Is a Matrix Symmetric if Row Space Equals Column Space?

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A matrix is not necessarily symmetric even if its row space equals its column space. The example given, the matrix \(\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}\), demonstrates this, as its transpose does not equal the original matrix. Both the row space and column space have a dimension of 2, indicating they span the same space, yet the matrix remains non-symmetric. Therefore, having equal row and column spaces does not imply symmetry. This reasoning clarifies that the condition is insufficient for establishing matrix symmetry.
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I had this question on a test and I was wondering why it is false:

If the row space equals teh column space then AT=A.
 
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Consider the matrix
\left(\begin{array}{cc}1 & 1\\0 & 1\end{array}\right).
 
Ok, so I had a rationale why it was false but I'm not sure if I am close.

Obviously here the AT does not equal A, but the dimension of the row space equals column space equals 2. Is this the right reasoning?
 
chaotixmonjuish said:
Ok, so I had a rationale why it was false but I'm not sure if I am close.

Obviously here the AT does not equal A, but the dimension of the row space equals column space equals 2. Is this the right reasoning?

Well, the row space and the column space are both R2, but the matrix is not symmetric.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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