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.

  • 2
    $\begingroup$ What is the question? $\endgroup$
    – lcv
    Mar 21, 2020 at 4:56
  • $\begingroup$ What if $V$ is Banach space that is not an inner product space? $\endgroup$ Mar 21, 2020 at 4:59

1 Answer 1


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 $\lambda_v(w) = \left<v,w\right>$. 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 $V$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.