Linear-Algebra-Done-Right · Unit 20 · Video 4 · Interactive Practice

Cholesky Factorization: Positive Definite $B = R^*R$

IKey Formulas

Formula	Name	What it says
$T \in \mathcal{L}(V)$ self-adjoint: $T$ positive and invertible $\iff \langle Tv, v\rangle > 0$ for every $v \neq 0$	Positive invertible test	Uses: $T$ positive and $\langle Tv, v\rangle = 0$ $\Rightarrow$ $Tv = 0$ ; injective operators on finite-dimensional $V$ are invertible
$B \in \mathbf{F}^{n,n}$ , $B^* = B$ and $\langle Bx, x\rangle > 0$ for every $x \neq 0$	Positive definite matrix	A positive invertible operator on $\mathbf{F}^n$ (Euclidean inner product), written as a matrix; the signs of the entries decide nothing
$B$ positive definite $\Rightarrow$ $B = R^*R$ for exactly one upper-triangular $R \in \mathbf{F}^{n,n}$ with positive diagonal	Cholesky factorization	$R^*$ is lower triangular, so $B$ is a lower-triangular matrix times its own conjugate transpose
$B = A^A$ , $A = QR$ ( $Q$ unitary) $\Rightarrow$ $B = R^Q^QR = R^R$	Existence proof	Any $A$ with $A^*A = B$ works; $A$ is invertible, so its QR factorization exists, and then $\langle Bx, x\rangle = \\|Rx\\|^2$

Key Insight: If $D$ is diagonal with $|d_k| = 1$ , then $(DR)^*(DR) = R^*D^*DR = R^*R$ , so triangular factors with some negative (or non-real) diagonal entries always exist; requiring a positive diagonal forces $D = I$ , and that is what makes $R$ unique.

IIPositive Definite Means $\langle Bx, x\rangle = \|Rx\|^2 > 0$

For real symmetric $B = \begin{pmatrix} a & b \\ b & c \end{pmatrix}$ , completing the square either writes $\langle Bx, x\rangle$ as $\|Rx\|^2$ or exhibits an $x \neq 0$ with $\langle Bx, x\rangle \le 0$ .

a

b

c

IIIEvery $A$ with $A^*A = B$ Yields the Same $R$

Each $A = QR$ with $Q$ unitary satisfies $A^*A = B$ , and Gram–Schmidt on its columns always returns the same $R$ .

Unitary

Q

θ 90°

IVQuiz Questions

Problem 1 · A $2 \times 2$ Cholesky Factor

Given: $B = \begin{pmatrix} 9 & 6 \\ 6 & 5 \end{pmatrix}$ , which is positive definite. Find its Cholesky factor: the upper-triangular $R$ with positive diagonal and $R^*R = B$ .

✅ Correct!

R^*R = \begin{pmatrix} 3 & 0 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} 3 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 9 & 6 \\ 6 & 5 \end{pmatrix}

, and the diagonal

3, 1

is positive.

❌ Not the Cholesky factor. This matrix does satisfy

R^*R = B

(it is

DR

with

D = \operatorname{diag}(1, -1)

), but its diagonal entry

-1

is negative; uniqueness holds only among factors with positive diagonal.

❌ Not quite. This is lower triangular; it is

R^*

, not

R

. Its own product

\begin{pmatrix} 3 & 2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 3 & 0 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 13 & 2 \\ 2 & 1 \end{pmatrix} \neq B

❌ Not quite. The

(2,2)

entry of

R^*R

r_{12}^2 + r_{22}^2

, not

r_{22}^2

alone: with

r_{22} = \sqrt{5}

it is

4 + 5 = 9 \neq 5

Show solution

Write $R = \begin{pmatrix} r_{11} & r_{12} \\ 0 & r_{22} \end{pmatrix}$ with $r_{11}, r_{22} > 0$ . Then

R^*R = \begin{pmatrix} r_{11} & 0 \\ r_{12} & r_{22} \end{pmatrix}\begin{pmatrix} r_{11} & r_{12} \\ 0 & r_{22} \end{pmatrix} = \begin{pmatrix} r_{11}^2 & r_{11}r_{12} \\ r_{11}r_{12} & r_{12}^2 + r_{22}^2 \end{pmatrix}.

Matching with $B$ : $r_{11}^2 = 9$ gives $r_{11} = 3$ ; $3r_{12} = 6$ gives $r_{12} = 2$ ; $4 + r_{22}^2 = 5$ gives $r_{22} = 1$ .

