5
$\begingroup$

I'm trying to write up solutions for my Linear Algebra students, and I can't seem to figure out this one part of a proof. They are told that a matrix $A$ is called Gramian if $$A=B^tB$$ for some real, square matrix $B$.

They are then asked to prove that $A$ is symmetric (trivial) and that all of its eigenvalues are non-negative (which is the part I'm stuck on).

I've tried looking up properties about Gramian matrices, but everything mentioned relates them to positive semidefinite matrices, which my students have not read anything about. I also know that $B$ and $B^t$ have the same eigenvalues, but I don't think they would have to have the same eigenvectors.

Maybe I'm just tired, but any help proving this without mentioning "positive semidefinite" is much appreciated.

$\endgroup$
  • 1
    $\begingroup$ $$ x^T A x = x^T B^T B x = (Bx)^T (Bx) = Bx \cdot Bx = \parallel Bx \parallel^2 $$ $\endgroup$ – Will Jagy Nov 21 '13 at 6:06
4
$\begingroup$

Use the inner product. For an eigenvalue $\lambda$ and eigenvector $x$ of $B^*B$

$\lambda ||x||^2=\langle B^*Bx,x\rangle=||Bx||^2$.

Hence $\lambda$ is real and nonnegative.

$\endgroup$
  • $\begingroup$ Thanks! I knew it must have been simple. $\endgroup$ – Patch Nov 21 '13 at 12:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.