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:

If $X$ is symmetric, show $k(X^2)$ = $k(X)^2$

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:

Condition number of a product of two matrices

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.

  • $\begingroup$ In general, we indeed have $\kappa(A^TA) = \kappa(A)^2$, assuming you mean the condition number with respect to the Euclidean norm. The only special thing with symmetric matrices is that $A^TA = A^2$. $\endgroup$ Dec 22, 2016 at 0:14

1 Answer 1


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 $A$.


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