Show that for any square matrix A the matrix is symmetric an

Show that for any square matrix A the matrix is symmetric and the matrix is skew-symmetric. Let A and B be two square matrices of the same size n Times n. If A is invertible then AT invertible and (AR)-1 = (A-1)T Let A and B be two square matrices of the same size n Times n. If A and B are invertible matrices then AB is invertible and (AB)-1=B-1A-1 Show that if A is skew-symmetric then the entries on the main diagonal are 0.

Solution

1. (a) Let B=A+A^T ,   Bij =Aij +Aji =B^Tij .
    (b) Let B=AA^T ,  Bij =Aij Aji =Aji (Aij )=(Aji Aij )= B^Tij

2.

(A^T)(A^-1)^T  = (A^-1A)^T =   I^T = I

This proves that the inverse of A^T is (A^-1)^T .

3.

Not really. Matrices do not follow exponential laws. In fact, (AB)^-1 = B^-1A^-1 .

Here is the proof:

Let I be a 3 by 3 identity matrix. If A and B are 3 by 3 invertible matrices, then:

                                    (AB)(AB)^-1   = I

                                   A^-1 (AB) (AB)^-1 = A^-1I

                                  (A^-1AB) (AB)^-1 = A^-1I

                                        (IB)(AB)^-1     =   A^-1

                                                             B(AB)^-1     =   A^-1

                                    B^-1B(AB)^-1     =   B^-1A^-1

                                    (B^-1B)(AB)^-1     =   B^-1A^-1

                                                               I (AB)^-1   = B^-1A^-1

                                                                 (AB)^-1   = B^-1A^-1
Hence Proved

4.

If a matrix is skew symmetric then A^T = - A,

that is the transpose of A is equal to negative A.

This implies that if A = a_ij , then a_ji = a_ij  .

If we\'re referring to diagonal entries, we can say a_jj = -a_jj .

The only way for this to be true is if a_jj = 0.

So therefore all the diagonal entries of a skew symmetric matrix are 0.