Let A and B be n n square matrices with A O and B O Prove

Let A and B be n × n square matrices with A = O and B = O. Prove that if A is symmetric and B is skew-symmetric (BT = B), then {A, B} is a linearly independent set.

Solution

we know that a square matrix A is symmetric matrix if A = A^T that is A(i,j)=A(j,i) and a square matrix B is skew - symmetric if B = -B^T, that is B(i,j)=-B(j,i)

we also know that If B is skew symmetric then diagonal element of B are all zero.

so B(i,i)=-B(i,i) implies that B(i,i)= 0 .

Now consider that {A,B} is a linearly dependent set, then exist a nonzero scalar k such that A = k*B

so we can say that,

A(i,j) = k*B(i,j) and A(j,i) = k*B(j,i)

Since B is skew-symmetric,

A(i,j) = k*[-B(j,i)] = -k*B(j,i)

Then, for all (i,j):

-k*B(j,i) = k*B(i,j) = A(i,j) = A(j,i) = k*B(j,i)

It is only possible when, k*B(j,i) = 0 for all (i,j)

But not all B(i,j) are zero means that k must be 0, this contradict the initial asumtion that k is not zero.

Hence we can say that set {A,B} is linearly independence.