Linear-Algebra-Done-Right · Unit 19 · Video 3 · Interactive Practice

Positive Operators: Six Equivalent Characterizations

IKey Formulas

Formula	Name	What it says
$T = T^*$ and $\langle Tv, v\rangle \ge 0$ for all $v \in V$	Positive operator (1)	Both requirements are needed over $\mathbf{R}$ ; equality is allowed for nonzero $v$
$T = T^*$ and every eigenvalue $\lambda \ge 0$	Eigenvalue test (2)	Equivalently (3): $\mathcal{M}(T)$ is diagonal with entries $\ge 0$ in some orthonormal basis
$T = R^2$ with $R$ positive, or with $R$ self-adjoint	Square roots (4), (5)	$R$ is a square root of $T$ when $R^2 = T$
$\langle R^*Rv, v\rangle = \langle Rv, Rv\rangle = \\|Rv\\|^2 \ge 0$	Condition (6): $T = R^*R$	Any $R \in \mathcal{L}(V)$ works; $R$ need not be a square root of $T$

Key Insight: Positive operators play the part of the nonnegative numbers: $z = s^2$ with $s \ge 0$ , $z = s^2$ with $s \in \mathbf{R}$ , and $z = \bar{w}w$ become a positive square root, a self-adjoint square root, and $T = R^*R$ .

IIThe Inequality in Every Direction

Since $\langle T(tv), tv\rangle = t^2\langle Tv, v\rangle$ for real $t$ , the sign of $\langle Tv, v\rangle$ depends only on the direction of $v$ .

\mathcal{M}(T)

💡 Over $\mathbf{R}$ , $\langle Tv, v\rangle$ only sees $\tfrac{1}{2}(T + T^*)$ , so an operator that is not self-adjoint can satisfy the inequality and still fail to be positive.

IIIEigenvalues Bound $\langle Tv, v\rangle$

For a unit vector $v = \cos\theta\, e_1 + \sin\theta\, e_2$ in an orthonormal eigenbasis, $\langle Tv, v\rangle = \lambda_1\cos^2\theta + \lambda_2\sin^2\theta$ .

\mathcal{M}(T)

θ 75°

IVCondition (6): $T = R^*R$

The entries of $R^*R$ are inner products of the columns of $R$ , and $\langle R^*Rv, v\rangle = \|Rv\|^2$ .

VQuiz Questions

Problem 1 · The Eigenvalue Test

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ with $\mathcal{M}(T) = \begin{pmatrix} 4 & 2 \\ 2 & 1 \end{pmatrix}$ in the standard basis. Which statement is correct?

✅ Correct!

T(2, 1) = (10, 5) = 5(2, 1)

and

T(1, -2) = (0, 0)

, so the eigenvalues are

5

and

0

, both

\ge 0

; together with

T = T^*

this is condition (2).

❌ Positive allows equality. The definition asks for

\langle Tv, v\rangle \ge 0

, not

\gt 0

. A nonzero

v

with

\langle Tv, v\rangle = 0

is allowed, and condition (2) permits the eigenvalue

0

❌ Those are not the eigenvalues.

T(1, 0) = (4, 2)

is not a multiple of

(1, 0)

, so

4

is not read off the diagonal here. The eigenvalues are

5

and

0

, with eigenvectors

(2, 1)

and

(1, -2)

❌ The theorem makes eigenvalues enough. For a self-adjoint

T

, condition (2), every eigenvalue

\ge 0

, is equivalent to positivity, so the two eigenvalues

5

and

0

settle the question.

Show solution

Self-adjoint: the matrix equals its transpose and the standard basis is orthonormal, so $T = T^*$ .

Eigenvalues:

T(2, 1) = (8 + 2,\; 4 + 1) = 5(2, 1), \qquad T(1, -2) = (4 - 4,\; 2 - 2) = (0, 0).

The eigenvalues are $5$ and $0$ , both $\ge 0$ , so condition (2) holds and $T$ is positive.

Direct check: $\langle T(x, y), (x, y)\rangle = (4x + 2y)x + (2x + y)y = 4x^2 + 4xy + y^2 = (2x + y)^2 \ge 0$ . It equals $0$ on the line spanned by $(1, -2)$ , which the definition allows.

Problem 2 · Positive Entries Are Not Enough

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ with $\mathcal{M}(T) = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}$ in the standard basis. Which statement is correct?

✅ Correct!

\mathcal{M}(T)^* = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} \ne \mathcal{M}(T)

, and

T(1, -1) = (-2, -1)

gives

\langle (-2, -1), (1, -1)\rangle = -2 + 1 = -1 \lt 0

. Either failure alone rules out positivity.

❌ Condition (2) has two parts. The only eigenvalue is indeed

1

, but (2) also requires

T = T^*

, and

\mathcal{M}(T)^* = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} \ne \mathcal{M}(T)

❌ The signs of the entries do not decide positivity.

T(1, -1) = (-2, -1)

, and

\langle (-2, -1), (1, -1)\rangle = -1 \lt 0

❌ The inequality fails too.

\langle T(x, y), (x, y)\rangle = x^2 + 3xy + y^2

, which is

1 - 3 + 1 = -1

(x, y) = (1, -1)

Show solution

Self-adjointness: the standard basis is orthonormal, so $\mathcal{M}(T^*)$ is the transpose $\begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} \ne \mathcal{M}(T)$ . Hence $T \ne T^*$ and $T$ is not positive.

The inequality: $T(x, y) = (x + 3y,\; y)$ , so

\langle T(x, y), (x, y)\rangle = (x + 3y)x + y^2 = x^2 + 3xy + y^2.

