Disambiguation and Clarification of Adjoint Operators

Question

I want talk about something that has been bugging me today regarding adjoint operators on finite dimensional inner product spaces. So here has been my understanding, let $T$ be an operator on $V$; $V$ a finite dimensional inner product space over $\mathbb{C}$ or $\mathbb{R}$ .

Now if $V$ $=$ $\mathbb{C}$$^{n}$ or $\mathbb{R}$$^{n}$, using the standard inner product defined by

$\langle$$w$ ,$u$ $\rangle$ = $\sum_{i} w_i\bar u_i$

we can show that if $M$ is the matrix of $T$ with respect to the standard orthonormal basis of $\mathbb{C}$$^{n}$ then the matrix of the adjoint of $T$, $M^*$ must be equal to the conjugate transpose of $M$. In particular, we have the following scenario:

$\langle$$Tv$,$u$$\rangle$ is equivalent to $\langle$$M$v$,$u$\rangle$ where $M$=($a_i$$_j$) is the matrix of T with respect to the standard orthonormal basis. Assuming $\langle$$M$v$,$u$\rangle$ = $\langle$v$,$$M$u$\rangle$ for some $v$ $=$ ($v_1$,...,$v_n$), $u$ $=$ ($u_1$,...,$u_n$) $\in$ $V$. We find that

$\sum_{i} (Mv)_i\bar u_i$ $=$ $\sum_{j}$$\sum_{k} a_j{_k} v_k \bar u_j$ and $\sum_{i} v_i \bar{(Mu)_i}$ $=$ $\sum_{j}$$\sum_{k} \bar a_k{_j} v_k \bar u_j$

And so since $\langle$$M$v$,$u$\rangle$ = $\langle$v$,$$M$u$\rangle$ must hold for all $v$ and $u$, it must be true the case that $M^*$ is equal to the conjugate transpose of $M$.

But questions arise. In the case of $V$ $=$ $\mathbb{C}$$^{n}$ or $\mathbb{R}$$^{n}$, I chose a particular inner product. Is it the case that this inner product only "makes sense" when we use an orthonormal basis? I mean, I believe that the axioms of the inner product will be satisfied no matter which basis we choose since we're just comparing lists of numbers here. On the other hand, every inner product induces a norm, so if we are particularly perverse in our choice of a basis, e.g., consider a collection of $n$-tuples consisting numbers which are both very large and very small in absolute value, it seems that all sorts of weird things would begin to happen with our norm.

So is it that because we used an inner product that requires an orthonormal basis to be used that we were able to conclude $M^*$ is the conjugate transpose of $M$? Or is there some more pure linear algebraic thing happening here? What if we used a different inner product on $V$? What if we consider a different inner product space. With an orthonormal basis, will the adjoint of $M$ still be equal to the conjugate transpose?

Answer 1

The fact that $M^*=\overline {M ^\top}$ doesn't depend on the inner product. In addition, it doesn't rely on the input and output basis being the same. To see this, suppose that $e_1,\dots,e_n$ and $f_1,\dots,f_n$ are orthonormal bases and let $M$ be the matrix of $T$ with respect to these bases and let $M^*$ be the matrix of the adjoint with respect to the $f$s as input and $e$s for output. Then $Te_k=\newcommand\ip[2]{\left\langle #1, #2\right\rangle}\ip{Te_k}{f_1}f_1+\dots+\ip{Te_k}{f_n}f_n$ and so the $j$th row and the $k$th column of $M$ is $\ip{ Te_k}{f_j}$. A similar argument shows that the entry at the $j$th row and $k$th column of $M^*$ is $\ip{T^*f_k}{e_j}=\ip{f_k}{Te_j}=\overline{\ip{Te_j}{f_k}}$ which is the complex conjugate of the $k$th row and $j$th column of $M$ and so $M^*=\overline {M ^\top}$.

However, you also ask if this will hold if the bases aren't orthonormal and in fact, the result doesn't always hold if the basis isn't orthonormal. For example, let's use $(2,0)$ and $(0,.5)$ as a basis for $\Bbb C^2$. If $T(x,y)=(y,.5x)$ Then $T^*(x,y)=(.5y,x)$ (which you can check independently). Then $M$ is $\begin{bmatrix}0 & .25 \\ 2 & 0\end{bmatrix}$ However, $M^*=\begin{bmatrix}0 & .125\\4 & 0\end{bmatrix}$.

Answer 2

Your argument shows (or can be trivially generalized to show) that for any pair of IPS's $V$ and $W$, with respect to any orthonormal bases on $V$ and $W$, the matrix of $T^*:W\to V$ is the conjugate transpose of the matrix for $T:V\to W$ for any linear transformation $T$ from $V$ to $W$. Which actually answers some of your questions I think, as in no the inner product doesn't matter, although $T^*$ will change if you change the inner product. Actually I don't understand your questions entirely. In fact there are entire paragraphs of your question that don't make a whole lot of sense to me.

