Linear-Algebra-Done-Right · Unit 19 · Video 4 · Interactive Practice

The Unique Positive Square Root $\sqrt{T}$

IKey Formulas

Formula	Name	What it says
$R$ positive, $R^2 = T$ , $Tv = \lambda v$ $\Rightarrow$ $Rv = \sqrt{\lambda}\, v$	Uniqueness	Every positive square root of $T$ acts the same way on each eigenvector of $T$ , so there is only one
$a_k(\lambda - \lambda_k) = 0$ for $k = 1, \dots, n$	Key step	Write $v = a_1 e_1 + \cdots + a_n e_n$ in an orthonormal eigenbasis of $R$ with $Re_k = \sqrt{\lambda_k}\, e_k$ and compare $\lambda v = R^2 v$ : only terms with $\lambda_k = \lambda$ survive
$Te_k = \lambda_k e_k \Rightarrow \sqrt{T}\, e_k = \sqrt{\lambda_k}\, e_k$	Recipe for $\sqrt{T}$	Keep an orthonormal eigenbasis of $T$ , replace each eigenvalue by its nonnegative square root
$\langle Tv, v\rangle = \\|\sqrt{T}\, v\\|^2$	Application	For $T$ positive, $\langle Tv, v\rangle = 0 \Rightarrow \sqrt{T}\, v = 0 \Rightarrow Tv = 0$

Key Insight: Square roots need not be unique: every reflection $R_\theta = \begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{pmatrix}$ satisfies $R_\theta^2 = I$ . Positivity is what singles out one: $\sqrt{I} = I$ .

IIMany Square Roots of $I$ , One Positive

Every reflection of $\mathbf{R}^2$ , and also $-I$ , squares to $I$ , but each has eigenvalue $-1$ ; only $I$ is positive.

Square root of

I

IIIThe Recipe: Keep the Eigenvectors, Take Roots of the Eigenvalues

$\sqrt{T}$ shares the orthonormal eigenvectors of $T$ and stretches by $\sqrt{\lambda_k}$ where $T$ stretches by $\lambda_k$ .

Positive operator

\lambda_1

\lambda_2

💡 Entrywise square roots work only when $\mathcal{M}(T)$ is diagonal; in general $\sqrt{T}$ is computed through eigenvectors and eigenvalues, never through entries.

IV $\langle Tv, v\rangle = \|\sqrt{T}\, v\|^2$

For positive $T$ , $\langle Tv, v\rangle$ is a squared norm, so it vanishes only where $Tv = 0$ ; self-adjointness alone is not enough.

Operator on

\mathbf{R}^2

VQuiz Questions

Problem 1 · The Square Root of a Diagonal Operator

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ , $T(x, y) = (9x, 4y)$ , with the Euclidean inner product. Find the matrix of $\sqrt{T}$ in the standard basis.

✅ Correct! The standard basis is an orthonormal eigenbasis with eigenvalues

9

and

4

; replacing them by

3

and

2

gives the unique positive square root.

❌ Not quite. This matrix does square to

\begin{pmatrix} 9 & 0 \\ 0 & 4 \end{pmatrix}

, but it has eigenvalue

-2

, so it sends

(0,1)

(0,-2)

, giving

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

: it is not positive.

\sqrt{T}

uses the nonnegative root of each eigenvalue.

❌ Not quite. This is the matrix of

T^2

, not of

\sqrt{T}

: its square is

\begin{pmatrix} 6561 & 0 \\ 0 & 256 \end{pmatrix}

, not

\mathcal{M}(T)

. Each eigenvalue is replaced by its square root, not its square.

❌ Not quite. This is

-\sqrt{T}

. It squares to

\mathcal{M}(T)

, but its eigenvalues

-3

and

-2

are negative, so it is not a positive operator.

Show solution

The standard basis $(1,0), (0,1)$ is orthonormal and consists of eigenvectors of $T$ : $T(1,0) = 9 \cdot (1,0)$ and $T(0,1) = 4 \cdot (0,1)$ . The recipe keeps these eigenvectors and replaces each eigenvalue by its nonnegative square root:

\sqrt{T}(1,0) = 3 \cdot (1,0), \qquad \sqrt{T}(0,1) = 2 \cdot (0,1), \qquad \mathcal{M}(\sqrt{T}) = \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix}.

