MHB True or False Question about Square Matrices

Thread starter TheScienceAlliance
Start date Apr 28, 2022
Tags

Matrices Square

AI Thread Summary

For every square matrix A, the expression C = A(A^t) + (A^t)A is confirmed to be symmetric. This can be demonstrated by calculating the transpose of C and applying the properties of transposes, specifically that the transpose of a sum is the sum of the transposes and the transpose of a product reverses the order of multiplication. The discussion emphasizes the importance of these rules in proving the symmetry of the matrix C. Ultimately, the conclusion is that the statement about C being symmetric is true for all square matrices A.

Apr 28, 2022

TheScienceAlliance

[MHB thread moved to the PF schoolwork forums by a PF Mentor]

For every square matrix A, C=A(A^t)+(A^t)A is symmetric.

Last edited by a moderator: Sep 11, 2022

Physics news on Phys.org

Sep 11, 2022

pasmith

Science Advisor

Homework Helper

TheScienceAlliance said:

For every square matrix A, C=A(A^t)+(A^t)A is symmetric.

Have you considered calculating the transpose of C, using the rules <br /> \begin{split}<br /> (A + B)^T &= A^T + B^T, \\<br /> (AB)^T &= B^TA^T? \end{split}

Thread 'Finding the number of ways to arrange identical balls in a circle (3 different colors)'

I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks

View full post »

Thread 'Making the denominator of a certain fraction real'

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

View full post »