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I am working on below statement: A is an $m \times n$ matrix, and $\lambda$ is an eigenvalue of $(A^T)A$ which eigenvector $X$ doesn't equal to zero. Show that $\lambda$ is greater than or equal to zero. I started from computing $||AX||^2$, then ended up with $\lambda\cdot (X^T)\cdot X$ is greater than or equal to zero. Then how can I know $(X^T)\cdot X$ is greater than zero?

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$X^T\cdot X$ is a sum of squares.

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Totally. I forgot X is an eigenvalue. I was thinking X could be a matrix. Thank you. – Shannon Feb 13 '12 at 0:09
@Shannon: $X$ is an eigen_vector_. An eigen_value_ is a scalar. – Henning Makholm Feb 13 '12 at 1:18
@Henning: Thanks; you beat me to it. – Brian M. Scott Feb 13 '12 at 1:19
@HenningMakholm Yes, it's an eigenvector, but the result of (X^T)X is still a sum of squares. – Shannon Feb 13 '12 at 4:18

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