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Self-adjoint endomorphisms

⌚ ~51 min ▢ 6 blocks ✓ 15 exercises ➣ Prerequisites : Real pre-Hilbert spaces, Reduction: eigen-elements and diagonalisation, Isometries of a Euclidean space

The chapter Isometries of a Euclidean space attached to every endomorphism $u$ of a Euclidean space its adjoint $u^*$, the endomorphism that moves the inner product from one side to the other: $\langle u(x) \mid y\rangle = \langle x \mid u^*(y)\rangle$. Two families of endomorphisms then stand out. The isometries, characterised by $u^* = u^{-1}$, were the subject of that chapter. The endomorphisms equal to their own adjoint, $u^* = u$, are the subject of this one --- the self-adjoint endomorphisms.
The chapter has three sections. The first defines the self-adjoint endomorphism and reads it off a matrix: in an orthonormal basis, self-adjoint means symmetric. The second proves the headline result, the spectral theorem --- a self-adjoint endomorphism is diagonalisable in an orthonormal basis, its eigenvalues are real, its eigenspaces are orthogonal. The third singles out the self-adjoint endomorphisms whose values $\langle u(x) \mid x\rangle$ keep a sign --- the positive ones --- and reads positivity off the spectrum.
Standing notation. $E$ is a Euclidean space of dimension $n \geq 1$, with inner product $\langle\cdot\mid\cdot\rangle$ and norm $\|\cdot\|$; « BON » abbreviates « orthonormal basis ». For $u \in \mathcal{L}(E)$, $u^*$ is its adjoint, $\operatorname{Sp}(u)$ its spectrum, $\chi_u$ its characteristic polynomial, $E_\lambda(u) = \ker(u - \lambda\operatorname{Id}_E)$ its eigenspace for $\lambda$, $m(\lambda)$ the multiplicity of $\lambda$ as a root of $\chi_u$. The Euclidean space, the orthonormal basis, the orthogonal complement $F^\perp$ and the adjoint $u^*$ are those of Isometries of a Euclidean space; the eigen-elements, the characteristic polynomial and the diagonalisability criteria are those of Reduction: eigen-elements and diagonalisation.

I Self-adjoint endomorphisms

I.1 Self-adjoint endomorphisms

Recall from Isometries of a Euclidean space the adjoint: every $u \in \mathcal{L}(E)$ has a unique adjoint $u^* \in \mathcal{L}(E)$ satisfying $\langle u(x) \mid y\rangle = \langle x \mid u^*(y)\rangle$ for all $x, y$. The self-adjoint endomorphism is simply the one that coincides with its adjoint.

Definition — Self-adjoint endomorphism

An endomorphism $u \in \mathcal{L}(E)$ is self-adjoint when $u^* = u$, that is $\langle u(x) \mid y\rangle = \langle x \mid u(y)\rangle$ for all $x, y \in E$. The set of self-adjoint endomorphisms of $E$ is written $\mathcal{S}(E)$.

Example — An orthogonal projection

Let $F$ be a subspace of $E$ and $p_F$ the orthogonal projection onto $F$. For $x, y \in E$, write $x = x_F + x_\perp$ and $y = y_F + y_\perp$ along $E = F \oplus F^\perp$. Then $\langle p_F(x) \mid y\rangle = \langle x_F \mid y_F + y_\perp\rangle = \langle x_F \mid y_F\rangle$, and symmetrically $\langle x \mid p_F(y)\rangle = \langle x_F \mid y_F\rangle$. The two are equal, so $p_F$ is self-adjoint.

Example — An orthogonal symmetry

The orthogonal symmetry $s_F$ with respect to a subspace $F$ satisfies $s_F = 2p_F - \operatorname{Id}_E$. Recalled from Isometries of a Euclidean space, an orthogonal symmetry equals its own adjoint; read through the present definition, $s_F$ is a self-adjoint endomorphism. (It is also an isometry: $s_F^* = s_F = s_F^{-1}$.)

