# symmetric/skew matrices

• Dec 12th 2010, 10:54 AM
khughes
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)
• Dec 12th 2010, 11:07 AM
HallsofIvy
Quote:

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.
• Dec 12th 2010, 11:17 AM
khughes
edited, sorry for the confusion.
• Dec 13th 2010, 04:22 AM
HallsofIvy
I don't see anything new!

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

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

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

Not at all the same thing!
• Dec 14th 2010, 02:39 PM
khughes
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