# What is the motivation behind the definition of adjoint of a linear operator?

Given $$T:V\to V$$ linear and $$V$$ being an inner product space, we define $$T^*$$ by a linear operator on $$V$$ such that $$\langle Tx,y\rangle=\langle x,T^*y\rangle$$ for each $$x,y\in V$$.

We later see that, for finite-dimensional inner product spaces, adjoint exists and, in fact, whenever an adjoint exists is unique.

For finite-dimensional space we can draw a correspondence between conjugate transpose of a matrix and adjoint of a linear operator, taking into consideration an orthonormal basis of $$V$$ and representing $$T$$ with respect to that basis.

Now I think the motivation behind such a definition is finding a linear transformation version of the conjugate transpose. We know that if $$T$$ is a linear operator on inner product space $$V$$, and $$\beta$$ be an orthonormal basis of $$V$$, then the corresponding matrix $$A_{ij}=\langle T\alpha_j,\alpha_i\rangle$$.

So, naturally if $$B$$ is the conjugate transpose of $$A$$, then $$B_{ij}=\overline{ A}_{ji}=\overline{\langle T\alpha_i,\alpha_j\rangle}$$. So, naturally, a question arises if there exists $$U$$ linear on $$V$$ such that $$B$$ is the matrix of $$U$$ with respect to the given orthonormal basis. Then $$B_{ij}=\langle T^*\alpha_j,\alpha_i\rangle=\overline{\langle T\alpha_i,\alpha_j\rangle}=\overline {A}_{ji}$$.

This, I think, motivated the definition of adjoint.

• What is the question?
– lcv
Mar 21, 2020 at 4:56
• What if $V$ is Banach space that is not an inner product space? Mar 21, 2020 at 4:59

A slightly different, but essentially equivalent, way of thinking of it is that the adjoint is the linear operator on the dual space $$V^*$$ induced by $$T$$.
Consider the dual space $$V^*$$, consisting of linear maps $$V \to \Bbb{R}$$. Then there is a map $$T^*$$ on $$V^*$$ defined as follows: for $$\lambda \in V^*$$, $$T^*(\lambda)$$ is the linear functional given by: $$\Big( T^*(\lambda) \Big)(v) = \lambda \Big( T(v) \Big)$$ If you choose a basis for $$V$$, for which $$A$$ is the matrix of $$T$$ in that basis, then the adjoint matrix (the transpose) is the matrix of $$T^*$$ in the dual basis.
The connection with your explanation is this: if you have a nondegenerate inner product on $$V$$ and $$V$$ is finite-dimensional, then there is an isomorphism between $$V$$ and $$V^*$$ using the inner product. For a vector $$v$$, define a linear functional $$\lambda_v$$ by the formula $$\lambda_v(w) = \left$$. If you use this to identify $$V$$ and $$V^*$$, (and if the basis is orthonormal), then the basis and the dual basis are the same, and you can think of $$T^*$$ as an operator on $$V$$.