Check: its square is $\begin{pmatrix} 9 & 0 \\ 0 & 4 \end{pmatrix}$ , it is symmetric, and its eigenvalues $3, 2$ are nonnegative, so it is positive. The matrices $\begin{pmatrix} \pm 3 & 0 \\ 0 & \pm 2 \end{pmatrix}$ with a minus sign also square to $\mathcal{M}(T)$ , but none of them is positive.

Problem 2 · Entrywise Roots Are a Trap

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ with standard matrix $\begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix}$ (Euclidean inner product). It has eigenvalues $9$ and $1$ . Find $\mathcal{M}(\sqrt{T})$ .

✅ Correct!

\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}^2 = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix}

, and the matrix is symmetric with eigenvalues

3

and

1

, so it is the positive square root.

❌ Not quite. Taking square roots entry by entry fails off the diagonal: this matrix squares to

\begin{pmatrix} 9 & 4\sqrt{5} \\ 4\sqrt{5} & 9 \end{pmatrix} \ne \mathcal{M}(T)

. The square root acts on eigenvalues, not entries.

❌ Not quite. The eigenvalue roots

3

and

1

are right, but they belong to the eigenvectors

(1,1)

and

(1,-1)

, not to the standard basis. This diagonal matrix squares to

\begin{pmatrix} 9 & 0 \\ 0 & 1 \end{pmatrix} \ne \mathcal{M}(T)

❌ Not quite. This matrix does square to

\mathcal{M}(T)

, but its eigenvalues are

-3

and

-1

, so it sends

(1,1)

(-3,-3)

and

\langle (-3,-3), (1,1)\rangle = -6 \lt 0

: it is a square root of

T

that is not positive.

Show solution

Eigenvectors. $T(1,1) = (9,9) = 9(1,1)$ and $T(1,-1) = (1,-1)$ . Normalized, $u_1 = \tfrac{1}{\sqrt{2}}(1,1)$ and $u_2 = \tfrac{1}{\sqrt{2}}(1,-1)$ form an orthonormal eigenbasis with eigenvalues $9$ and $1$ .

Recipe. $\sqrt{T}\, u_1 = 3u_1$ and $\sqrt{T}\, u_2 = u_2$ . Since $(1,0) = \tfrac{1}{\sqrt{2}}(u_1 + u_2)$ and $(0,1) = \tfrac{1}{\sqrt{2}}(u_1 - u_2)$ ,

\sqrt{T}(1,0) = \tfrac{1}{\sqrt{2}}(3u_1 + u_2) = \tfrac{1}{2}(3,3) + \tfrac{1}{2}(1,-1) = (2, 1),

\sqrt{T}(0,1) = \tfrac{1}{\sqrt{2}}(3u_1 - u_2) = \tfrac{1}{2}(3,3) - \tfrac{1}{2}(1,-1) = (1, 2).

So $\mathcal{M}(\sqrt{T}) = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ . Check: $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}^2 = \begin{pmatrix} 4 + 1 & 2 + 2 \\ 2 + 2 & 1 + 4 \end{pmatrix} = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix}$ .

Problem 3 · From Eigenvectors to the Matrix of $\sqrt{T}$

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ with standard matrix $\begin{pmatrix} 10 & 6 \\ 6 & 10 \end{pmatrix}$ (Euclidean inner product).

Which eigenvalue goes with which eigenvector?

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

✅ Correct!

\sqrt{T}

multiplies

(1,1)

4

and

(1,-1)

2

, which is the matrix

\begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}

, and its square is

\begin{pmatrix} 10 & 6 \\ 6 & 10 \end{pmatrix}

❌ Not quite. The pairing is swapped:

T(1,1) = (10 + 6, 6 + 10) = (16, 16)

, so

(1,1)

has eigenvalue

16

, and

T(1,-1) = (4, -4)

❌ Not quite. The diagonal entries are not eigenvalues of a non-diagonal matrix:

T(1,0) = (10, 6)

, which is not a multiple of

(1,0)

❌ Not quite. This matrix multiplies

(1,1)

2

and

(1,-1)

4

, so the eigenvalue roots were attached to the wrong eigenvectors; its square is

\begin{pmatrix} 10 & -6 \\ -6 & 10 \end{pmatrix} \ne \mathcal{M}(T)

❌ Not quite.

