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How to show that the singular values of $A^{T}A$ are $σ_{1}^{2},...,σ_{n}^{2}$, and thus $∥A^{T}A∥_{2}$ = $σ_{max}^{2}$. In order to do this I have to find a singular vale decomposition for $A^{T}A$ ?

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    Can you work out $A^TA$, using the fact that $U$ has orthonormal columns? Then look at the result carefully. – Hans Engler Nov 13 '20 at 17:45
  • To help you further : you can compute $A^TA$ explicitely in terms of the matrices in the SVD decomposition of $A$. – nicomezi Nov 13 '20 at 17:53
  • since U has orthonormal columns would it mean that it would be $Σ^{T}ΣV^{T}V$ –  Nov 13 '20 at 17:58
  • How do you achieve this ? You have the good matrices left but the order is compltely wrong. Recall that $(CD)^T=D^TC^T$ and $CD \ne DC$ in general. – nicomezi Nov 13 '20 at 18:00
  • it would be this $A^{T}A=V(Σ^{T}Σ)V^{T}$? –  Nov 13 '20 at 18:14

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Note that $$ A^TA=(U\Sigma V^T)^T(U\Sigma V^T=V\Sigma^TU^TU\Sigma V^T=V(\Sigma^T\Sigma)V^T $$ and that $\Sigma^T\Sigma$ is a diagonal matrix with nonnegative entries on the diagonal. Thus this is a singular value decomposition of $A^TA$. The entries on the diagonal of $\Sigma^T\Sigma$ are $\sigma_i^2$, where $\sigma_1,\dots,\sigma_n$ are the singular values of $A$.

Recall that the singular value decomposition is essentially unique, in the sense that if $A=U_1\Sigma_1V_1^T=U_2\Sigma_2V_2^T$, with $\Sigma$ (pseudo)-diagonal with nonnegative entries and $U_1,U_2,V_1,V_2$ are orthogonal, then $\Sigma_1$ equals $\Sigma_2$ up to permutation of the (pseudo-)diagonal elements. The orthogonal matrices are not unique, though.

egreg
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