2b. Linear operators

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Let us get now into the study of linear operators [math]T:H\to H[/math]. Before anything, we should mention that things are quite tricky with respect to quantum mechanics, and physics in general. Indeed, if there is a central operator in physics, this is the Laplace operator on the smooth functions [math]f:\mathbb R^N\to\mathbb C[/math], given by:

[[math]] \Delta f(x)=\sum_i\frac{d^2f}{dx_i^2} [[/math]]

And the problem is that what we have here is an operator [math]\Delta:C^\infty(\mathbb R^N)\to C^\infty(\mathbb R^N)[/math], which does not extend into an operator [math]\Delta:L^2(\mathbb R^N)\to L^2(\mathbb R^N)[/math]. Thus, we should perhaps look at operators [math]T:H\to H[/math] which are densely defined, instead of looking at operators [math]T:H\to H[/math] which are everywhere defined. We will not do so, for two reasons:

(1) Tactical retreat. When physics looks too complicated, as it is the case now, you can always declare that mathematics comes first. So, let us be pure mathematicians, simply looking in generalizing linear algebra to infinite dimensions. And from this viewpoint, it is a no-brainer to look at everywhere defined operators [math]T:H\to H[/math].

(2) Modern physics. We will see later, towards the middle of the present book, when talking about various mathematical physics findings of Connes, Jones, Voiculescu and others, that a lot of interesting mathematics, which is definitely related to modern physics, can be developed by using the everywhere defined operators [math]T:H\to H[/math].

In short, you'll have to trust me here. And hang on, we are not done yet, because with this choice made, there is one more problem, mathematical this time. The problem comes from the fact that in infinite dimensions the everywhere defined operators [math]T:H\to H[/math] can be bounded or not, and for reasons which are mathematically intuitive and obvious, and physically acceptable too, we want to deal with the bounded case only.

Long story short, let us avoid too much thinking, and start in a simple way, with:

Proposition

For a linear operator [math]T:H\to H[/math], the following are equivalent:

• [math]T[/math] is continuous.
• [math]T[/math] is continuous at [math]0[/math].
• [math]T(B)\subset cB[/math] for some [math]c \lt \infty[/math], where [math]B\subset H[/math] is the unit ball.
• [math]T[/math] is bounded, in the sense that [math]||T||=\sup_{||x||\leq1}||Tx||[/math] satisfies [math]||T|| \lt \infty[/math].

Show Proof

This is elementary, with [math](1)\iff(2)[/math] coming from the linearity of [math]T[/math], then [math](2)\iff(3)[/math] coming from definitions, and finally [math](3)\iff(4)[/math] coming from the fact that the number [math]||T||[/math] from (4) is the infimum of the numbers [math]c[/math] making (3) work.

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Regarding such operators, we have the following result:

Theorem

The linear operators [math]T:H\to H[/math] which are bounded,

[[math]] ||T||=\sup_{||x||\leq1}||Tx|| \lt \infty [[/math]]

form a complex algebra with unit [math]B(H)[/math], having the property

[[math]] ||ST||\leq||S||\cdot||T|| [[/math]]

and which is complete with respect to the norm.

Show Proof

The fact that we have indeed an algebra, satisfying the product condition in the statement, follows from the following estimates, which are all elementary:

[[math]] ||S+T||\leq||S||+||T|| [[/math]]

[[math]] ||\lambda T||=|\lambda|\cdot||T|| [[/math]]

[[math]] ||ST||\leq||S||\cdot||T|| [[/math]]

Regarding now the last assertion, if [math]\{T_n\}\subset B(H)[/math] is Cauchy then [math]\{T_nx\}[/math] is Cauchy for any [math]x\in H[/math], so we can define the limit [math]T=\lim_{n\to\infty}T_n[/math] by setting:

[[math]] Tx=\lim_{n\to\infty}T_nx [[/math]]

Let us first check that the application [math]x\to Tx[/math] is linear. We have:

[[math]] \begin{eqnarray*} T(x+y) &=&\lim_{n\to\infty}T_n(x+y)\\ &=&\lim_{n\to\infty}T_n(x)+T_n(y)\\ &=&\lim_{n\to\infty}T_n(x)+\lim_{n\to\infty}T_n(y)\\ &=&T(x)+T(y) \end{eqnarray*} [[/math]]

Similarly, we have as well the following computation:

[[math]] \begin{eqnarray*} T(\lambda x) &=&\lim_{n\to\infty}T_n(\lambda x)\\ &=&\lambda\lim_{n\to\infty}T_n(x)\\ &=&\lambda T(x) \end{eqnarray*} [[/math]]