Example — Homotheties

For $\lambda \in \mathbb{R}$, the homothety $\lambda\operatorname{Id}_E$ is self-adjoint: $\langle \lambda x \mid y\rangle = \lambda\langle x \mid y\rangle = \langle x \mid \lambda y\rangle$. In particular $0$ and $\operatorname{Id}_E$ lie in $\mathcal{S}(E)$.

Proposition — The set of self-adjoint endomorphisms

$\mathcal{S}(E)$ is a \textcolor{colorprop}{linear subspace} of $\mathcal{L}(E)$.

Proof

The map $\varphi : u \mapsto u^* - u$ is an endomorphism of $\mathcal{L}(E)$: the adjunction $u \mapsto u^*$ is linear (recalled from Isometries of a Euclidean space), and so is the identity. Now $\mathcal{S}(E) = \{u : u^* = u\} = \ker\varphi$, the kernel of a linear map, hence a linear subspace of $\mathcal{L}(E)$.

Method — Proving an endomorphism is self-adjoint

Three routes. Directly: check $\langle u(x) \mid y\rangle = \langle x \mid u(y)\rangle$ for all $x, y$. Through the adjoint: compute $u^*$ and compare it with $u$. Through a matrix: by the criterion of the next subsection, verify that the matrix of $u$ in an orthonormal basis is symmetric.

The defining identity says the inner product does not see which of the two slots $u$ acts on.

For a self-adjoint $u$, the number $\langle u(x) \mid y\rangle$ is unchanged when $u$ is moved from $x$ to $y$.

Skills to practice

Recognising a self-adjoint endomorphism

I.2 Matrix characterisation and stable subspaces

In an orthonormal basis the adjoint is computed by transposing the matrix (recalled from Isometries of a Euclidean space). A self-adjoint endomorphism is therefore exactly one with a symmetric matrix --- provided the basis is orthonormal.

Definition — Symmetric matrix

A matrix $M \in \mathcal{M}_n(\mathbb{R})$ is symmetric when $M^\mathsf{T} = M$. The set of symmetric matrices of size $n$ is written $\mathcal{S}_n(\mathbb{R})$.

Theorem — Matrix characterisation

An endomorphism $u$ is self-adjoint if and only if its matrix in \textcolor{colorprop}{one} (equivalently, in \textcolor{colorprop}{every}) orthonormal basis is symmetric.

Proof

Let $\mathcal{B}$ be an orthonormal basis and $M = \operatorname{Mat}_\mathcal{B}(u)$. By the adjoint-in-a-BON formula (recalled), $\operatorname{Mat}_\mathcal{B}(u^*) = M^\mathsf{T}$. Hence $u^* = u \Leftrightarrow M^\mathsf{T} = M$: self-adjointness read in one orthonormal basis is the symmetry of $M$. If this holds in one orthonormal basis it holds in every one, since the equality $u^* = u$ is intrinsic; equivalently, a change-of-basis matrix between two orthonormal bases is orthogonal, and $P^\mathsf{T} M P$ is symmetric whenever $M$ is.

Example — Reading self-adjointness off a matrix

Equip $\mathbb{R}^3$ with its canonical inner product, for which the canonical basis is orthonormal. The endomorphism of matrix $\begin{pmatrix} 1 & 4 & 0 \\ 4 & 2 & -3 \\ 0 & -3 & 5 \end{pmatrix}$ is self-adjoint, that matrix being symmetric. On $\mathbb{R}^2$ with its canonical inner product, the endomorphism of matrix $\begin{pmatrix} 1 & 4 \\ 0 & 2 \end{pmatrix}$ is not self-adjoint, since $\begin{psmallmatrix} 1 & 4 \\ 0 & 2 \end{psmallmatrix}^\mathsf{T} \neq \begin{psmallmatrix} 1 & 4 \\ 0 & 2 \end{psmallmatrix}$.

Proposition — Dimension of $\mathcal{S}(E)$

$\dim\mathcal{S}(E) = \dim\mathcal{S}_n(\mathbb{R}) = \textcolor{colorprop}{\dfrac{n(n+1)}{2}}$.

