I'm trying to compare the condition number of $A^TA$ if $A$ is symmetric or not. I know that the condition number of $A^T A$ is equal to $k(A)^2$, if $A$ is symmetric and invertible.
You have a proof, for example, here:
Now I was wondering what can we say about $k(A^T A)$ if $A$ is not symmetric compared to when $A$ is symmetric.
In general, if $A$ and $B$ are a square matrix (of the same dimensions) we have $$cond(AB) \leq cond(A) cond(B)$$
See here for an explanation:
So in my particular case I would have $cond(A^T A) \leq cond(A^T) cond(A)$
If $A$ is not symmetric, could in this case $cond(A^T A) \leq cond(A^T) cond(A)$ still hold because of equality or can we say for sure that we have an inequality? A proof would be nice.
Best Answer
The result still holds.
Let $A = \sum_{i=1}^{n} \sigma_i u_i v_i^T$ be the singular value decomposition of the real invertible matrix $n\times n$ matrix $A$ here $u_i$'s are orthogonal and $v_i$'s are orthogonal and $\sigma_1 \geq \sigma_2 \geq \dots \sigma_n > 0$. The condition number of $A$ is $\dfrac{\sigma_1}{\sigma_n}$.
Note $A^TA = \sum_{i=1}^{n} \sigma_i^2 v_i v_i^T$, this is a spectral and singular value decomposition of $A^TA$ and the condition number of $A^TA$ is given by $\dfrac{\sigma_1^2}{\sigma_n^2}$.
So the condition number of $A^TA$ is always the square of the condition number of $A$.