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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.

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    $\begingroup$ $$ x^T A x = x^T B^T B x = (Bx)^T (Bx) = Bx \cdot Bx = \parallel Bx \parallel^2 $$ $\endgroup$
    – Will Jagy
    Nov 21, 2013 at 6:06

1 Answer 1

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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.

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  • $\begingroup$ Thanks! I knew it must have been simple. $\endgroup$
    – Patch
    Nov 21, 2013 at 12:11

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