Proof

Fix an orthonormal basis $\mathcal{B}$. By the matrix characterisation, $u \mapsto \operatorname{Mat}_\mathcal{B}(u)$ restricts to a linear isomorphism from $\mathcal{S}(E)$ onto $\mathcal{S}_n(\mathbb{R})$, so the two spaces have the same dimension. A symmetric matrix is determined by its diagonal and its strict upper triangle, so the family $(E_{ii})_{1 \leq i \leq n} \cup (E_{ij} + E_{ji})_{1 \leq i < j \leq n}$ is a basis of $\mathcal{S}_n(\mathbb{R})$. It has $n + \dfrac{n(n-1)}{2} = \dfrac{n(n+1)}{2}$ elements.

Proposition — Stability of the orthogonal complement

Let $u$ be self-adjoint and $F$ a subspace stable by $u$. Then $F^\perp$ is \textcolor{colorprop}{also stable by $u$}.

Proof

By a result of Isometries of a Euclidean space, if $F$ is stable by $u$ then $F^\perp$ is stable by the adjoint $u^*$. Here $u$ is self-adjoint, so $u^* = u$, and the statement reads: $F^\perp$ is stable by $u$.

Example — A stable subspace and its orthogonal

Let $u$ be self-adjoint and $x$ an eigenvector of $u$. The line $F = \mathbb{R}x$ is stable by $u$, so the hyperplane $F^\perp$ is stable by $u$ as well. Iterating this remark on $F^\perp$ is the engine of the spectral theorem of the next section.

Method — Using the matrix criterion

To recognise a self-adjoint endomorphism, read its matrix in an orthonormal basis and test the symmetry $M^\mathsf{T} = M$. Warning: the test is valid only in an orthonormal basis --- a symmetric matrix in a non-orthonormal basis does not signal a self-adjoint endomorphism. To prepare a reduction, look for a stable subspace $F$: its orthogonal $F^\perp$ is then stable too, and $u$ splits along $E = F \oplus F^\perp$.

For a self-adjoint endomorphism, a stable subspace $F$ drags its orthogonal $F^\perp$ along: both are stable, and the study of $u$ splits in two.

The orthogonal decomposition $E = F \oplus F^\perp$ is respected by $u$.

Skills to practice

Using the matrix criterion and stable subspaces

II The spectral theorem

II.1 Real eigenvalues and orthogonal eigenspaces

Two facts prepare the spectral theorem: a self-adjoint endomorphism has only real eigenvalues --- in fact its characteristic polynomial is split over $\mathbb{R}$ --- and its eigenspaces for distinct eigenvalues are orthogonal.

Proposition — Real eigenvalues

Let $u$ be self-adjoint. Every complex root of $\chi_u$ is \textcolor{colorprop}{real}; equivalently, $\chi_u$ is \textcolor{colorprop}{split over $\mathbb{R}$}. In particular $u$ has at least one eigenvalue.

Proof

Let $S$ be the matrix of $u$ in an orthonormal basis: $S \in \mathcal{S}_n(\mathbb{R})$. The polynomial $\chi_u = \chi_S$ lies in $\mathbb{R}[X]$ and, by d'Alembert--Gauss, has a complex root $\lambda$. Viewing $S$ as an element of $\mathcal{M}_n(\mathbb{C})$, the eigenvalue--root equivalence of Reduction: eigen-elements and diagonalisation, applied over $\mathbb{C}$, furnishes a complex eigenvector $X \in \mathbb{C}^n$, $X \neq 0$, with $SX = \lambda X$.
Compute $\overline{X}^\mathsf{T} S X$ in two ways. On one hand $\overline{X}^\mathsf{T} S X = \overline{X}^\mathsf{T}(\lambda X) = \lambda\,\overline{X}^\mathsf{T} X$. On the other, since $S$ is real symmetric, $\overline{X}^\mathsf{T} S X = (\overline{S X})^\mathsf{T} X = (\overline{\lambda X})^\mathsf{T} X = \overline{\lambda}\,\overline{X}^\mathsf{T} X$. As $\overline{X}^\mathsf{T} X = \sum_{i} |x_i|^2 > 0$, comparing gives $\lambda = \overline{\lambda}$, so $\lambda \in \mathbb{R}$. The same argument applies to every complex root of $\chi_u$, so all its roots are real and $\chi_u$ is split over $\mathbb{R}$.

