$\text{cond}_2^2(M)= \text{cond}_2(M^T M)$ for non square matrix

Question

Let $M \in \mathbb{R}^{m \times n}$ be a matrix with full colum rank. Proof

$$ \text{cond}_2^2(M)= \text{cond}_2(M^T M). $$

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What I got so far:

Denote the pseudo-inverse (Moore–Penrose pseudoinverse) of $M$ with $M^+$.

$$\text{cond}_2^2(M) = ||M^{+}||_2^2 \cdot||M||_2^2 = \lambda_{\text{max}} ((M^+)^T M^+) \cdot \lambda_{\text{max}}(M^T M)$$

where $\lambda_{\text{max}}(\cdot)$ denotes the largest eigenvalue of the matrix. We also have

$$\text{cond}_2(M^T M) = ||(M^T M)^{-1}||_2 \cdot ||M^T M||_2 \\= \left( \lambda_{\text{max}}(((M^T M)^{-1})^T(M^T M)^{-1}) \cdot \lambda_{\text{max}}((M^T M)^T (M^T M)) \right)^{\frac{1}{2}} \\ = \lambda_{\text{max}}((M^T M)^{-1}) \cdot \lambda_{\text{max}}(M^T M)$$

So, if we compare the two equations, we see the right hand side is the same, hence we are left to show the following identity.

$$ ||M^+||_2^2 = ||(M^TM)^{-1}||_2 $$

or

$$\lambda_{\text{max}} ((M^+)^T M^+) = \lambda_{\text{max}}((M^T M)^{-1}) $$

But I have no idea how to show this identity. Any help would be greatly appreciated.

Answer 1

If $M$ has full column rank, then $M^+=(M^TM)^{-1}M^T$. So $$ \begin{split} \|M^+\|_2^2&=\lambda_\max((M^+)^TM^+) \\&=\lambda_\max(M^+(M^+)^T) \\&=\lambda_\max((M^TM)^{-1}M^TM(M^TM)^{-1}) \\&=\lambda_\max((M^TM)^{-1}). \end{split} $$