Thus we have a linear map [math]T:A\to A[/math]. It remains to prove that we have [math]T\in B(H)[/math], and that we have [math]T_n\to T[/math] in norm. For this purpose, observe that we have:

[[math]] \begin{eqnarray*} &&||T_n-T_m||\leq\varepsilon\ ,\ \forall n,m\geq N\\ &\implies&||T_nx-T_mx||\leq\varepsilon\ ,\ \forall||x||=1\ ,\ \forall n,m\geq N\\ &\implies&||T_nx-Tx||\leq\varepsilon\ ,\ \forall||x||=1\ ,\ \forall n\geq N\\ &\implies&||T_Nx-Tx||\leq\varepsilon\ ,\ \forall||x||=1\\ &\implies&||T_N-T||\leq\varepsilon \end{eqnarray*} [[/math]]

As a first consequence, we obtain [math]T\in B(H)[/math], because we have:

[[math]] \begin{eqnarray*} ||T|| &=&||T_N+(T-T_N)||\\ &\leq&||T_N||+||T-T_N||\\ &\leq&||T_N||+\varepsilon\\ & \lt &\infty \end{eqnarray*} [[/math]]

As a second consequence, we obtain [math]T_N\to T[/math] in norm, and we are done.

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In the case where [math]H[/math] comes with a basis [math]\{e_i\}_{i\in I}[/math], we can talk about the infinite matrices [math]M\in M_I(\mathbb C)[/math], with the remark that the multiplication of such matrices is not always defined, in the case [math]|I|=\infty[/math]. In this context, we have the following result:

Theorem

Let [math]H[/math] be a Hilbert space, with orthonormal basis [math]\{e_i\}_{i\in I}[/math]. The bounded operators [math]T\in B(H)[/math] can be then identified with matrices [math]M\in M_I(\mathbb C)[/math] via

[[math]] Tx=Mx\quad,\quad M_{ij}= \lt Te_j,e_i \gt [[/math]]

and we obtain in this way an embedding as follows, which is multiplicative:

[[math]] B(H)\subset M_I(\mathbb C) [[/math]]

In the case [math]H=\mathbb C^N[/math] we obtain in this way the usual isomorphism [math]B(H)\simeq M_N(\mathbb C)[/math]. In the separable case we obtain in this way a proper embedding [math]B(H)\subset M_\infty(\mathbb C)[/math].

Show Proof

We have several assertions to be proved, the idea being as follows:

(1) Regarding the first assertion, given a bounded operator [math]T:H\to H[/math], let us associate to it a matrix [math]M\in M_I(\mathbb C)[/math] as in the statement, by the following formula:

[[math]] M_{ij}= \lt Te_j,e_i \gt [[/math]]

It is clear that this correspondence [math]T\to M[/math] is linear, and also that its kernel is [math]\{0\}[/math]. Thus, we have an embedding of linear spaces [math]B(H)\subset M_I(\mathbb C)[/math]. (2) Our claim now is that this embedding is multiplicative. But this is clear too, because if we denote by [math]T\to M_T[/math] our correspondence, we have:

[[math]] \begin{eqnarray*} (M_{ST})_{ij} &=& \lt STe_j,e_i \gt \\ &=&\left \lt S\sum_k \lt Te_j,e_k \gt e_k,e_i\right \gt \\ &=&\sum_k \lt Se_k,e_i \gt \lt Te_j,e_k \gt \\ &=&\sum_k(M_S)_{ik}(M_T)_{kj}\\ &=&(M_SM_T)_{ij} \end{eqnarray*} [[/math]]

(3) Finally, we must prove that the original operator [math]T:H\to H[/math] can be recovered from its matrix [math]M\in M_I(\mathbb C)[/math] via the formula in the statement, namely [math]Tx=Mx[/math]. But this latter formula holds for the vectors of the basis, [math]x=e_j[/math], because we have:

[[math]] \begin{eqnarray*} (Te_j)_i &=& \lt Te_j,e_i \gt \\ &=&M_{ij}\\ &=&(Me_j)_i \end{eqnarray*} [[/math]]

Now by linearity we obtain from this that the formula [math]Tx=Mx[/math] holds everywhere, on any vector [math]x\in H[/math], and this finishes the proof of the first assertion.

(4) In finite dimensions we obtain an isomorphism, because any matrix [math]M\in M_N(\mathbb C)[/math] determines an operator [math]T:\mathbb C^N\to\mathbb C^N[/math], according to the formula [math] \lt Te_j,e_i \gt =M_{ij}[/math]. In infinite dimensions, however, we do not have an isomorphism. For instance on [math]H=l^2(\mathbb N)[/math] the following matrix does not define an operator:

[[math]] M=\begin{pmatrix}1&1&\ldots\\ 1&1&\ldots\\ \vdots&\vdots \end{pmatrix} [[/math]]

Indeed, [math]T(e_1)[/math] should be the all-one vector, which is not square-summable.

