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

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  • $\begingroup$ How did you construct $A$? Because $9.3e-16$ is pretty close to the machine epsilon. $\endgroup$ – aziiri Mar 13 at 2:16
  • $\begingroup$ @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. $\endgroup$ – M B Mar 13 at 2:18
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    $\begingroup$ 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 $A=A^T$. $\endgroup$ – aziiri Mar 13 at 2:27
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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.

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  • $\begingroup$ My thoughts exactly. I agree that $A$ is symmetric. $\endgroup$ – M B Mar 13 at 2:18
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    $\begingroup$ As a side comment, MATLAB checks whether A - A' is equal to 0 before using a eigensolver for symmetric matrices or an eigensolver for general non-symmetric matrices. You can observe this switch by looking at the ordering of the output for eig(A). $\endgroup$ – user7440 Mar 13 at 2:20
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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.

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