Linear-Algebra-Done-Right · Unit 20 · Video 3 · Interactive Practice

QR Factorization: Gram–Schmidt Recorded as A = QR

IKey Formulas

Formula	Name	When it holds
$Q^Q = QQ^ = I$	Unitary matrix	$Q$ is $n$ -by- $n$ with orthonormal columns (equivalently, orthonormal rows)
$R_{j,k} = \langle v_k, e_j\rangle$	Entries of $R$	$v_1, \dots, v_n$ are the linearly independent columns of $A$ ; $e_1, \dots, e_n$ is their Gram–Schmidt output and the columns of $Q$
$R_{k,k} = \\|f_k\\| > 0$ , where $f_k = v_k - \sum_{j<k} \langle v_k, e_j\rangle e_j$	Positive diagonal	Same hypotheses; $f_k \neq 0$ because $v_k \notin \operatorname{span}(v_1, \dots, v_{k-1})$
$Ax = b \iff Rx = Q^*b$	Solving with QR	$A = QR$ with $Q$ unitary and $R$ upper triangular with positive diagonal

Key Insight: Gram–Schmidt and $A = QR$ are one computation: $Q$ stores the vectors $e_k$ , $R$ stores the coefficients, and requiring $R_{k,k} > 0$ makes the pair unique.

IIGram–Schmidt Fills In $R$

Gram–Schmidt turns the columns $v_1, v_2$ of $A$ into $e_1, e_2$ ; the coefficients it computes are the entries of $R$ .

IIIWhy the Diagonal Must Be Positive

Every unitary $\widehat{Q}$ gives $A = \widehat{Q}\widehat{R}$ with $\widehat{R} = \widehat{Q}^*A$ ; only one makes $\widehat{R}$ upper triangular with positive diagonal.

Direction of

q_1

0°

💡 Over $\mathbf{C}$ the freedom is larger, $q_k = \zeta_k e_k$ with $|\zeta_k| = 1$ , and $\widehat{R}_{k,k} > 0$ still forces every $\zeta_k = 1$ .

IVSolving $Ax = b$ by Back Substitution

For the video's matrix $A = QR$ , the system $Ax = b$ becomes the triangular system $Rx = Q^*b$ , solved from the last row up.

b_1

b_2

−5

b_3

−1 Step 0

VQuiz Questions

Problem 1 · The Factor $R$ of a $2 \times 2$ Matrix

Given: $A = \begin{pmatrix} 3 & 1 \\ 4 & 2 \end{pmatrix}$ — find the upper triangular factor $R$ in its QR factorization.

✅ Correct!

R_{1,1} = \|v_1\| = 5

R_{1,2} = \langle v_2, e_1\rangle = \tfrac{11}{5}

and

R_{2,2} = \|f_2\| = \tfrac{2}{5}

❌ Rows and columns swapped.

R_{j,k} = \langle v_k, e_j\rangle

is zero when

j > k

, so the nonzero off-diagonal entry

\langle v_2, e_1\rangle

sits in row 1, column 2:

R

is upper triangular.

❌ Not normalized.

11 = \langle v_2, v_1\rangle

, but

R_{1,2} = \langle v_2, e_1\rangle

with the unit vector

e_1 = v_1/\|v_1\| = \tfrac{1}{5}(3, 4)

, so

R_{1,2} = \tfrac{11}{5}

❌ Wrong sign on the diagonal. Taking

e_2 = (\tfrac{4}{5}, -\tfrac{3}{5})

still gives a unitary

Q

, but then

R_{2,2} = -\tfrac{2}{5} < 0

. Gram–Schmidt's

e_2 = f_2/\|f_2\|

makes

R_{2,2} = \|f_2\| > 0

Show solution

The columns are $v_1 = (3, 4)$ and $v_2 = (1, 2)$ . Since $\|v_1\| = 5$ , $e_1 = \tfrac{1}{5}(3, 4)$ and $R_{1,1} = \langle v_1, e_1\rangle = 5$ .

