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

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, $$\|A-A^T\|_2$$.

• How did you construct $A$? Because $9.3e-16$ is pretty close to the machine epsilon. – aziiri Mar 13 at 2:16
• @aziiri This was all the information provided in the question. I think as the value is so close to machine epsilon, it is reasonable to assume that $A-A'=0$, and thus $A$ is symmetric. – M B Mar 13 at 2:18
• If the matrix is generated through some numerical procedure then there is a very good chance that the difference between $A$ and $A^T$ is due to some numerical error and it is reasonable to assume that $A=A^T$. – aziiri Mar 13 at 2:27

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.
• My thoughts exactly. I agree that $A$ is symmetric. – M B Mar 13 at 2:18
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.