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In connection with our previous comments on bases, the above result is something quite theoretical, because for basic Hilbert spaces like [math]L^2[0,1][/math], which do not have a simple orthonormal basis, the embedding [math]B(H)\subset M_\infty(\mathbb C)[/math] that we obtain is not something very useful. In short, while the bounded operators [math]T:H\to H[/math] are basically some infinite matrices, it is better to think of these operators as being objects on their own.

As another comment, the construction [math]T\to M[/math] makes sense for any linear operator [math]T:H\to H[/math], but when [math]\dim H=\infty[/math], we do not obtain an embedding [math]\mathcal L(H)\subset M_I(\mathbb C)[/math] in this way. Indeed, set [math]H=l^2(\mathbb N)[/math], let [math]E=span(e_i)[/math] be the linear space spanned by the standard basis, and pick an algebraic complement [math]F[/math] of this space [math]E[/math], so that we have [math]H=E\oplus F[/math], as an algebraic direct sum. Then any linear operator [math]S:F\to F[/math] gives rise to a linear operator [math]T:H\to H[/math], given by [math]T(e,f)=(0,S(f))[/math], whose associated matrix is [math]0[/math]. And, restrospectively speaking, it is in order to avoid such pathologies that we decided some time ago to restrict the attention to the bounded case, [math]T\in B(H)[/math].

As in the finite dimensional case, we can talk about adjoint operators, in this setting, the definition and main properties of the construction [math]T\to T^*[/math] being as follows:

Theorem

Given a bounded operator [math]T\in B(H)[/math], the following formula defines a bounded operator [math]T^*\in B(H)[/math], called adjoint of [math]H[/math]:

[[math]] \lt Tx,y \gt = \lt x,T^*y \gt [[/math]]

The correspondence [math]T\to T^*[/math] is antilinear, antimultiplicative, and is an involution, and an isometry. In finite dimensions, we recover the usual adjoint operator.

Show Proof

There are several things to be done here, the idea being as follows:

(1) We will need a standard functional analysis result, stating that the continuous linear forms [math]\varphi:H\to\mathbb C[/math] appear as scalar products, as follows, with [math]z\in H[/math]:

[[math]] \varphi(x)= \lt x,z \gt [[/math]]

Indeed, in one sense this is clear, because given [math]z\in H[/math], the application [math]\varphi(x)= \lt x,z \gt [/math] is linear, and continuous as well, because by Cauchy-Schwarz we have:

[[math]] |\varphi(x)|\leq||x||\cdot||z|| [[/math]]

Conversely now, by using a basis we can assume [math]H=l^2(\mathbb N)[/math], and our linear form [math]\varphi:H\to\mathbb C[/math] must be then, by linearity, given by a formula of the following type:

[[math]] \varphi(x)=\sum_ix_i\bar{z}_i [[/math]]

But, again by Cauchy-Schwarz, in order for such a formula to define indeed a continuous linear form [math]\varphi:H\to\mathbb C[/math] we must have [math]z\in l^2(\mathbb N)[/math], and so [math]z\in H[/math], as desired.

(2) With this in hand, we can now construct the adjoint [math]T^*[/math], by the formula in the statement. Indeed, given [math]y\in H[/math], the formula [math]\varphi(x)= \lt Tx,y \gt [/math] defines a linear map [math]H\to\mathbb C[/math]. Thus, we must have a formula as follows, for a certain vector [math]T^*y\in H[/math]:

[[math]] \varphi(x)= \lt x,T^*y \gt [[/math]]

Moreover, this vector [math]T^*y\in H[/math] is unique with this property, and we conclude from this that the formula [math]y\to T^*y[/math] defines a certain map [math]T^*:H\to H[/math], which is unique with the property in the statement, namely [math] \lt Tx,y \gt = \lt x,T^*y \gt [/math] for any [math]x,y[/math].

