I have the following exercise:
Relate the 2-norm condition of $X\in \Bbb R^{m\times n}\ (m\geq n)$ to the 2-norm condition of the matrices: $$B=\begin{equation}
\begin{bmatrix}
I_m & X\\
0 & I_n
\end{bmatrix}
\end{equation}$$ and $$C=\begin{equation}
\begin{bmatrix}
X\\
I_n
\end{bmatrix}
\end{equation} $$
My attemp:
We know that the condition number of a matrix for the 2-norm is given by the following expression:
$$\kappa_2(A)=||A||_2 ||A^{-1}||_2$$
Also I have a property that says if $B$ is a submatrix of $A$ then
$$||B||_2\leq ||A||_2$$
So using this I have:
$$\kappa_2(B)=||B||_2 ||B^{-1}||_2\geq ||X||_2||X^{-1}||_2=\kappa_2(X)$$
But I am not sure that it is necessarily true that if $X$ is a submatrix of $B$ then necessarily $ X^{-1} $ is a submatrix of $B^{-1}$
If this works then for the $ C $ it would be worth the same, but doing numerical experiments this fails, numerically I get:
$$\kappa_2(C)\leq \kappa_2(X) \leq \kappa_2(B)$$
I appreciate any help you can give me, regards
Edit:
I have seen that $X$ is not square so the inverse does not exist and therefore kappa will not exist either. For this, in Golub's book it is said that it can also be calculated as the maximum singular value between the minimum singular value, that is:
$$\kappa_2(A)=\dfrac{\sigma_{max}}{\sigma_{min}}$$
And in this case, as the matrix $X$ is given, for any matrix the 2-norm condition can be obtained, applying the singular value decomposition theorem
Best Answer
$$C=\begin{bmatrix} X \\ I_n\end{bmatrix}$$
$$C^TC=X^TX+I$$
Hence $$\lambda_{\max}(C^TC)=\lambda_{\max}(X^TX)+1$$ $$\lambda_{\min}(C^TC)=\lambda_{\min}(X^TX)+1$$
That is $$\kappa(C)=\frac{\sigma_{\max}(C)}{\sigma_{\min}(C)}=\sqrt{\frac{\lambda_{\max}(X^TX)+1}{\lambda_{\min}(X^TX)+1}}\le \sqrt{\frac{\lambda_{\max}(X^TX)}{\lambda_{\min}(X^TX)}}=\kappa(X)$$