Determing if a matrix is symmetric by knowing norm$(A-A^T)$

Question

Consider the linear system $Ax = b$, where $A$ is known and you are given the following Matlab outputs:

norm(A-A') = 9.3e-16
min(eig(A)) = 1.3e-7 
max(eig(A)) = 2.7e+6 

I am wondering if it possible to determine if the matrix $A$ is symmetric.

A square matrix, $A$, is symmetric iff $A=A^T$ (where there transpose of a matrix $A$ is denoted by $A'$ in the Matlab outputs). If $A$ is symmetric, we expect $A-A'=0$. In the Matlab output, the norm used is the $2$-norm by default.

I am wondering what can be deduced by knowing norm$(A-A')$, or more conventionally, $\|A-A^T\|_2$.

Answer 1

If the exact value of $\|A - A^\top\|$ is not zero, then $A$ is not exactly equal to $A^\top$ and so $A$ is not perfectly symmetric.

But, the fact that $\|A - A^\top\|$ is very small ($9.3 \times 10^{-16}$) suggests that $A$ is very close to being equal to $A^\top$. Presuming that the norm of $A$ is much larger, it would be fair to guess that this is just numerical error, and that $A$ is indeed symmetric. This is not a proof, but merely numerical evidence of this hypothesis.

Answer 2

It is best to use all exercise data. $A$ is close to a symmetric matrix $S$ that has the same spectrum as $A$. Since we consider the $2$-norm and the transpose, we may assume that $S$ is diagonal ($S=diag(1.3.10^{-7},\cdots,2.7.10^6$)) and $A=S+N$, where $N=[n_{i,j}]$ is any small matrix. $||A-A^T||_2=\rho(A-A^T)=\rho(N-N^T)=r\approx 10^{-15}$.

In general, $||N||_2\approx r$ and, about the eigenvalues, $|\lambda(A)-\lambda(S)|=|\Delta\lambda|\approx r$ is negligible compared to $\lambda(A)$ (since $r$ is negligible compared to $1.3.10^{−7}$).

Now, we can conclude that $A$ is almost symmetric.