The computation above uses a complex eigenvector $X \in \mathbb{C}^n$ and the quantity $\overline{X}^\mathsf{T} X = \sum_i |x_i|^2$. This is a coordinate identity in $\mathbb{C}^n$; no hermitian inner-product structure is invoked --- the hermitian inner product is out of the MP program.

Example — The $2 \times 2$ case

A symmetric matrix $\begin{pmatrix} a & b \\ b & c \end{pmatrix}$ has characteristic polynomial $X^2 - (a+c)X + (ac - b^2)$, of discriminant $(a+c)^2 - 4(ac - b^2) = (a - c)^2 + 4b^2 \geq 0$. The two roots are real --- as the Proposition predicts --- and they are equal exactly when $a = c$ and $b = 0$, that is when the matrix is $a I_2$.

Proposition — Orthogonal eigenspaces

Let $u$ be self-adjoint and $\lambda \neq \mu$ two eigenvalues of $u$. The eigenspaces $E_\lambda(u)$ and $E_\mu(u)$ are \textcolor{colorprop}{orthogonal}.

Proof

Take $x \in E_\lambda(u)$ and $y \in E_\mu(u)$. Self-adjointness gives $\langle u(x) \mid y\rangle = \langle x \mid u(y)\rangle$, that is $\langle \lambda x \mid y\rangle = \langle x \mid \mu y\rangle$, hence $\lambda\langle x \mid y\rangle = \mu\langle x \mid y\rangle$. So $(\lambda - \mu)\langle x \mid y\rangle = 0$, and since $\lambda \neq \mu$ we get $\langle x \mid y\rangle = 0$. Every vector of $E_\lambda(u)$ is thus orthogonal to every vector of $E_\mu(u)$.

Example — Orthogonal eigenlines

The endomorphism of $\mathbb{R}^2$ with matrix $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ in the canonical (orthonormal) basis is self-adjoint. Its eigenvalues are $1$ and $3$, with eigenvectors $(1, -1)$ and $(1, 1)$. The two eigenvectors satisfy $\langle (1,-1) \mid (1,1)\rangle = 1 - 1 = 0$: the eigenlines are orthogonal, as the Proposition guarantees.

Method — Locating eigenvalues and testing orthogonality

For a self-adjoint endomorphism, the eigenvalues are the roots of $\chi_u$ and they are all real --- no complex root to discard. Once eigenvectors of distinct eigenvalues are known, their orthogonality is automatic; it serves as a numerical check on a computation.

When a self-adjoint endomorphism of the plane has two distinct eigenvalues, its two eigenlines meet at a right angle.

For distinct eigenvalues $\lambda \neq \mu$, the eigenspaces $E_\lambda(u)$ and $E_\mu(u)$ are orthogonal.

Skills to practice

Eigenvalues and orthogonality of eigenspaces

II.2 The spectral theorem

Everything is in place for the headline result: a self-adjoint endomorphism is not merely diagonalisable, it is diagonalisable in an orthonormal basis. Read on matrices, this says that a real symmetric matrix is orthogonally diagonalisable.

Theorem — Spectral theorem

Let $u$ be a self-adjoint endomorphism of $E$. Then $u$ is \textcolor{colorprop}{diagonalisable in an orthonormal basis}: $E$ is the orthogonal direct sum of the eigenspaces of $u$, and there is an orthonormal basis of $E$ made of eigenvectors of $u$.

