1. ## symmetric/skew matrices

So I understand that a matrix A = sym(A) + skew(A) = 1/2(A + Atranspose) + 1/2(A - Atranspose)

I cant seem to understand why multiplying any matrix A by a symmetric or skew matrix B is equal to B.sym(A) or B.skew(A), respectively. in other words:

B.A = B.sym(A) when B is symmetric and
B.A = B.skew(A) when B is a skew matrix

any help is appreciated. (B.A is the inner product of B and A in case there is any confusion)

2. Originally Posted by khughes
So I understand that a matrix A = sym(A) + skew(A) = 1/2(A + Atranspose) + 1/2(A - Atranspose)

I cant seem to understand why multiplying any matrix A by a symmetric or skew matrix B is equal to B.sym(A) or B.skew(A), respectively. in other words:

B.A = B.sym(A) when B is symmetric and
B.A = B.skew(A) when B is a skew matrix

any help is appreciated. (B.A is the dot product of B and A in case there is any confusion)
Actually, I wasn't confused until you said "B.A is the dot product of B and A"! B and A are matrices- what do you mean by the "dot product" of two matrices?
If you simply mean the matrix product, then what you have written is NOT true.
BA is NOT equal to B(symm(A)) when B is a symmetric matrix.

3. edited, sorry for the confusion.

4. I don't see anything new!

And, if A.B means simply matrix multiplication, again, the statement is NOT true.

For example, suppose $A= \begin{bmatrix}3 & 2 \\ 1 & 2\end{bmatrix}$ and B is the symmetric matrix $B= \begin{bmatrix}2 & 1 & 1 & 3\end{bmatrix}$
Then $symm(A)= \begin{bmatrix}3 & \frac{3}{2} \\ \frac{3}{2} & 2\end{bmatrix}$

$BA= \begin{bmatrix}7 & 6 \\ 6 & 8\end{bmatrix}$
$B symm(A)= \begin{bmatrix}\frac{15}{2} & 5\\ \frac{15}{2} & \frac{15}{2}\end{bmatrix}$

Not at all the same thing!

5. Gah! im sorry, i reread my book and A.B is the scalar product, so A.B = tr(Atranspose*B).

that made life easier. managed to figure it out.

B.sym(A) = tr(B(.5(A+Atranspose)) = .5[ tr(BA) + tr(BtransposeAtranspose)] = .5(2tr(BA)) = tr(BA) = B.A

similar for other one