Is AA^T Always a Symmetric Matrix?

[kq6up]

Homework Statement

Mary Boas 3.9.5

Show that the product AA^T is a symmetric matrix.

Homework Equations

$${ \left( AB \right) }_{ ij }=\sum _{ k }^{ }{ { A }_{ ik } } { B }_{ kj }$$

The Attempt at a Solution

I do not have a solution for this one, so could someone please check my work? I never took geometry or was required to do any proofs in calculus, so go easy on me.

If $${ \left( AA \right) }_{ ij }=\sum _{ k }^{ }{ { A }_{ ik } } A_{ kj }$$, then $${ \left( A{ A }^{ T } \right) }_{ ij }=\sum _{ k }^{ }{ { A }_{ ik } } A_{ jk }$$.

∴ $${ \left( A{ A }^{ T } \right) }_{ ij }=\sum _{ k }^{ }{ { A }_{ jk } } A_{ ik }=\sum _{ k }^{ }{ { A }_{ jk } } { A }_{ ki }^{ T }={ \left( A{ A }^{ T } \right) }_{ ji }$$

Since, $$ { \left( A{ A }^{ T } \right) }_{ ij }={ \left( A{ A }^{ T } \right) }_{ ji }$$, $$A{ A }^{ T } $$ is a symmetric matrix.

 

[jbunniii]

Yes, it looks fine to me. Using exactly the same method of proof, you can show that $(XY)^T = Y^T X^T$ for any matrices $X$ and $Y$ of the appropriate size for the product to make sense. The symmetry of $AA^T$ is a special case of this result: set $X = A$ and $Y = A^T$.

 

[xiavatar]

You cannot have $A_{ki}^T$ because $A_{ki}$ is a term in the matrix. Note however that the terms commute so

$A_{jk}A_{ik}=A_{ik}A_{jk}$

$(AA^T)_{ij}=\sum_kA_{jk}A_{ik}=\sum_kA_{ik}A_{jk}=(AA^T)_{ji}$

 

[jbunniii]

You cannot have $A_{ki}^T$ because $A_{ki}$ is a term in the matrix.

I assumed he meant $(A^T)_{ki}$, the element in row $k$ and column $i$ of $A^T$, which is of course the same as $A_{ik}$.

 

[xiavatar]

Okay, then its fine.

 

[kq6up]

Yes, indeed that is what I meant, but I did not know how to express it.

Chris

 

[kq6up]

A friend of mine suggested $${ { (AA }^{ T }) }^{ T }={ A }^{ T }A$$ which is incorrect, but it led me to the MUCH more elegant proof using the "Relevant EQ" above to get $${ { (AA }^{ T }) }^{ T }={ AA }^{ T }$$. I think that does the job very nicely.

Regards,
Chris Maness

 

[D H]

$${ { (AA }^{ T }) }^{ T }={ AA }^{ T }$$. I think that does the job very nicely.

Both correct and nice. Very nicely indeed.

 

[kq6up]

Thank you for the encouragement. I have been working my butt off in this book. I started chapter 1 in the middle of February and I am only in Chapter 3, but I am doing every problem I have a solution and/or answer for. Chapter 3 is hard work because I have never had linear algebra.

Chris