Proof

We prove, by induction on $n = \dim E \geq 0$, that every self-adjoint endomorphism of a Euclidean space of dimension $n$ admits an orthonormal basis of eigenvectors. For $n \leq 1$ the statement is immediate (for $n = 0$, the empty basis).
Let $n \geq 2$ and $u$ self-adjoint on $E$ of dimension $n$. By the « real eigenvalues » Proposition, $u$ has a real eigenvalue $\lambda$; the eigenspace $E_\lambda(u)$ is non-zero and stable by $u$, so $E_\lambda(u)^\perp$ is stable by $u$ (stability of the orthogonal complement). Equipped with the restriction of $\langle\cdot\mid\cdot\rangle$, the subspace $E_\lambda(u)^\perp$ is a Euclidean space, and the induced endomorphism $u_{|E_\lambda(u)^\perp}$ is self-adjoint for that structure. Its dimension is $< n$, so the induction hypothesis gives an orthonormal basis of $E_\lambda(u)^\perp$ made of eigenvectors of $u$.
Concatenate an orthonormal basis of $E_\lambda(u)$ with that orthonormal eigenbasis of $E_\lambda(u)^\perp$. The two blocks are orthogonal, $E_\lambda(u)^\perp$ being by construction orthogonal to $E_\lambda(u)$; together they form an orthonormal basis of $E = E_\lambda(u) \oplus E_\lambda(u)^\perp$ made of eigenvectors of $u$. Grouping the basis vectors by eigenvalue exhibits $E$ as the orthogonal direct sum of the eigenspaces.

Theorem — Spectral theorem$\virgule$ matrix form

Every real symmetric matrix $M \in \mathcal{S}_n(\mathbb{R})$ is \textcolor{colorprop}{orthogonally diagonalisable}: there is an orthogonal matrix $P \in \operatorname{O}_n(\mathbb{R})$ such that $P^\mathsf{T} M P = P^{-1} M P$ is diagonal with real entries.

Proof

Equip $\mathbb{R}^n$ with its canonical inner product, for which the canonical basis $\mathcal{B}_0$ is orthonormal, and let $u$ be the endomorphism of matrix $M$ in $\mathcal{B}_0$. As $M$ is symmetric, $u$ is self-adjoint (matrix characterisation). The spectral theorem gives an orthonormal basis $\mathcal{B}$ of eigenvectors of $u$; in $\mathcal{B}$ the matrix of $u$ is diagonal with real entries. The change-of-basis matrix $P$ from $\mathcal{B}_0$ to $\mathcal{B}$ passes between two orthonormal bases, hence is orthogonal (recalled from Isometries of a Euclidean space), and $P^\mathsf{T} M P = P^{-1} M P$ is that diagonal matrix.

Example — Orthogonally diagonalising a $3 \times 3$ matrix

Take $M = \begin{pmatrix} 6 & -2 & 2 \\ -2 & 5 & 0 \\ 2 & 0 & 7 \end{pmatrix} \in \mathcal{S}_3(\mathbb{R})$. Its characteristic polynomial is $\chi_M = (X - 3)(X - 6)(X - 9)$, so $\operatorname{Sp}(M) = \{3, 6, 9\}$. Unit eigenvectors are $\tfrac13(2, 2, -1)$ for $3$, $\tfrac13(1, -2, -2)$ for $6$, $\tfrac13(2, -1, 2)$ for $9$; they are pairwise orthogonal, as the spectral theorem guarantees. With $$ P = \frac13\begin{pmatrix} 2 & 1 & 2 \\ 2 & -2 & -1 \\ -1 & -2 & 2 \end{pmatrix} \in \operatorname{O}_3(\mathbb{R}), \qquad P^\mathsf{T} M P = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 9 \end{pmatrix}. $$

Example — A $2 \times 2$ reduction

For $M = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$, the eigenvalues are $1$ and $3$ with orthonormal eigenvectors $\tfrac{1}{\sqrt2}(1, -1)$ and $\tfrac{1}{\sqrt2}(1, 1)$. With $P = \tfrac{1}{\sqrt2}\begin{psmallmatrix} 1 & 1 \\ -1 & 1 \end{psmallmatrix} \in \operatorname{O}_2(\mathbb{R})$, $P^\mathsf{T} M P = \begin{psmallmatrix} 1 & 0 \\ 0 & 3 \end{psmallmatrix}$.