Nonetheless given this set up, I think your questions are the following: Why does this generalize to any inner product space, and is it necessary to have an inner product space, or is there something more linear algebraic going on here, and also is it necessary to choose an orthonormal basis rather than a general basis?

So let's begin with the most general form of this theorem.

Let $V$ and $W$ be vector spaces over a field $k$ with $v_1,\ldots,v_n$ a basis for $V$ and $w_1,\ldots,w_m$ a basis for $W$. Let $T:V\to W$ be a linear transformation. Then $T$ induces a map $T^*:W^*\to V^*$ defined by $\omega\mapsto \omega \circ T$. The matrix for $T$ with respect to the bases on $V$ and $W$ is the transpose of the matrix for $T^*$ with respect to the dual bases $w_1^*,\ldots,w_m^*$ and $v_1^*,\ldots,v_n^*$ of the bases on $W$ and $V$.

Proof:

The proof is the same as that given in your question. We use $\langle\alpha,v\rangle$ to denote evaluation of a linear functional $\alpha$ at a vector $v$ for simplicity of notation. Then the $i,j$th component of the matrix for $T$ is essentially by definition $\langle w_i^*,Tv_j\rangle$ ($w_i^*$ picks out the coefficient of $w_i$ in the expansion of $T(v_j)$ in terms of the basis on $W$). However $\langle w_i^*,Tv_j\rangle=\langle T^*w_i^*,v_j\rangle$ by definition of $T^*$, and $\langle T^*w_i^*,v_j\rangle$ is the coefficient of $v_j^*$ in the expansion of $T^*w_i^*$, which is the $j,i$th component of the matrix for $T^*$ with respect to the dual bases. Hence the matrices for $T$ and $T^*$ are transposes of each other. $\blacksquare$

How is this result related?

Now how does this relate to inner product spaces? Well, first let's consider the case where we have a nondegenerate bilinear form. What is a nondegenerate bilinear form (on a finite dimensional vector space)? It's an isomorphism between a vector space and its dual space. Thus if $V$ and $W$ have nondegenerate bilinear forms on them $B_V$ and $B_W$, we can identify $T^*$ with a map from $W$ to $V$ and the same theorem as above will hold for the matrix of this map, as long as we remember that the bases we write the matrix for $T^*$ with respect to will be the dual bases of the bases we started with on $V$ and $W$.

This last part is the key part. We don't like having to write $T$ and $T^*$ with respect to different bases. It's not nice and not how we usually do things. Thus we want bilinear forms with a self dual basis. I.e., a basis with respect to which the bilinear form is just the identity matrix.

Well, in the case of $\Bbb{R}$ or $\Bbb{C}$, that means we have (almost) an inner product. The difference of course is that inner products are sesquilinear rather than bilinear, but that's simple to resolve. Let $\bar{V}$ denote the vector space with the same underlying abelian group as $V$, but where the product is $(\lambda,v)\mapsto \bar{\lambda}v$, where the product on the right hand side is the usual product on $V$. There is a conjugate linear map $\bar{V}\to V$ which is the identity map on the underlying set, and composing this with one of the variables of the bilinear form, we get a sesquilinear form on $V$. (Note that this conjugate linear map is the reason we get the conjugate transpose for inner product spaces rather than the transpose as above.) Since the bilinear form has a self dual basis, this sesquilinear form on $V$ is isomorphic to the usual inner product on $\Bbb{R}^n$ or $\Bbb{C}^n$. Thus if a bilinear form has a self dual basis, then it can be regarded as an inner product space.

Finally this is where we point out that self dual bases (with respect to an inner product) are precisely the orthonormal bases. Thus our original general theorem becomes your result:

If $V$ and $W$ are inner product spaces over $\Bbb{R}$ or $\Bbb{C}$ with orthonormal bases $v_1,\ldots,v_n$ and $w_1,\ldots,w_m$, then the matrix of $T$ with respect to these bases is the conjugate transpose of the matrix of $T^*$ with respect to these bases.

Summary:

Key things to point out. The process of moving from general to specific should hopefully illuminate why the bases need to be orthonormal. In the original theorem, we needed to choose dual bases for the result to hold, and if we'd chosen another basis on the dual it wouldn't have worked. Thus if one wants to use the same bases on $V$ and $W$ for both $T$ and $T^*$ you have to use a self dual basis (i.e. and orthonormal basis).

Answer 3

Consider an inner product of the form $\langle x, y \rangle = y^HGx, \, x,y\in\mathbb{C}^n$ where $y^H = \bar{y}^T$ is the conjugate transpose and $G^H=G$ positive definite, then the adjoint $M^*$ of $M\in\mathbb{C}^{n\times n}$ can be derived using:

$$\langle Mx,y \rangle = y^HGMx = y^H(M^*)^HGx = (M^*y)^HGx = \langle x, M^*y\rangle.$$

Then $GM = (M^*)^HG$, and thus $M^*=G^{-1}M^HG$ and $M=G^{-1}(M^*)^HG$. Then it is clear that $M^*=M^H$ for any $M$ holds only if $G=I$, that is, your metric tensor $G$ arises from an orthonormal system.