How to find $\textsf T^t$ on $\mathbb R^2$, equipped with the standard inner product?

Question

When given a linear operator $\textsf T: \mathbb R^2 \to \mathbb R^2$ defined by $$\textsf T(x,y) = (x+y,x+2y)$$ How can I find $\textsf T^t$ on the standard inner product space $\mathbb R^2$?

Answer 1

I suppose that by transpose of a linear operator $\textsf T$, on a inner product space $( \textsf V , \langle \cdot , \cdot \rangle )$, you mean that the linear operator $\textsf T^t$ such that $$\langle x,\textsf T^t(y) \rangle = \langle \textsf T(x),y \rangle$$ for all $x$ and $y$ on $\textsf V$.

In your concrete example, following the definition, fix $(y_1,y_2) \in \mathbb R^2$ and for all $(x_1,x_2) \in \mathbb R^2$ we have that $$\langle (x_1,x_2),\textsf{T}^t(y_1,y_2) \rangle = \langle \textsf{T}(x_1,x_2),(y_1,y_2) \rangle = \langle (x_1+x_2,x_1+2x_2),(y_1,y_2) \rangle = (x_1+x_2)y_1+(x_1+2x_2)y_2 = x_1(y_1+y_2)+x_2(y_1+2y_2) = \langle (x_1,x_2),(y_1+y_2,y_1+2y_2) \rangle$$ Thus, $$\textsf{T}^t(y_1,y_2) = (y_1+y_2,y_1+2y_2)$$ As you can see, in this particular case, $\textsf T^t = \textsf T$.