R_{1,2} = \langle v_2, e_1\rangle = \tfrac{1}{5}(3 + 8) = \tfrac{11}{5}

f_2 = v_2 - \tfrac{11}{5}e_1 = (1, 2) - \big(\tfrac{33}{25}, \tfrac{44}{25}\big) = \big(-\tfrac{8}{25}, \tfrac{6}{25}\big), \qquad \|f_2\| = \tfrac{10}{25} = \tfrac{2}{5}

So $e_2 = f_2/\|f_2\| = (-\tfrac{4}{5}, \tfrac{3}{5})$ , $R_{2,1} = 0$ and $R_{2,2} = \tfrac{2}{5}$ :

Q = \begin{pmatrix} \tfrac{3}{5} & -\tfrac{4}{5} \\ \tfrac{4}{5} & \tfrac{3}{5} \end{pmatrix}, \qquad R = \begin{pmatrix} 5 & \tfrac{11}{5} \\ 0 & \tfrac{2}{5} \end{pmatrix}.

Check: column 2 of $QR$ is $\tfrac{11}{5}e_1 + \tfrac{2}{5}e_2 = \big(\tfrac{33}{25} - \tfrac{8}{25}, \tfrac{44}{25} + \tfrac{6}{25}\big) = (1, 2) = v_2$ .

Problem 2 · Already Triangular Is Not Enough

Given: $A = \begin{pmatrix} 2 & 1 \\ 0 & -5 \end{pmatrix}$ , which is already upper triangular — find the $Q$ and $R$ of its QR factorization.

✅ Correct!

Q

is unitary,

R

is upper triangular with diagonal

2, 5 > 0

, and

QR = A

; by uniqueness this is the only such pair.

❌ The diagonal is not positive.

IA = A

is a factorization, but

R_{2,2} = -5 < 0

, so it is not the QR factorization the theorem describes.

❌ $QR \neq A$ . This product is

\begin{pmatrix} 2 & -1 \\ 0 & -5 \end{pmatrix}

. Negating column 2 of

Q

must be matched by negating only row 2 of

R

;

R_{1,2} = \langle v_2, e_1\rangle = 1

is unchanged.

❌ Wrong sign in the first entry.

(-I)(-A) = A

, but now

R_{1,1} = -2 < 0

, while

R_{2,2} = 5

is fine. Only the second column of

Q

needs negating.

Show solution

Run Gram–Schmidt on $v_1 = (2, 0)$ , $v_2 = (1, -5)$ . First $e_1 = (1, 0)$ and $R_{1,1} = 2$ . Then $R_{1,2} = \langle v_2, e_1\rangle = 1$ and

f_2 = v_2 - 1 \cdot e_1 = (0, -5), \qquad \|f_2\| = 5, \qquad e_2 = (0, -1).

Q = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \qquad R = \begin{pmatrix} 2 & 1 \\ 0 & 5 \end{pmatrix}, \qquad QR = \begin{pmatrix} 2 & 1 \\ 0 & -5 \end{pmatrix} = A.

Taking $Q = I$ , $R = A$ is the factorization $(QD)(DR)$ with $D = \operatorname{diag}(1, -1)$ : still a unitary times an upper triangular matrix, but with a negative diagonal entry.

Problem 3 · A $3 \times 3$ Factorization

Given: $A = \begin{pmatrix} 1 & 3 & 5 \\ 2 & 3 & -5 \\ 2 & 0 & -2 \end{pmatrix}$ with columns $v_1, v_2, v_3$ , and $e_1 = \tfrac{1}{3}(1, 2, 2)$ — find $e_2$ and the third column of $R$ .

What is $e_2$ ?

What is the third column $(R_{1,3}, R_{2,3}, R_{3,3})$ of $R$ ?