Example — A repeated eigenvalue

The matrix $M = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{pmatrix}$ has eigenvalues $2$, $2$, $0$: the eigenvalue $2$ has multiplicity $2$, with eigenplane $E_2(M) = \operatorname{Vect}\big((1,0,0),(0,1,1)\big)$. The two listed vectors are already orthogonal; normalising gives $(1,0,0)$ and $\tfrac{1}{\sqrt2}(0,1,1)$. Together with $\tfrac{1}{\sqrt2}(0,1,-1)$ spanning $E_0(M)$, they form an orthonormal eigenbasis. Inside an eigenspace of multiplicity $\geq 2$, the Gram--Schmidt process produces the orthonormal vectors needed.

Method — Orthogonally diagonalising a symmetric matrix

Given $M \in \mathcal{S}_n(\mathbb{R})$: compute the eigenvalues from $\chi_M$ (all real). For each eigenvalue, find a basis of the eigenspace and orthonormalise it by Gram--Schmidt --- needed only inside an eigenspace of multiplicity $\geq 2$, since eigenspaces of distinct eigenvalues are already orthogonal. Stacking these orthonormal eigenvectors as columns yields $P \in \operatorname{O}_n(\mathbb{R})$ with $P^\mathsf{T} M P$ diagonal.

The orthonormal eigenbasis of a self-adjoint endomorphism gives the principal axes: along them $u$ acts by simple scalings $\lambda_i$.

In the orthonormal eigenbasis $(e_1, \dots, e_n)$, the endomorphism $u$ is the diagonal scaling $e_i \mapsto \lambda_i e_i$.

Skills to practice

Orthogonally diagonalising a symmetric matrix

III Positive self-adjoint endomorphisms

III.1 Positive and positive-definite self-adjoint endomorphisms

Among the self-adjoint endomorphisms, those for which the number $\langle u(x) \mid x\rangle$ never goes negative deserve a name. They are the positive ones, and --- when it is moreover strictly positive off the origin --- the positive-definite ones.

Definition — Positive and positive-definite endomorphisms

A self-adjoint endomorphism $u$ is positive when $\langle u(x) \mid x\rangle \geq 0$ for every $x \in E$, and positive-definite when $\langle u(x) \mid x\rangle > 0$ for every $x \neq 0$. Their sets are written $\mathcal{S}^+(E)$ and $\mathcal{S}^{++}(E)$. The matching matrix sets are $\mathcal{S}_n^+(\mathbb{R})$ --- symmetric $A$ with $X^\mathsf{T} A X \geq 0$ for all $X$ --- and $\mathcal{S}_n^{++}(\mathbb{R})$ --- symmetric $A$ with $X^\mathsf{T} A X > 0$ for all $X \neq 0$.

Example — An orthogonal projection

The orthogonal projection $p_F$ onto a subspace $F$ is self-adjoint, and $\langle p_F(x) \mid x\rangle = \langle p_F(x) \mid p_F(x) + (x - p_F(x))\rangle = \|p_F(x)\|^2 \geq 0$ (the second term is orthogonal to $F$). So $p_F \in \mathcal{S}^+(E)$. It lies in $\mathcal{S}^{++}(E)$ only when $F = E$: for $F \neq E$, a non-zero $x \in F^\perp$ gives $\langle p_F(x) \mid x\rangle = 0$.

Example — The endomorphism $v^* \circ v$

For any $v \in \mathcal{L}(E)$, the endomorphism $v^* \circ v$ is self-adjoint, since $(v^* v)^* = v^* v^{**} = v^* v$, and positive: $\langle v^* v(x) \mid x\rangle = \langle v(x) \mid v(x)\rangle = \|v(x)\|^2 \geq 0$. So $v^* \circ v \in \mathcal{S}^+(E)$ for every $v$.

Example — Homotheties and the inclusions

For $\lambda > 0$, the homothety $\lambda\operatorname{Id}_E$ satisfies $\langle \lambda x \mid x\rangle = \lambda\|x\|^2 > 0$ for $x \neq 0$, so $\lambda\operatorname{Id}_E \in \mathcal{S}^{++}(E)$; in particular $\operatorname{Id}_E \in \mathcal{S}^{++}(E)$. From the definitions, $\mathcal{S}^{++}(E) \subset \mathcal{S}^+(E) \subset \mathcal{S}(E)$.

