Let us define a relation $\rho$ on the set $A=\{1,2,3 \} $ by $ \rho=\{(1,1)\}$. Is the relation symmetric, transitive or not?
I have confusion abut symmetry or transitive. If $(a, b) \in \rho\implies (b, a) \in \rho$. Then $\rho$ is symmetric. It seems that this relation is symmetric. Am I correct?
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Yes, the relation is symmetric. You can construct a 3-by-3 matrix $A$ of zeros and ones, where $A_{ij} = 1$ if $(i,j)$ belongs to the relation, and $0$ otherwise. Thus, of the 9 entries in the matrix, only $A_{11}=1$, and the remaining eight entries are zeros. This is the matrix representation of the relation.
A relation is symmetric if and only if its matrix is equal to its tranpose (i.e. when you make the rows of the matrix its columns, and vice versa, you must get back the same matrix). This means if any off-diagonal element (say, above the main diagonal) is a 1, then its mirror image (below the main diagonal) must also be a 1.
Your definition for symmetry is correct; and indeed the relation you gave is symmetric. But if you are ever confused about the properties like symmetry and transitivity, you may interpret their definitions like "as long as there is no element (or two elements in transitivity case) that does not break symmetry (or transitivity), that relation is symmetric (or transitive)".
As an example, $\{(1,2),(2,1),(1,3)\}$ is not symmetric because non-existence of $(3,1)$ breaks it. Or $\{(1,2),(2,1),(1,1),(1,3)\}$ is not transitive because when we consider $(2,1),(1,3)$ notice that $(2,3)$ is not in the relation so non-existence of $(2,3)$ breaks it. In your example, $\{(1,1)\}$ is both symmetric and transitive since there is no element that breaks those properties.
Long story short, you can interpret the definitions of symmetry and transitivity by not considering existence of the elements in relation but non-existence of the elements that would satisfy those properties if they were in the relation.