4

and

2

are the right eigenvalues of

\sqrt{T}

, but this matrix puts them on the standard basis instead of on

(1,1)

and

(1,-1)

; its square is

\begin{pmatrix} 16 & 0 \\ 0 & 4 \end{pmatrix} \ne \mathcal{M}(T)

❌ Not quite. Entrywise square roots fail here: the top-left entry of the square of this matrix is

(\sqrt{10})^2 + (\sqrt{6})^2 = 10 + 6 = 16 \ne 10

. The square root acts on eigenvalues, not entries.

Show solution

Eigenvectors. $T(1,1) = (16,16) = 16(1,1)$ and $T(1,-1) = (4,-4) = 4(1,-1)$ . The unit vectors $u_1 = \tfrac{1}{\sqrt{2}}(1,1)$ and $u_2 = \tfrac{1}{\sqrt{2}}(1,-1)$ are orthogonal, so they form an orthonormal eigenbasis.

Recipe. $\sqrt{T}\, u_1 = 4u_1$ and $\sqrt{T}\, u_2 = 2u_2$ . With $(1,0) = \tfrac{1}{\sqrt{2}}(u_1 + u_2)$ and $(0,1) = \tfrac{1}{\sqrt{2}}(u_1 - u_2)$ :

\sqrt{T}(1,0) = \tfrac{1}{2}(4,4) + \tfrac{1}{2}(2,-2) = (3, 1), \qquad \sqrt{T}(0,1) = \tfrac{1}{2}(4,4) - \tfrac{1}{2}(2,-2) = (1, 3).

So $\mathcal{M}(\sqrt{T}) = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$ . Check: $\begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}^2 = \begin{pmatrix} 9 + 1 & 3 + 3 \\ 3 + 3 & 1 + 9 \end{pmatrix} = \begin{pmatrix} 10 & 6 \\ 6 & 10 \end{pmatrix}$ , and the matrix is symmetric with eigenvalues $4, 2 \geq 0$ .

Problem 4 · A Square Root of $I$ That Is Not $\sqrt{I}$

Given: $R \in \mathcal{L}(\mathbf{R}^2)$ , $R(x, y) = (y, x)$ , the reflection across the line $y = x$ . Then $R^2(x,y) = R(y,x) = (x,y)$ , so $R^2 = I$ . Which statement is correct?

✅ Correct!

R

is self-adjoint with eigenvalues

1

(1,1)

and

-1

(1,-1)

; the eigenvalue

-1

rules out positivity, and

\sqrt{I} = I

because every eigenvalue of

I

1

❌ Not quite.

\sqrt{I}

is the unique positive square root, and self-adjoint is weaker than positive:

R(1,-1) = (-1,1) = -(1,-1)

, so

R

has eigenvalue

-1

❌ Not quite.

\mathcal{M}(R) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

equals its transpose, so

R

is self-adjoint. What fails is the inequality

\langle Rv, v\rangle \geq 0

❌ Not quite.

I

does have infinitely many square roots (every reflection is one), but exactly one of them is positive. The uniqueness theorem is about positive square roots, so

\sqrt{I}

is well defined.

Show solution

$\mathcal{M}(R) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ is symmetric, so $R$ is self-adjoint, and $R^2 = I$ . Its eigenvectors are $(1,1)$ with eigenvalue $1$ and $(1,-1)$ with eigenvalue $-1$ , and

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

so $R$ is not positive. By the theorem, $I$ has exactly one positive square root; since every eigenvalue of $I$ is $1$ and $\sqrt{1} = 1$ , the recipe gives $\sqrt{I} = I$ . The reflection $R$ (like $-I$ and every other reflection) is one of the infinitely many square roots of $I$ that are not positive.

Solved: 0 / 4

The Unique Positive Square Root T\sqrt{T}T​

IKey Formulas

IIMany Square Roots of III, One Positive

IIIThe Recipe: Keep the Eigenvectors, Take Roots of the Eigenvalues

IV⟨Tv,v⟩=∥T v∥2\langle Tv, v\rangle = \|\sqrt{T}\, v\|^2⟨Tv,v⟩=∥T​v∥2

VQuiz Questions

The Unique Positive Square Root $\sqrt{T}$

IIMany Square Roots of $I$ , One Positive

IV $\langle Tv, v\rangle = \|\sqrt{T}\, v\|^2$