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Prove that $||A^{-1}||=\frac{1}{\sigma_n}$ is the largest singular value (in singular value decomposition)

My attempt

Considering $A= U \Sigma V^H$ then $A^{-1}=V \Sigma^{-1}U^H=\sum_{j=1}^{r}\frac{1}{\sigma_i}v_iu_i^h$

Therefore the singular values are of the form

$\frac{1}{\sigma_1},\frac{1}{\sigma_2}, \cdots , \frac{1}{\sigma_n}$

but I don't know how to continue, please help me.

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We know that the singular values of $A$ are in the order $$ \sigma_1 \geq \sigma_2 \geq \cdots \geq \sigma_n. $$ It follows that the singular values of $A^{-1}$ are in the order $$ \frac{1}{\sigma_1} \leq \frac{1}{\sigma_2} \leq \cdots \leq \frac{1}{\sigma_n}. $$ It follows that $\frac 1{\sigma_n}$ is the largest singular value of $A^{-1}$, which is what we wanted.

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