any n-order matrix A, a + A ^ t and symmetric matrix, A-A ^ t is antisymmetric

(a + a ')' = a '+ a = a + a', so a + a 'is symmetric
(A-A ')' = a '- a = - (A-A'), so A-A 'is antisymmetric

Let a and B be matrices of order n, and a be symmetric. It is proved that B ＾ tab is also symmetric

Because a is a symmetric matrix, a '= a (a' is the transpose of a)
So (b'ab) '= b'a' (B ')' = b'ab
So AB's symmetric

It is proved that all symmetric matrices of order n form (n ^ 2 + 2n) / 2-dimensional linear space, so anti symmetric matrices of order n form (n ^ 2-N) / 2-dimensional linear space;