(3) Let us prove that we have [math]T^*\in B(H)[/math]. By using once again the uniqueness of [math]T^*[/math], we conclude that we have the following formulae, which show that [math]T^*[/math] is linear:

[[math]] T^*(x+y)=T^*x+T^*y\quad,\quad T^*(\lambda x)=\lambda T^*x [[/math]]

Observe also that [math]T^*[/math] is bounded as well, because we have:

[[math]] \begin{eqnarray*} ||T|| &=&\sup_{||x||=1}\sup_{||y||=1} \lt Tx,y \gt \\ &=&\sup_{||y||=1}\sup_{||x||=1} \lt x,T^*y \gt \\ &=&||T^*|| \end{eqnarray*} [[/math]]

(4) The fact that the correspondence [math]T\to T^*[/math] is antilinear, antimultiplicative, and is an involution comes from the following formulae, coming from uniqueness:

[[math]] (S+T)^*=S^*+T^*\quad,\quad (\lambda T)^*=\bar{\lambda}T^* [[/math]]

[[math]] (ST)^*=T^*S^*\quad,\quad (T^*)^*=T [[/math]]

As for the isometry property with respect to the operator norm, [math]||T||=||T^*||[/math], this is something that we already know, from the proof of (3) above.

(5) Regarding finite dimensions, let us first examine the general case where our Hilbert space comes with a basis, [math]H=l^2(I)[/math]. We can compute the matrix [math]M^*\in M_I(\mathbb C)[/math] associated to the operator [math]T^*\in B(H)[/math], by using [math] \lt Tx,y \gt = \lt x,T^*y \gt [/math], in the following way:

[[math]] \begin{eqnarray*} (M^*)_{ij} &=& \lt T^*e_j,e_i \gt \\ &=&\overline{ \lt e_i,T^*e_j \gt }\\ &=&\overline{ \lt Te_i,e_j \gt }\\ &=&\overline{M}_{ji} \end{eqnarray*} [[/math]]

Thus, we have reached to the usual formula for the adjoints of matrices, and in the particular case [math]H=\mathbb C^N[/math], we conclude that [math]T^*[/math] comes indeed from the usual [math]M^*[/math].

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As in finite dimensions, the operators [math]T,T^*[/math] can be thought of as being “twin brothers”, and there is a lot of interesting mathematics connecting them. We first have:

Proposition

Given a bounded operator [math]T\in B(H)[/math], the following happen:

• [math]\ker T^*=(Im T)^\perp[/math].
• [math]\overline{Im T^*}=(\ker T)^\perp[/math].

Show Proof

Both these assertions are elementary, as follows:

(1) Let us first prove “[math]\subset[/math]”. Assuming [math]T^*x=0[/math], we have indeed [math]x\perp ImT[/math], because:

[[math]] \lt x,Ty \gt = \lt T^*x,y \gt =0 [[/math]]

As for “[math]\supset[/math]”, assuming [math] \lt x,Ty \gt =0[/math] for any [math]y[/math], we have [math]T^*x=0[/math], because:

[[math]] \lt T^*x,y \gt = \lt x,Ty \gt =0 [[/math]]

(2) This can be deduced from (1), applied to the operator [math]T^*[/math], as follows:

[[math]] (\ker T)^\perp =(Im T^*)^{\perp\perp} =\overline{Im T^*} [[/math]]

Here we have used the formula [math]K^{\perp\perp}=\bar{K}[/math], valid for any linear subspace [math]K\subset H[/math] of a Hilbert space, which for [math]K[/math] closed reads [math]K^{\perp\perp}=K[/math], and comes from [math]H=K\oplus K^\perp[/math], and which in general follows from [math]K^{\perp\perp}\subset\bar{K}^{\perp\perp}=\bar{K}[/math], the reverse inclusion being clear.

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Let us record as well the following useful formula, relating [math]T[/math] and [math]T^*[/math]:

Theorem

We have the following formula,

[[math]] ||TT^*||=||T||^2 [[/math]]

valid for any operator [math]T\in B(H)[/math].

Show Proof

We recall from Theorem 2.16 that the correspondence [math]T\to T^*[/math] is an isometry with respect to the operator norm, in the sense that we have:

[[math]] ||T||=||T^*|| [[/math]]

In order to prove now the formula in the statement, observe first that we have:

[[math]] ||TT^*|| \leq||T||\cdot||T^*|| =||T||^2 [[/math]]

On the other hand, we have as well the following estimate:

[[math]] \begin{eqnarray*} ||T||^2 &=&\sup_{||x||=1}| \lt Tx,Tx \gt |\\ &=&\sup_{||x||=1}| \lt x,T^*Tx \gt |\\ &\leq&||T^*T|| \end{eqnarray*} [[/math]]

By replacing [math]T\to T^*[/math] we obtain from this that we have:

[[math]] ||T||^2\leq||TT^*|| [[/math]]

Thus, we have obtained the needed inequality, and we are done.

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General references

Banica, Teo (2024). "Principles of operator algebras". arXiv:2208.03600 [math.OA].