R = \begin{pmatrix} 3 & 2 \\ 0 & 1 \end{pmatrix}.

Equivalently, $\langle Bx, x\rangle = 9x_1^2 + 12x_1x_2 + 5x_2^2 = (3x_1 + 2x_2)^2 + x_2^2 = \|Rx\|^2$ .

Problem 2 · Which One Is Positive Definite?

Given: four real $2 \times 2$ matrices, with the Euclidean inner product on $\mathbf{R}^2$ . Which one is positive definite?

✅ Correct!

\langle Bx, x\rangle = 5x_1^2 - 8x_1x_2 + 5x_2^2 = 4(x_1 - x_2)^2 + x_1^2 + x_2^2 > 0

for

x \neq 0

, and

B^* = B

; negative entries do not matter.

❌ Positive entries are not enough. For

x = (1, -1)

Bx = (-3, 1)

and

\langle Bx, x\rangle = -3 - 1 = -4 < 0

❌ Not quite.

\langle Bx, x\rangle = (2x_1 + 3x_2)^2 \ge 0

, so this matrix is positive, but

x = (3, -2) \neq 0

gives

\langle Bx, x\rangle = 0

; it is not invertible, so it is not positive definite.

❌ Check the first condition. Here

\langle Bx, x\rangle = 3x_1^2 + 3x_1x_2 + 3x_2^2 > 0

for real

x \neq 0

, but

B^* = \begin{pmatrix} 3 & 2 \\ 1 & 3 \end{pmatrix} \neq B

; positive definite requires

B^* = B

Show solution

Positive definite needs both $B^* = B$ and $\langle Bx, x\rangle > 0$ for every $x \neq 0$ .

$\begin{pmatrix} 5 & -4 \\ -4 & 5 \end{pmatrix}$ : symmetric, and $5x_1^2 - 8x_1x_2 + 5x_2^2 = 4(x_1 - x_2)^2 + x_1^2 + x_2^2$ , which is $0$ only when $x_1 = x_2 = 0$ . Positive definite.
$\begin{pmatrix} 1 & 4 \\ 4 & 3 \end{pmatrix}$ : $x = (1, -1)$ gives $1 - 8 + 3 = -4$ . Fails.
$\begin{pmatrix} 4 & 6 \\ 6 & 9 \end{pmatrix}$ : $x = (3, -2)$ gives $36 - 72 + 36 = 0$ . Fails (positive, not invertible).
$\begin{pmatrix} 3 & 1 \\ 2 & 3 \end{pmatrix}$ : not self-adjoint. Fails.

Problem 3 · A $3 \times 3$ Cholesky Factor

Given: $B = \begin{pmatrix} 9 & 3 & 3 \\ 3 & 5 & 1 \\ 3 & 1 & 6 \end{pmatrix}$ and its Cholesky factor $R = \begin{pmatrix} r_{11} & r_{12} & r_{13} \\ 0 & r_{22} & r_{23} \\ 0 & 0 & r_{33} \end{pmatrix}$ , upper triangular with positive diagonal and $R^*R = B$ . The first row is $r_{11} = 3$ , $r_{12} = 1$ , $r_{13} = 1$ , and $r_{22} = 2$ . Find $r_{23}$ and $r_{33}$ .

What is $r_{23}$ ?

What is $r_{33}$ ?

✅ Correct!

R = \begin{pmatrix} 3 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 0 & \sqrt{5} \end{pmatrix}

, and multiplying out

R^*R

returns

B

❌ Not quite. The

(2,3)

entry of

R^*R

r_{12}r_{13} + r_{22}r_{23}

, not

r_{22}r_{23}

alone: subtract

r_{12}r_{13} = 1

from

b_{23} = 1

before dividing by

r_{22}

❌ Not quite.

r_{23} = b_{23}

ignores both the term

r_{12}r_{13}

and the factor

r_{22}

in the

(2,3)

entry

r_{12}r_{13} + r_{22}r_{23}

R^*R

❌ Not quite. Dividing

b_{23}

r_{11}

is the rule for the first row only; the

(2,3)

entry of

R^*R

r_{12}r_{13} + r_{22}r_{23}

❌ Check $r_{23}$ . Match the

(2,3)

entry:

r_{12}r_{13} + r_{22}r_{23} = b_{23}

❌ Not quite. The

(3,3)

entry of

R^*R

r_{13}^2 + r_{23}^2 + r_{33}^2

, so

r_{33}^2 = 6 - r_{13}^2 - r_{23}^2

, not

6

❌ Not quite. This uses

r_{23} = \tfrac{1}{2}

6 - 1 - \tfrac{1}{4} = \tfrac{19}{4}

. Recompute

r_{23}

from

r_{12}r_{13} + r_{22}r_{23} = 1

❌ Not quite.

