If is a square matrix and for all , then prove is a symmetric matrix. ie. prove

I cant just go rite? So what should i do instead?

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- Apr 16th 2010, 02:26 AMvuze88Symmetric matrix
If is a square matrix and for all , then prove is a symmetric matrix. ie. prove

I cant just go rite? So what should i do instead? - Apr 17th 2010, 12:26 AMBobP
Use the matrix property

- Apr 17th 2010, 02:01 AMvuze88
yeh i know that property but how do i do it? ive just used the property and now im at . But i cant just cancel them out to get because they are matrices...

- Apr 17th 2010, 04:08 AMBobP
You're told that the relationship is true for all and , so it's certainly true for specific and

Choose to consist of entirely of zeros apart from a 1 in the ith position and to consist entirely of zeros apart from 1 in the jth position. Then evaluate both sides of the equation - Apr 18th 2010, 12:11 AMNoxide
You started with:

A is nxn

x^TAy= (Ax)^Ty

so

x^TAy= x^TA^Ty

x^TAy - x^TA^Ty = 0

factor...

x^T(A - A^T)y = 0

this is true for all x,y in R^n

consider x = 0, and y = anything in R^n we get x^T(A - A^T)y = 0

consider y = 0, and x = anything in R^n we get x^T(A - A^T)y = 0

consider y=x=0 we get x^T(A - A^T)y = 0

we know the statement is true for all x,y not just x, y, being zero vectors

the only way x^T(A - A^T)y = 0 is true for any x and y is if A- A^T = zero matrix

hope this helped