A linear operator $T$ on an inner product space $V$ is said to have an adjoint operator $T^*$ on $V$ if $⟨T(u),v⟩=⟨u,T^*(v)⟩$ for every $u,v\in V$

I know how to proof "why this operator exist$(\text{Uniquely})$" using the fact that

If $V$ be a finite dimentional inner product space and $f$ be a linear functional on $V$ then there exist a unique $y\in V$ such that $f(x)=⟨x,y⟩,\forall x\in V$

But $\color{red}{\text{I don't have an intuitive understanding of adjoint operator}}$. What actually it does$?$ And is there any relation between $T^*$ with the transpose$(\text{or dual})$
Thanks for your time. Thanks in advance .

  • $\begingroup$ A Hilbert space is just a complete inner product space. If you are working with finite-dimensional spaces, then all inner product spaces are Hilbert spaces. $\endgroup$
    – Math1000
    Commented Nov 11, 2019 at 20:13
  • $\begingroup$ It is also possible to define an adjoint for operators on Banach spaces (complete normed vector spaces without an inner product). Are you interested in that case, or only for inner product spaces? $\endgroup$
    – Math1000
    Commented Nov 11, 2019 at 20:16
  • $\begingroup$ I am only interested in inner product space because I am not introduced with those stuff$(\text{Hilbert spaces, Banach spaces})$ yet @Math1000 Sir $\endgroup$
    – emonHR
    Commented Nov 12, 2019 at 3:53
  • $\begingroup$ @Math1000 Sir I am waiting for your response. If you need any farther query please ask me. I really stuck on this topic for a while Sir. $\endgroup$
    – emonHR
    Commented Nov 12, 2019 at 18:18
  • $\begingroup$ I don't have a great intuition for the notion of an adjoint operator, sorry. There is a question with some good answers on Math Overflow though: mathoverflow.net/questions/6552/… $\endgroup$
    – Math1000
    Commented Nov 12, 2019 at 18:20

1 Answer 1


Let me first recall the definition of the transpose. Let $T: \ \ X \longrightarrow Y$ be a linear operator from $X$ to $Y$. The dual $X^*$ is the space of (continuous) linear functional on $X$, i.e., linear operators from $X \longrightarrow \mathbb{C}$. So if $x\in X^*, \ v \in X$, $x(v) \in \mathbb{C}$. We may even denote the composition of $x$ with $v$ as

$$ x(v):= (x,v). \ \ \quad \ \ (0) $$

Now the transpose of $T$ is the linear operator $ T^t : Y^* \longrightarrow X^*$ defined as

$$ (y, Tv) = ( T^t y, v) \ \ \quad \ \ (1) $$

for all $ v\in X, \ y \in Y^*$. Take now bases of $X,Y$ (say $e_i, \ f_j$) and natural bases of $X^*,Y^*$ (satisfying $ (\eta_i , e_j) = \delta_{i,j}$ and so on). Assume $\dim X = n$ and $\dim Y = m$. So for example $(0)$ becomes

$$ (x,v) = \sum_{i} x_i v_i \ \ \quad \ \ (2) $$

The matrix of $T$ in these bases is defined as

$$ Te_i = \sum_{j=1}^m T_{j,i} f_j $$

(for $i=1,2,\ldots, n$). Using the convention above we have

$$ Tv = T\sum_i v_i e_i = \sum_{i,j} v_i T_{j,i} f_j $$

so that the components of $(Tv)_k$ are given by $ (Tv)_k = \sum_i T_{k,i} v_i$. These are the usual matrix-vector multiplication rules.

You can see that the transpose is associated to the matrix $(T^t)_{i,j}=T_{j,i}$. Note that $T$ corresponds to a $m\times n$ matrix.

However what ultimately allows this to work, or alternatively, what makes definition $(1)$ works, is the following simple fact.

Once you have a matrix of $m\times n$ numbers $T_{i,j}$, you can form $n$-dimensional vectors via $\sum_{j=1}^m T_{k,j} v_j$ ($k=1,\ldots n$) but also $m$-dimensional vectors via $\sum_{i=1}^n T_{i,k} w_i$ ($k=1,\ldots m$).

The latter operation corresponds (using the usual matrix-vector multiplications rules) to

$$ (w_1,w_2, \ldots w_m) \left(\begin{array}{cccc} T_{1,1} & T_{1,2} & \cdots & T_{1,n}\\ T_{2,1}\\ \vdots\\ T_{m,1} & & \cdots & T_{m,n} \end{array}\right) $$

according to the usual rules (row-vector times matrix).

To form the adjoint we proceed in a very similar way. However now we use the scalar product to put vectors in correspondence with linear functionals. That is, instead of $(2)$ we use the scalar product $\langle \bullet, \bullet \rangle$:

$$ \langle x, v \rangle = \sum_{j=1}^n x_j^* v_j $$

Then the construction of the adjoint is the same as that for the transpose, but you get the additional complex conjugate terms.

In component of course, the matrix associated to the adjoint satisfies:

$$ (T^*)_{i,j} = T_{j,i}^* $$

  • 2
    $\begingroup$ I still didn't find What actually it does? @lcv but thanks for your response $\endgroup$
    – emonHR
    Commented Nov 29, 2019 at 19:42
  • $\begingroup$ @emonHR Is It clearer now? $\endgroup$
    – lcv
    Commented Nov 30, 2019 at 19:22
  • $\begingroup$ $(1)\textbf{By defining a basis on V correspondence with }n^2\textbf{ numbers}$ I didn't get that part. <br>$(2)\textbf{The operator that..essentially the transpose}.$ But what I mean by transpose is if $f:V\rightarrow W$ is a linear map then the transpose is defined by ${}^tf:W^*\rightarrow V^*$. Then how $(x_1,x_2,\cdots x_n)$ is equivalent with linear function$?$ I never saw any transformation with left multiplication even I don't know what's it's meaning$?$ $\endgroup$
    – emonHR
    Commented Dec 1, 2019 at 6:33
  • 1
    $\begingroup$ I think if that part is cleared then all of your telling will be meaningful to me.<br>Sorry for asking too many things but I really want to understand your answer because it teach me something interesting. Thank you again(+1) @lch $\endgroup$
    – emonHR
    Commented Dec 1, 2019 at 6:33
  • 1
    $\begingroup$ I'll be adding some details. By the way, I think it is a good question. $\endgroup$
    – lcv
    Commented Dec 2, 2019 at 16:35

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