5

r_{33}^2

; take the positive square root.

❌ Check $r_{33}$ . Match the

(3,3)

entry:

r_{13}^2 + r_{23}^2 + r_{33}^2 = b_{33}

Show solution

Column $j$ of $R$ is $(r_{1j}, \dots, r_{jj}, 0, \dots)$ , and the $(i,j)$ entry of $R^*R$ is the inner product of columns $i$ and $j$ of $R$ :

(R^*R)_{23} = r_{12}r_{13} + r_{22}r_{23}, \qquad (R^*R)_{33} = r_{13}^2 + r_{23}^2 + r_{33}^2.

Matching $b_{23} = 1$ : $1 \cdot 1 + 2r_{23} = 1$ , so $r_{23} = 0$ .

Matching $b_{33} = 6$ : $1 + 0 + r_{33}^2 = 6$ , so $r_{33}^2 = 5$ and $r_{33} = \sqrt{5}$ (positive diagonal).

R = \begin{pmatrix} 3 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 0 & \sqrt{5} \end{pmatrix}, \qquad R^*R = \begin{pmatrix} 9 & 3 & 3 \\ 3 & 5 & 1 \\ 3 & 1 & 6 \end{pmatrix} = B.

Since $R$ is upper triangular with nonzero diagonal it is invertible, so $\langle Bx, x\rangle = \|Rx\|^2 > 0$ for $x \neq 0$ : $B$ is indeed positive definite.

Problem 4 · How Many Triangular Factors?

Given: the matrix $B = \begin{pmatrix} 9 & 3 & 3 \\ 3 & 5 & 1 \\ 3 & 1 & 6 \end{pmatrix}$ from Problem 3. How many upper-triangular matrices $S$ with real entries satisfy $S^*S = B$ , if the sign condition on the diagonal is dropped?

✅ Correct! They are exactly

S = DR

with

D = \operatorname{diag}(\pm 1, \pm 1, \pm 1)

2^3 = 8

matrices.

❌ Not quite. Uniqueness needs the positive diagonal. Without it, flipping the sign of any row of

R

gives another factor, e.g.

\operatorname{diag}(-1, 1, 1)\,R

❌ Not quite. Each of the three rows can be flipped independently, and any combination of flips works, including none and all three:

2 \cdot 2 \cdot 2

choices.

❌ Over $\mathbf{C}$ that would be right (any

|d_k| = 1

works), but a real

S

forces

d_k = \pm 1

, which leaves finitely many.

Show solution

Every $DR$ works. If $D$ is diagonal with entries $\pm 1$ , then $DR$ is upper triangular and $(DR)^*(DR) = R^*D^*DR = R^*R = B$ because $D^*D = I$ .

Nothing else works. Let $S$ be real upper triangular with $S^*S = B$ . Since $B$ is invertible, so is $S$ , so its diagonal entries are nonzero. Let $D$ be diagonal with $d_k = \pm 1$ the sign of $s_{kk}$ . Then $DS$ is upper triangular with positive diagonal and $(DS)^*(DS) = S^*D^*DS = S^*S = B$ . By uniqueness of the Cholesky factor, $DS = R$ , so $S = D^{-1}R = DR$ .

Three independent sign choices give $2^3 = 8$ matrices $S$ , all distinct because their diagonals differ.

Solved: 0 / 4

Cholesky Factorization: Positive Definite B=R∗RB = R^*RB=R∗R

IKey Formulas

IIPositive Definite Means ⟨Bx,x⟩=∥Rx∥2>0\langle Bx, x\rangle = \|Rx\|^2 > 0⟨Bx,x⟩=∥Rx∥2>0

IIIEvery AAA with A∗A=BA^*A = BA∗A=B Yields the Same RRR

IVQuiz Questions

Cholesky Factorization: Positive Definite $B = R^*R$

IIPositive Definite Means $\langle Bx, x\rangle = \|Rx\|^2 > 0$

IIIEvery $A$ with $A^*A = B$ Yields the Same $R$