Gramian matrix

**namelessguy** · November 15th 2007, 12:26 AM

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.

**kalagota** · November 15th 2007, 06:27 AM

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.

$( \implies )$
suppose $A_{nxn}$ is a Gramian Matrix.
Then there exists a matrix $B_{nxn}$ such that $A = B^T B$

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

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

**namelessguy** · November 15th 2007, 07:07 AM

Originally Posted by kalagota

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

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

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

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

**kalagota** · November 15th 2007, 07:19 AM

Originally Posted by namelessguy

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

oww, what was your definition of a symmetric matrix?

here it is.. A is symmetric if and only if $a_{ij} = a_{ji}$ 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 $A=[a_{ij}]$ is $A^T = [b_{ij}]$ where $b_{ij} = a_{ji}$

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