Method — Testing positivity through the quadratic form

To decide whether a self-adjoint $u$ is positive, study the sign of $x \mapsto \langle u(x) \mid x\rangle$: it is positive when this number is $\geq 0$ everywhere, positive-definite when it is $> 0$ off the origin. On matrices, with the canonical Euclidean structure of $\mathbb{R}^n$, this is the sign of $X \mapsto X^\mathsf{T} A X$. The spectral characterisation of the next subsection usually settles the question faster.

For a positive-definite $u$ and $c > 0$, the level sets $\langle u(x) \mid x\rangle = c$ are ellipses (ellipsoids in higher dimension), nested around the origin.

The quadratic form $x \mapsto \langle u(x) \mid x\rangle$ of a positive-definite endomorphism has elliptic level curves.

Skills to practice

Recognising a positive endomorphism

III.2 Spectral characterisation

The spectral theorem turns positivity into a condition on the spectrum: a self-adjoint endomorphism is positive exactly when its eigenvalues are $\geq 0$.

Proposition — Spectral characterisation of positivity

Let $u$ be a self-adjoint endomorphism. Then $$ u \in \mathcal{S}^+(E) \iff \textcolor{colorprop}{\operatorname{Sp}(u) \subset \mathbb{R}_+}, \qquad u \in \mathcal{S}^{++}(E) \iff \textcolor{colorprop}{\operatorname{Sp}(u) \subset \mathbb{R}_+^*}. $$

Proof

Suppose $u \in \mathcal{S}^+(E)$ and let $\lambda$ be an eigenvalue, $u(x) = \lambda x$ with $x \neq 0$. Then $0 \leq \langle u(x) \mid x\rangle = \lambda\|x\|^2$, so $\lambda \geq 0$.
Conversely, suppose $\operatorname{Sp}(u) \subset \mathbb{R}_+$. By the spectral theorem there is an orthonormal eigenbasis $(e_1, \dots, e_n)$, $u(e_i) = \lambda_i e_i$ with each $\lambda_i \geq 0$. Writing $x = \sum_i x_i e_i$, the basis being orthonormal gives $\langle u(x) \mid x\rangle = \sum_i \lambda_i x_i^2 \geq 0$, so $u \in \mathcal{S}^+(E)$.
For the strict version: if $u \in \mathcal{S}^{++}(E)$ the same first step gives $\lambda\|x\|^2 > 0$, so $\lambda > 0$. Conversely if every $\lambda_i > 0$, then for $x \neq 0$ some $x_i \neq 0$ and $\langle u(x) \mid x\rangle = \sum_i \lambda_i x_i^2 > 0$.

Example — Deciding positivity from the eigenvalues

The matrix $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$, of eigenvalues $1$ and $3$, both $> 0$, lies in $\mathcal{S}_2^{++}(\mathbb{R})$. The matrix $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, of eigenvalues $0$ and $2$, lies in $\mathcal{S}_2^+(\mathbb{R})$ but not in $\mathcal{S}_2^{++}(\mathbb{R})$. The matrix $\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$, of eigenvalues $-1$ and $3$, is in neither.

Proposition — Positive square root

Every $u \in \mathcal{S}^+(E)$ has a \textcolor{colorprop}{unique} positive self-adjoint square root: a $w \in \mathcal{S}^+(E)$ with $w^2 = u$, written $\sqrt{u}$. It lies in $\mathcal{S}^{++}(E)$ exactly when $u$ does.

