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?
Is $(X^T)*X$ greater than zero if $X$ doesn't equal to zero?
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linear-algebra
1 Answers
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$X^T\cdot X$ is a sum of squares.
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0Totally. I forgot X is an eigenvalue. I was thinking X could be a matrix. Thank you. – 2012-02-13
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1@Shannon: $X$ is an eigen_vector_. An eigen_value_ is a scalar. – 2012-02-13
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0@Henning: Thanks; you beat me to it. – 2012-02-13
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0@HenningMakholm Yes, it's an eigenvector, but the result of (X^T)X is still a sum of squares. – 2012-02-13