Hi I am struggling to understand the following step in the explanation around singular values and SVD.

This is taken from the book "Linear Algebra - A modern Introduction" p613:

For any $m \times n$ matrix $A$, the $n \times n$ matrix $A^TA$ is symmetric and hence can be orthogonally diagonalized, by the Spectral Theorem. The eigenvalues of $A^TA$ are real, nonnegative. To show this, let $\lambda$ be an eigenvalue of $A^TA$ with corresponding unit eigenvector $\bf v$ then:

Step (1) $ \ \ \ \ 0 \leq ||Av||^2 = (Av)\cdot(Av) = (Av)^TAv = v^TA^TAV $

Step(2) $ \ \ \ \ = v^T \lambda v = \lambda (v \cdot v) = \lambda ||v||^2 = \lambda$

I don't see how to go from step (1) to step (2) in particular from $v^TA^TAv = v^T \lambda v $

--Edit-- thanks for the quick answer

Indeed if I let $A^TA = S$ knowing that $v$ is an eigenvector of $S$ and using the definition: $Sv = \lambda v$ this gives:

$v^TA^TAv = v^TSv=v^T \lambda v$


$v$ is an eigenvector of the matrix $A^TA$, therefore $$v^TA^TAv=v^T((A^TA)v) = v^T\lambda v$$


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