It is currently Mon Sep 25, 2017 7:18 am


All times are UTC [ DST ]




Post new topic Reply to topic  [ 2 posts ] 
Author Message
 Post subject: A symmetric matrix
PostPosted: Mon Mar 13, 2017 9:53 pm 

Joined: Sat Nov 14, 2015 6:32 am
Posts: 126
Location: Melbourne, Australia
Let $A$ be square matrix over a field $\mathbb{F}$. If
\begin{equation}A^2 = A A^\top \end{equation}
holds , then prove that $A$ is symmetric.

_________________
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$


Top
Offline Profile  
Reply with quote  

 Post subject: Re: A symmetric matrix
PostPosted: Thu Jun 22, 2017 9:40 am 
Administrator
Administrator
User avatar

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 792
Location: Larisa
Let $A$ be an $n \times n$ square matrix over a field $\mathbb{F}$ such that

\begin{equation} A^2 =AA^{\top} \end{equation}

Taking transposed matrices back at $(1)$ we get that

\begin{align*}
\left ( A^2 \right )^\top = \left ( A A^\top \right )^\top &\Rightarrow \left ( A^\top \right )^2 = \left ( A^\top \right )^\top A^\top
\\ &\Rightarrow \left ( A^2 \right )^2 = A A^\top
\end{align*}

and thus

\begin{equation} A^2 = \left( A^\top \right)^2 \end{equation}

On the other hand it holds that

$$\left ( A A^\top - A^\top A \right )^2 = \mathbb{O}_{n \times n}$$

since

\begin{align*}
\left ( A A^\top - A^\top A \right )^2 &= \left ( A A^\top - A^\top A \right ) \left ( A A^\top - A^\top A \right ) \\
&=A A^\top A A^\top - A A^\top A^\top A - \\
&\quad \quad -A^\top A A A^\top + A^\top A A^\top A \\
&\overset{(2)}{=} \cancel{A A A A - A A AA} - \\ &\quad \quad -A^\top AA A^\top +A^\top A A^\top A\\
&=-A^\top AA A^\top +A^\top A A^\top A \\
&\overset{(2)}{=} \cancel{A^\top A^\top A^\top A^\top - A^\top A^\top A^\top A^\top} \\ &=\mathbb{O}_{n \times n}
\end{align*}

Of course it holds that if a matrix $M$ is symmetric or antisymmetric and $M^2=\mathbb{O}_{n \times n}$ then $M=\mathbb{O}_{n \times n}$. The proof is left as an exercise to the reader.

We can safely conclude using the above observations that for our matrix $A$ it holds that

\begin{equation} A A^\top = A^\top A \end{equation}

But then for the matrix $A-A^\top$ it holds that

\begin{align*}
\left ( A - A^\top \right )^2 &= \left ( A - A^\top \right ) \left ( A - A^\top \right ) \\
&=A A - A A^\top - A^\top A + A^\top A^\top \\
&\mathop {=} \limits_{(3)}^{(2)} A A - A A^\top -A A^\top + A A \\
&\overset{(2)}{=} A A - AA - AA + AA\\
&= \mathbb{O}_{n \times n}
\end{align*}

and since the matrix $A- A^\top$ is antisymmetric we conclude that $A - A^\top = \mathbb{O}_{n \times n}$ and thus $A = A^\top $. Hence the result.

_________________
Imagination is much more important than knowledge.
Image


Top
Offline Profile  
Reply with quote  

Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 2 posts ] 

All times are UTC [ DST ]


Mathimatikoi Online

Users browsing this forum: No registered users and 1 guest


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
Powered by phpBB® Forum Software © phpBB Group Color scheme created with Colorize It.
Theme created StylerBB.net