I am studying linear algebra by myself, using Serge Lang's "Linear Algebra", and I'm having some difficuties undersending the relationship between transpose, symmetric, Hermitian and unitary operators and their matrices. I have a few questions but, seeing as they all seem very closely related, I decided to post them together. Here they are:

1) Let $V$ be an n-dimensional vector space over $\mathbb{R}$, let $\langle\cdot,\cdot\rangle$ be a scalar product defined on $V$, and let $A:V\rightarrow V$ be a linear operator$^{[1]}$. Moreover, let $\mathcal{B} = \{v_1,v_2,\dots,v_n\}$ be a base of $V$, and $[L]^{\mathcal{B}}_{\mathcal{B}}$ be the matrix representing a linear application $L$ with respect to $\mathcal{B}$. What conditions (non-degenerate, definite positive, etc) are needed on $\langle\cdot,\cdot\rangle$ and/or on $\mathcal{B}$, to make sure that, for each possible choice of $A$, the following hold:

  1. $[^tA]^{\mathcal{B}}_{\mathcal{B}}=([A]^{\mathcal{B}}_{\mathcal{B}})^T$, where $^tA$ is the only operator such that $\forall v,w \in V, \langle A(v),w\rangle = \langle v,^tA(w)\rangle$
  2. $^tA=A \iff [^tA]^{\mathcal{B}}_{\mathcal{B}}=([A]^{\mathcal{B}}_{\mathcal{B}})^T$
  3. $A$ is unitary (that is, $\langle A(v),A(w)\rangle = \langle v,w\rangle$) $\iff [A]^{\mathcal{B}}_{\mathcal{B}}*([A]^{\mathcal{B}}_{\mathcal{B}})^T=I$

2) What would change if, instead of on $\mathbb{R}$, we defined $V$ on any other field $\mathbb{K}$?

3)Let $V$ now be defined on $\mathbb{C}$ and $\langle\cdot,\cdot\rangle$ be an Hermitian product. Same as before, what conditions (non-degenerate, definite positive, etc) are needed on $\langle\cdot,\cdot\rangle$, or possibly on $\mathcal{B}$, to make sure that, for each possible choice of $A$, the following hold:

  1. $[A^*]^{\mathcal{B}}_{\mathcal{B}}=\overline{([A]^{\mathcal{B}}_{\mathcal{B}})}^T$, where $A^*$ is the only operator such that $\forall v,w \in V, \langle A(v),w\rangle = \langle v,A^*(w)\rangle$
  2. $A^*=A \iff [^tA]^{\mathcal{B}}_{\mathcal{B}}=\overline{([A]^{\mathcal{B}}_{\mathcal{B}})}^T$
  3. $A$ is complex unitary (that is, $\langle A(v),A(w)\rangle = \langle v,w\rangle$) $\iff [A]^{\mathcal{B}}_{\mathcal{B}}\ \overline{([A]^{\mathcal{B}}_{\mathcal{B}})}^T=I$

4) What would change if $V$ were not finite dimensional?

I admittedly found some similar questions here on stackexchange, but most of them seem to limit themselves to the standard dot product on $\mathbb{R^n}$ or the standard Hermitian product on $\mathbb{C^n}$, whereas I'm interested in the generic version. Working by myself I found that most of these seem to hold if the matrix $M$ representing the scalar or Hermitian product with respect to $\mathcal{B}$ is either $I$ or $-I$ (which, in a way, makes them isomorphic to the standard dot and Hermitian products), and the implications from matrix equalities to operator equalities also seem to hold if $M$ is something akin to: $$\left(\begin{matrix} \pm I & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 0\\ \end{matrix}\right).$$ but I'm really not sure if there are other cases which work.

I'd be grateful even for a partial answer, seeing as my post is quite long.

