If a real linear operator $T$ satisfies $T^{t}T = TT^{t}$, is it necessarily true that $T = T^{t}$? If $T^{t}T = TT^{t}$, does it imply that $T = T^{t}$? Here, $T$ is a linear operator on a real vector space.
 A: \begin{pmatrix}
 0&1\\
-1&0
\end{pmatrix}
fullfils your condition since the product is just the identity. But it is not symmetric.
A: No, it does not. For example, for any orthogonal matrix, $T^t T = I = T T^t$, but not all such matrices satisfy $T^t = T$.
Over $\Bbb R$, any $2 \times 2$ orthogonal matrix with positive determinant can be written as
$$T = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$$ for some angle $\theta \in \Bbb R$. Again, these satisfy $T^t T = TT^t$, but they do not satisfy $T^t = T$ unless $\theta = k \pi$ for some $k \in \Bbb Z$.
Without too much fuss, one can produce examples that are not in any of the classes of examples (symmetric, orthogonal, skew-symmetric, positive semidefinite) in xivavy's answer. For example,
$$A = \begin{pmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{pmatrix}$$
again satisfies $A^t A = A A^t$ but not $A^t = A$.
For the record, matrices $T$ that satisfy $T^t T = TT^t$ are called normal and those that satisfy $T^t = T$ are called symmetric. Symmetric matrices are always normal; the question here asks whether the converse is true, and the examples show that it is now.
A: $$T=\begin{pmatrix}\frac{1}{2}&\frac{\sqrt{3}}{2}\\-\frac{\sqrt{3}}{2}&\frac{1}{2}\end{pmatrix}$$
is a conterexample
A: No, that just means that the operator is normal. Some examples of matrices of normal operators are:


*

*Symmetric matrices $T^t=T$

*Orthogonal matrices $T^t=T^{-1}$

*Skew-symmetric matrices $T^t=-T$

*Positive semidefinite matrices $T=A^tA$.

