# Thread: Gramian matrix

1. ## Gramian matrix

Could anyone give me some help on this one problem?

An n by n matrix A is called Gramian matrix if there exists a real (square) matrix B such that A=tranpose of B times B. Prove that A is Gramian matrix if and only if A is symmetric and all of its eigenvalues are non-negative.

2. Originally Posted by namelessguy
Could anyone give me some help on this one problem?

An n by n matrix A is called Gramian matrix if there exists a real (square) matrix B such that A=tranpose of B times B. Prove that A is Gramian matrix if and only if A is symmetric and all of its eigenvalues are non-negative.
$\displaystyle ( \implies )$
suppose $\displaystyle A_{nxn}$ is a Gramian Matrix.
Then there exists a matrix $\displaystyle B_{nxn}$ such that $\displaystyle A = B^T B$

Let B be the matrix such that $\displaystyle A = B^T B$
$\displaystyle A^T = (B^T B)^T = B^T B = A$ *
hence, A is symmetric. (to show that all of its eigenvalues are non-negative, take an arbitrary symmetric matrix $\displaystyle A = B^T B$ and show that its eigenvalues are non-negative.)

* have you proven that $\displaystyle (AB)^T = B^T A^T$?

3. Originally Posted by kalagota
$\displaystyle $$\implies$$$
suppose $\displaystyle A_{nxn}$ is a Gramian Matrix.
Then there exists a matrix $\displaystyle B_{nxn}$ such that $\displaystyle A = B^T B$

Let B be the matrix such that $\displaystyle A = B^T B$
$\displaystyle A^T = (B^T B)^T = B^T B = A$ *
hence, A is symmetric. (to show that all of its eigenvalues are non-negative, take an arbitrary symmetric matrix $\displaystyle A = B^T B$ and show that its eigenvalues are non-negative.)

* have you proven that $\displaystyle (AB)^T = B^T A^T$?
Thanks for your help. Yes, we have prove that $\displaystyle (AB)^t = B^t A^t$ . I'm sorry that I'm kinda slow why if $\displaystyle A^t = A$ * then A is symmetric? I thought I had to prove $\displaystyle A^t = A$

4. Originally Posted by namelessguy
Thanks for your help. Yes, we have prove that $\displaystyle (AB)^t = B^t A^t$ . I'm sorry that I'm kinda slow why if $\displaystyle A^t = A$ * then A is symmetric? I thought I had to prove $\displaystyle A^t = A$
oww, what was your definition of a symmetric matrix?

here it is.. A is symmetric if and only if $\displaystyle a_{ij} = a_{ji}$ for all i,j..
and you will see that using that definition, and form the transpose of A, that transpose is same as A..

besides, the transpose of A, for $\displaystyle A=[a_{ij}]$ is $\displaystyle A^T = [b_{ij}]$ where $\displaystyle b_{ij} = a_{ji}$