What is common between adjoint operator and transpose of the matrix? I am confused. 
What is Connection between Adjoint Operator and transpose of the Matrix? 
I will be very grateful if someone can help me to clarify it.
 A: A linear operator $L : X\rightarrow X$ on a linear space $X$ over a field $\mathcal{F}$ does not require the notion of a basis in order to be defined. Such an $L$ is a function with the property that
$$
  L(\alpha x+\beta y ) =\alpha L(x)+\beta L(y),\;\;\; x,y \in X,
        \;\;\alpha,\beta \in \mathcal{F}.
$$
If $X$ is finite-dimensional, and if $\{ v_{j}\}_{j=1}^{N}$ is a basis for $X$, then you end up with a matrix representation of $L$ given by
$$
         [Lx]_{v} = [L]_{v}[x]_{v}
$$
Then you can define a new linear transformation $M_{v} : X\rightarrow X$ such that
$$
           [M_{v}x]_{v} = [L]_{v}^{T}[x]_{v},
$$
where $[L]_{v}^{T}$ is the transpose of $[L]$. A subscript of $v$ on the operator $M_{v}$ is necessary because the resulting linear operator definitely depends on the choice of basis; it is not uniquely determined by the linear operator $L$.
The adjoint of $L$ is also something which also requires a subscript, and it requires further specialization to the field $\mathcal{F}$ of real or complex numbers. For your situation, assume $\mathcal{F}$ is the field of real numbers. The general concept of adjoint in this case requires you to consider the dual space $X^{\star}$ consisting of linear functions $x^{\star} : X\rightarrow\mathbb{R}$. The adjoint $L^{\star} : X^{\star}\rightarrow X^{\star}$ maps $x^{\star}$ to $x^{\star}\circ L$. That is, $L^{\star}x^{\star}=x^{\star}\circ L$.
A connection between adjoint and transpose is found when choosing a basis $\{ v_{n}^{\star}\}_{n=1}^{N}$ for $X^{\star}$ such that $v_{j}^{\star}(v_{k})=\delta_{j,k}$. This is the so-called dual basis of $\{ v_{j}\}_{j=1}^{k}$ for $X^{\star}$. In this basis, notice that $L^{\star}v_{j}^{\star}$ is defined by its action on the basis $\{ v_{j}\}_{j=1}^{k}$, which is easily found from the definition of $L^{\star}$.
If $[L]_{v}=[l_{j,k}]$, then $Lv_{k}=l_{1,k}v_{1}+\cdots+l_{N,k}v_{N}$ and
$$
        (L^{\star}v_{j}^{\star})(v_{k})=(v_{j}^{\star}\circ L)(v_{k})=v_{j}^{\star}(Lv_{k})=v_{j}^{\star}(l_{1,k}v_{1}+\cdots l_{N,k}v_{N})=l_{j,k}.
$$
In other words, $L^{\star}v_{j}^{\star}$ is determined by its action on the basis $\{ v_{j}^{\star}\}$ to be 
$$
      L^{\star}v_{j}=l_{j,1}v_{1}^{\star}+\cdots+l_{j,N}v_{N}^{\star}.
$$
So the representation of $L^{\star}$ with respect to $\{ v_{j}^{\star}\}$ is the transpose of the matrix representation of $L$ with respect to $\{ v_{j}\}$. That is,
$$
                [L^{\star}]_{v^{\star}}=[L]_{v}^{T}.
$$
But this holds only if $v^{\star}$ is the dual basis of $v$.
If you have a real inner-product, then the dual basis $\{ e_{n}^{\star}\}_{n=1}^{N}$ of an orthonormal basis $\{ e_{n}\}_{n=1}^{N}$ is identified with the original basis because $e_{j}^{\star}(x)=(x,e_{j})$ does define $e_{j}^{\star} \in X^{\star}$ such that $e_{j}^{\star}(e_{k})=\delta_{j,k}$.
A: Suppose a real operator $T:\Bbb{R}^n \to \Bbb{R}^m$ is given by
$$
T(x) = Ax
$$
Where $x$ is a column vector and $A$ is an $m \times n$ matrix.  The adjoint of this operator will be given by
$$
T^*(y) = A^Ty
$$
A: Let $f$ an endomorphism in an euclidean  space $(E,\langle,\rangle)$ and let $A$ the matrix that represents $f$ relative to an orthonormal basis $\mathcal B$. Let $x,y\in E$ and $X=[x]_{\mathcal B}, Y=[y]_{\mathcal B}$ and $A^*$ the matrix of the adjoint  $f^*$ of  $f$ in $\mathcal B$ then
$$\langle f(x),y\rangle=\langle x,f^*(y)\rangle\iff(AX)^TY=X^TA^TY=X^TA^*Y$$
and the last equality imply:
$$A^*=A^T$$