At $(1, -1)$ this is $1 - 3 + 1 = -1 \lt 0$ , so $T$ fails the second requirement as well.

The eigenvalue trap: $T(x, y) = \lambda(x, y)$ forces $y = \lambda y$ and $x + 3y = \lambda x$ ; if $y \ne 0$ then $\lambda = 1$ and $3y = 0$ , a contradiction, so $y = 0$ and $\lambda = 1$ . A nonnegative eigenvalue says nothing until $T = T^*$ is known.

Problem 3 · Building $R^*R$

Given: $R \in \mathcal{L}(\mathbf{R}^2)$ with $\mathcal{M}(R) = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ in the standard basis, and $T = R^*R$ .

What is $\mathcal{M}(T)$ ?

What is $\langle T(3, -1), (3, -1)\rangle$ ?

✅ Correct!

\langle Tv, v\rangle = \|Rv\|^2 = \|(2, -1)\|^2 = 5

. In general

\|R(x, y)\|^2 = (x + y)^2 + y^2 \ge 0

, so

T

is positive by condition (6), even though

R

itself is not self-adjoint.

❌ That is $RR^*$ , the product in the wrong order.

\mathcal{M}(R)\mathcal{M}(R)^{\mathrm{t}} = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}

;

T = R^*R

puts the adjoint on the left.

❌ That is $R^2$ .

R^2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}

is not even self-adjoint; condition (6) uses

R^*R

, which never equals

R^2

here.

❌ That is $R^*$ alone.

\mathcal{M}(R^*)

is the transpose

\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}

; it still has to be multiplied by

\mathcal{M}(R)

❌ That uses $RR^*$ .

\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}(3, -1) = (5, 2)

and

\langle (5, 2), (3, -1)\rangle = 13

, but

T = R^*R

, so

T(3, -1) = (2, 1)

❌ That uses $R^2$ .

R^2(3, -1) = (1, -1)

gives

\langle (1, -1), (3, -1)\rangle = 4

; the operator here is

R^*R

, with

T(3, -1) = (2, 1)

❌ That is $\|v\|^2 = 9 + 1$ . It is

\langle Iv, v\rangle

; the question asks for

\langle Tv, v\rangle = \|Rv\|^2

with

Rv = (2, -1)

Show solution

Step 1: the matrix. In the orthonormal standard basis $\mathcal{M}(R^*)$ is the transpose of $\mathcal{M}(R)$ :

\mathcal{M}(T) = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}.

Each entry is an inner product of columns of $\mathcal{M}(R)$ : the columns are $(1, 0)$ and $(1, 1)$ , with $\|(1, 0)\|^2 = 1$ , $\langle (1, 1), (1, 0)\rangle = 1$ , $\|(1, 1)\|^2 = 2$ .

Step 2: the inner product. $T(3, -1) = (3 - 1,\; 3 - 2) = (2, 1)$ , so

\langle T(3, -1), (3, -1)\rangle = 6 - 1 = 5.

Check with condition (6): $R(3, -1) = (3 - 1,\; -1) = (2, -1)$ and $\|(2, -1)\|^2 = 4 + 1 = 5$ .

Since $\langle Tv, v\rangle = \|Rv\|^2 \ge 0$ for every $v$ and $T^* = (R^*R)^* = R^*R = T$ , the operator $T$ is positive. Note $R^2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \ne T$ , so $R$ is not a square root of $T$ .

Problem 4 · Which Condition Falls Short?

Given: $T$ is an operator on a finite-dimensional inner product space $V$ . Which statement does not guarantee that $T$ is positive?

✅ Correct! On

\mathbf{R}^2

take

R(x, y) = (-y, x)

. Then

R^2(x, y) = R(-y, x) = (-x, -y)

, so

R^2 = -I

, and

\langle -Iv, v\rangle = -\|v\|^2 \lt 0

for

v \ne 0

. A square root alone guarantees nothing; the theorem needs a positive or a self-adjoint one.

❌ That is condition (6).

(R^*R)^* = R^*R

and

\langle R^*Rv, v\rangle = \langle Rv, Rv\rangle = \|Rv\|^2 \ge 0

, so it does guarantee positivity.

❌ That is condition (5). With

R = R^*

(R^2)^* = R^2

and

\langle R^2v, v\rangle = \langle Rv, R^*v\rangle = \|Rv\|^2 \ge 0

, so

T

is positive.

❌ That is condition (3). A real diagonal matrix in an orthonormal basis gives

T = T^*

, and for

v = \sum c_k e_k

\langle Tv, v\rangle = \sum \lambda_k |c_k|^2 \ge 0

Show solution

Options (a), (c) and (d) are conditions (6), (5) and (3) of the theorem, each equivalent to $T$ being positive.

Option (b) drops the requirement on $R$ . The rotation $R(x, y) = (-y, x)$ of $\mathbf{R}^2$ by a right angle satisfies

R^2(x, y) = R(-y, x) = (-x, -y), \qquad \text{so } R^2 = -I.

Then $T = -I$ has a square root, yet $\langle Tv, v\rangle = -\|v\|^2 \lt 0$ for every $v \ne 0$ , so $T$ is not positive. The analogy with numbers: $-1 = i^2$ has a square root, just not a real or nonnegative one.

Solved: 0 / 4

Positive Operators: Six Equivalent Characterizations

IKey Formulas

IIThe Inequality in Every Direction

IIIEigenvalues Bound ⟨Tv,v⟩\langle Tv, v\rangle⟨Tv,v⟩

IVCondition (6): T=R∗RT = R^*RT=R∗R

VQuiz Questions

IIIEigenvalues Bound $\langle Tv, v\rangle$

IVCondition (6): $T = R^*R$