$^{[1]}$: The definition of operator I've learned up until now is simply "linear funcion from a space into itself", so there might be something I'm missing


1 Answer 1

  1. The necessary and sufficient condition is that the basis you use is orthonormal up to scale, that is, $\langle v_i, v_j \rangle = c \delta_{ij}$ for some $c \neq 0$. (Note that we need nondegeneracy in order for adjoints to be unique: for example adjoints clearly can't be unique if the scalar product is zero. This orthonormality condition then implies that the scalar product is symmetric.)

  2. Nothing except that "positive definite" no longer makes sense.

  3. The same conditions, but $c$ must be (positive and) real.

  4. In this case the interpretation of transposes is more complicated. Things don't behave much like the finite-dimensional case unless $V$ is assumed to be a Hilbert space.

Here is the relevant computation. Write

$$A v_i = \sum_j A_{ji} v_j$$

so that

$$\langle A v_i, v_j \rangle = \sum_k A_{ki} \langle v_k, v_j \rangle.$$

Similarly, writing $B$ for the adjoint of $A$ with respect to the scalar product (this is to make referring to the components of $B$ slightly less confusing), we have

$$B v_i = \sum_j B_{ji} v_j$$

so that

$$\langle v_i, B v_j \rangle = \sum_k B_{kj} \langle v_i, v_k \rangle.$$

We want the matrix of $B$ to be the transpose of the matrix of $A$, meaning that $B_{kj} = A_{jk}$. This means that we want

$$\sum_k A_{ki} \langle v_k, v_j \rangle = \sum_k A_{jk} \langle v_i, v_k \rangle$$

for all $i, j$ and all $A$. We can treat the entries of $A$ as variables. The two sides of this linear equation have no variables in common except $A_{ji}$, and equating their coefficients gives

$$\langle v_j, v_j \rangle = \langle v_i, v_i \rangle.$$

All of the other coefficients vanish, hence $\langle v_i, v_j \rangle = 0$ for $i \neq j$. Now nondegeneracy / the uniqueness of adjoints requires that the common value of $\langle v_i, v_i \rangle = c$ not vanish.

  • $\begingroup$ Thank you very much for your answer, though there are still a couple of things bugging me: in 1) wouldn't $\langle v_i,v_j\rangle=-\delta_{ij}$ also work? Also, how would I go about proving that orthonormality of the base is a necessary condition? Once we prove that the base has to be orthogonal it should be sufficient to carry out the matrix multiplication and verify that either $\langle v_i,v_j\rangle=\delta_{ij}$ or $\langle v_i,v_j\rangle=-\delta_{ij}$, but I'm kinda lost on proving that the base must be orthogonal $\endgroup$
    – NPG
    Commented Dec 29, 2017 at 22:58
  • $\begingroup$ Thank you again, same should go for question 3 right? $\endgroup$
    – NPG
    Commented Dec 29, 2017 at 23:09
  • $\begingroup$ In question 3 $\langle v_i, v_i \rangle$ has to be real or else the scalar product can't be Hermitian. $\endgroup$ Commented Dec 29, 2017 at 23:13
  • $\begingroup$ Sorry, I spoke too hastily again. What we actually need is $\langle v_i, v_j \rangle = c \delta_{ij}$ for $c \neq 0$. In practice people only work with orthonormal bases, so $c = 1$. $\endgroup$ Commented Dec 29, 2017 at 23:26
  • $\begingroup$ Actually, if you define the matrix $M$ of the Hermitian product to be such that $\langle v,w\rangle=v^TM\overline{w}$, and then impose $\langle v,w\rangle=\overline{\langle w,v\rangle}$, you should get $\overline{M}^T=M$, and thus the diagonal elements must be real; moreover as you said in an Hermitian product $\langle v,v\rangle \in \mathbb{R}\ \forall v$, else you'd violate $\langle v,w\rangle=\overline{\langle w,v\rangle}$ $\endgroup$
    – NPG
    Commented Dec 29, 2017 at 23:46

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