Proof

Existence. By the spectral theorem fix an orthonormal eigenbasis $(e_1, \dots, e_n)$, $u(e_i) = \lambda_i e_i$; the eigenvalues $\lambda_i \geq 0$ by the spectral characterisation. Let $w$ be the endomorphism with $w(e_i) = \sqrt{\lambda_i}\,e_i$. Its matrix in the orthonormal basis $(e_i)$ is diagonal, hence symmetric, so $w$ is self-adjoint; its eigenvalues $\sqrt{\lambda_i} \geq 0$ make $w \in \mathcal{S}^+(E)$; and $w^2(e_i) = \lambda_i e_i = u(e_i)$, so $w^2 = u$. It is invertible iff every $\lambda_i > 0$, that is iff $u \in \mathcal{S}^{++}(E)$.
Uniqueness. Let $w \in \mathcal{S}^+(E)$ with $w^2 = u$. Then $w$ commutes with $u = w^2$, so $w$ stabilises each eigenspace $E_\lambda(u)$. The restriction $w_{|E_\lambda(u)}$ is positive self-adjoint with $(w_{|E_\lambda(u)})^2 = \lambda\operatorname{Id}$; any eigenvalue $\mu \geq 0$ of this restriction satisfies $\mu^2 = \lambda$, forcing $\mu = \sqrt{\lambda}$. Being diagonalisable with sole eigenvalue $\sqrt{\lambda}$, the restriction equals $\sqrt{\lambda}\operatorname{Id}$ on $E_\lambda(u)$. So $w$ coincides with $\sqrt{u}$ on every eigenspace, hence on $E$.

Proposition — Factorisation $u = v^* \circ v$

A self-adjoint endomorphism $u$ lies in $\mathcal{S}^+(E)$ if and only if $u = v^* \circ v$ for \textcolor{colorprop}{some $v \in \mathcal{L}(E)$}; it lies in $\mathcal{S}^{++}(E)$ if and only if such a $v$ can be chosen in $\operatorname{GL}(E)$.

Proof

$(\Leftarrow)$ For any $v$, $v^* v$ is self-adjoint and $\langle v^* v(x) \mid x\rangle = \|v(x)\|^2 \geq 0$, so $v^* v \in \mathcal{S}^+(E)$. If moreover $v \in \operatorname{GL}(E)$, then $v(x) \neq 0$ for every $x \neq 0$, so $\|v(x)\|^2 > 0$ and $v^* v \in \mathcal{S}^{++}(E)$.
$(\Rightarrow)$ If $u \in \mathcal{S}^+(E)$, take $v = \sqrt{u}$: it is self-adjoint, so $v^* \circ v = v^2 = u$. By the previous Proposition $\sqrt{u}$ is invertible exactly when $u \in \mathcal{S}^{++}(E)$, which gives the strict case.

Example — The Gram bilinear form

Let $A \in \mathcal{S}_n(\mathbb{R})$. The bilinear form $(X, Y) \mapsto X^\mathsf{T} A Y$ on $\mathbb{R}^n$ is symmetric; it is an inner product if and only if it is positive-definite, that is $X^\mathsf{T} A X > 0$ for $X \neq 0$ --- exactly the condition $A \in \mathcal{S}_n^{++}(\mathbb{R})$. The positive-definite symmetric matrices are thus precisely the matrices that define an inner product on $\mathbb{R}^n$.

Method — Using the spectral characterisation

To decide positivity, compute the spectrum: $u$ is positive iff all eigenvalues are $\geq 0$, positive-definite iff all are $> 0$ --- faster than handling $\langle u(x) \mid x\rangle$ directly. To produce a positive endomorphism, write it as $v^* \circ v$; to extract a square root of a positive endomorphism, use $\sqrt{u}$.

The positive-definite endomorphisms sit inside the positive ones, themselves inside the self-adjoint ones.

The strict inclusions are $\mathcal{S}^{++}(E) \subset \mathcal{S}^+(E) \subset \mathcal{S}(E)$. Within $\mathcal{S}(E)$ --- where the spectrum is already real, $u$ being self-adjoint --- membership in the two inner sets is read on the spectrum by the spectral characterisation: $u \in \mathcal{S}^+(E)$ iff $\operatorname{Sp}(u) \subset \mathbb{R}_+$, and $u \in \mathcal{S}^{++}(E)$ iff $\operatorname{Sp}(u) \subset \mathbb{R}_+^*$.

Going further

The spectral theorem is the engine behind several constructions just out of reach of this chapter: the reduction of a quadratic form to a sum of squares, the polar decomposition $M = \Omega R$ of an invertible matrix into an orthogonal and a positive-definite factor, and the singular value decomposition of an arbitrary matrix. The exercise sheet's « Going further » section explores them.

Skills to practice

Using the spectral characterisation