What is the SVD of $A^{-1}$? Let $A\in R^{n\times n}$ with full SVD $U\Sigma V^T$ where $U$ and $V$ are orthogonal $n\times n$ matrices and $\Sigma$ is an $n\times n$ diagonal matrix with entries $\sigma_1 \geq\cdots\geq \sigma_n \geq 0$.
1- What is the SVD of $A^{-1}$ ?
2- Given that ||A|| = $\sigma_1$, how would we express ||$A^{-1}$|| in terms of the singular values of A?
3- What is the condition number of A?
 A: Hint:
If $A = U\Sigma V^T$, then 
$$
A^{-1} = (U\Sigma V^T)^{-1} = 
(V^T)^{-1} \Sigma^{-1} U^{-1}
$$
Keep in mind that if $A$ is invertible, then $\sigma_i > 0$ for each $i$. For 3, you should find that the condition number for $A^{-1}$ is identical to that of $A$ under the norm $\|A\| = \sigma_1(A)$.
A: The answer is posed as a special case of a general problem.
$(1)$ Singular Value Decomposition
Every matrix
$$
\mathbf{A} \in \mathbb{C}^{m\times n}_{\rho}
$$
has a singular value decomposition
$$
\begin{align}
  \mathbf{A} &=
  \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\
%
 &=
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S}_{\rho\times \rho} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
% V* 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
  \end{array} \right]  \\
%
%
 &=
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cccc|cc}
     \sigma_{1} & 0 & \dots &  &   & \dots &  0 \\
     0 & \sigma_{2}  \\
     \vdots && \ddots \\
       & & & \sigma_{\rho} \\\hline
       & & & & 0 & \\
     \vdots &&&&&\ddots \\
     0 & & &   &   &  & 0 \\
  \end{array} \right]
% V 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
  \end{array} \right]  \\
%
  & =
% U
   \left[ \begin{array}{cccccccc}
    \color{blue}{u_{1}} & \dots & \color{blue}{u_{\rho}} & \color{red}{u_{\rho+1}} & \dots & \color{red}{u_{n}}
  \end{array} \right]
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
% V
   \left[ \begin{array}{c}
    \color{blue}{v_{1}^{*}} \\ 
    \vdots \\
    \color{blue}{v_{\rho}^{*}} \\
    \color{red}{v_{\rho+1}^{*}} \\
    \vdots \\ 
    \color{red}{v_{n}^{*}}
  \end{array} \right]
%
\end{align}
$$
The $\rho$ singular values are ordered and satisfy
$$
  \sigma_{1} \ge \sigma_{2} \ge \dots \sigma_{\rho} > 0
$$
The column vectors are orthonormal basis vectors:
$$
\begin{align} 
% R A
\color{blue}{\mathcal{R} \left( \mathbf{A} \right)} &=
\text{span} \left\{
 \color{blue}{u_{1}}, \dots , \color{blue}{u_{\rho}}
\right\} \\
% R A*
\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} &=
\text{span} \left\{
 \color{blue}{v_{1}}, \dots , \color{blue}{v_{\rho}}
\right\} \\
% N A*
\color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)} &=
\text{span} \left\{
\color{red}{u_{\rho+1}}, \dots , \color{red}{u_{m}}
\right\} \\
% N A
\color{red}{\mathcal{N} \left( \mathbf{A} \right)} &=
\text{span} \left\{
\color{red}{v_{\rho+1}}, \dots , \color{red}{v_{n}}
\right\} \\
%
\end{align}
$$
Moore-Penrose Pseudoinverse Matrix
$$
\begin{align}
  \mathbf{A}^{+} &= \mathbf{V} \, \Sigma^{+} \mathbf{U}^{*} \\
%
&=
% V
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{V}_{\mathcal{R}}} & 
     \color{red}{\mathbf{V}_{\mathcal{N}}}
  \end{array} \right] 
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S}^{-1} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
%
% U
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{U}_{\mathcal{N}}}^{*}
  \end{array} \right]  \\
%
\end{align}
%
$$
Special Case: $m=n=\rho$
Your question is about the special case of a square matrix with full rank. In this instance the pseudoinverse is equivalent to the classic inverse:
$$
\mathbf{A}^{+} = \mathbf{A}^{-1}
Both null spaces are trivial:
$$
  \color{red}{\mathcal{N} \left( \mathbf{A} \right)} = \color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)} = \mathbf{0}
$$
The SVD is
$$
\mathbf{A} = 
%
  \color{blue}{\mathbf{U}_{\mathcal{R}}}\,  
% Sigma
  \mathbf{S} \,
% V
  \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*}  
%
$$
and the inverse is
$$
\mathbf{A}^{+} = 
%
  \color{blue}{\mathbf{V}_{\mathcal{R}}}\,  
% Sigma
  \mathbf{S}^{-1} \,
% V
  \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*}  
%
=   \left( \color{blue}{\mathbf{U}_{\mathcal{R}}}\,  
% Sigma
  \mathbf{S} \,
% V
  \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*}  \right)^{-1}
%
= \mathbf{A}^{-1}
%
$$

More about classification of matrix inverses: [What forms does the Moore-Penrose inverse take under systems with full rank, full column rank, and full row rank?][1]

## $(2)$ Matrix Norm
$$
 \lVert \mathbf{A} \rVert_{2} = \sigma_{1}, 
\qquad \Rightarrow \qquad
 \lVert \mathbf{A}^{-1} \rVert_{2} = \frac{1}{\sigma_{\rho}}
$$
$(3)$ Condition Number
$$
 \kappa_{p}  = \lVert \mathbf{A}^{-1} \rVert_{p} \lVert \mathbf{A}_{p} \rVert
  \qquad \Rightarrow \qquad
 \kappa_{2}  = \lVert \mathbf{A}^{-1} \rVert_{2} \lVert \mathbf{A} \rVert_{2} = \frac{\sigma_{1}}{\sigma_{\rho}}
$$