✅ Correct!

e_2 = \tfrac{1}{3}(2, 1, -2)

and column 3 of

R

(\langle v_3, e_1\rangle, \langle v_3, e_2\rangle, \|f_3\|) = (-3, 3, 6)

❌ That is $f_2$ .

f_2 = v_2 - 3e_1 = (2, 1, -2)

has norm

3

; divide by it to get the unit vector

e_2

❌ The projection was not removed. This is

v_2/\|v_2\|

, which is not orthogonal to

e_1

\langle v_2, e_1\rangle = 3 \neq 0

. Subtract

3e_1

from

v_2

first.

❌ Wrong sign.

-e_2

is also a unit vector orthogonal to

e_1

, but then

R_{2,2} = \langle v_2, -e_2\rangle = -3 < 0

. Gram–Schmidt's

e_2 = f_2/\|f_2\|

points along

f_2 = (2, 1, -2)

❌ Used $f_2$ instead of $e_2$ .

\langle v_3, f_2\rangle = 9

, but

R_{2,3} = \langle v_3, e_2\rangle = 9/\|f_2\| = 3

❌ $R_{3,3}$ is $\|f_3\|$ , not $\|v_3\|$ .

\|v_3\| = \sqrt{54} = 3\sqrt{6}

; after removing the components along

e_1, e_2

f_3 = (4, -4, 2)

has norm

6

❌ That is the third row. Row 3 of

R

(0, 0, 6)

; the third column holds the coefficients of

v_3

along

e_1, e_2, e_3

Show solution

The columns are $v_1 = (1, 2, 2)$ , $v_2 = (3, 3, 0)$ , $v_3 = (5, -5, -2)$ , and $\|v_1\| = 3$ gives $e_1 = \tfrac{1}{3}(1, 2, 2)$ .

\langle v_2, e_1\rangle = \tfrac{1}{3}(3 + 6 + 0) = 3, \qquad f_2 = v_2 - 3e_1 = (2, 1, -2), \qquad \|f_2\| = 3, \qquad e_2 = \tfrac{1}{3}(2, 1, -2).

\langle v_3, e_1\rangle = \tfrac{1}{3}(5 - 10 - 4) = -3, \qquad \langle v_3, e_2\rangle = \tfrac{1}{3}(10 - 5 + 4) = 3,

f_3 = v_3 + 3e_1 - 3e_2 = (5, -5, -2) + (1, 2, 2) - (2, 1, -2) = (4, -4, 2), \qquad \|f_3\| = 6.

Q = \tfrac{1}{3}\begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{pmatrix}, \qquad R = \begin{pmatrix} 3 & 3 & -3 \\ 0 & 3 & 3 \\ 0 & 0 & 6 \end{pmatrix}.

Check column 3: $-3e_1 + 3e_2 + 6e_3 = (-1, -2, -2) + (2, 1, -2) + (4, -4, 2) = (5, -5, -2) = v_3$ , with $e_3 = \tfrac{1}{3}(2, -2, 1)$ .

Problem 4 · Solve Without Elimination

Given: the factorization $A = QR$ from Problem 3, $Q = \tfrac{1}{3}\begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{pmatrix}, \qquad R = \begin{pmatrix} 3 & 3 & -3 \\ 0 & 3 & 3 \\ 0 & 0 & 6 \end{pmatrix},$ and $b = (0, -9, 0)$ — find $Q^*b$ and the solution $x$ of $Ax = b$ .

What is $Q^*b$ ?

What is $x$ ?

✅ Correct!

Rx = Q^*b = (-6, -3, 6)

gives

x_3 = 1

x_2 = -2

x_1 = 1

, and indeed

A(1, -2, 1) = (0, -9, 0)

❌ That is $b$ itself.

Q^*b

lists the inner products

\langle b, e_1\rangle, \langle b, e_2\rangle, \langle b, e_3\rangle

❌ The factor $\tfrac{1}{3}$ was dropped. Every entry of

Q

carries

\tfrac{1}{3}

, so

\langle b, e_1\rangle = \tfrac{1}{3}(0 - 18 + 0) = -6

❌ Divided by $3$ twice. The columns

\tfrac{1}{3}(1, 2, 2)

, … are already unit vectors;

\langle b, e_1\rangle = \tfrac{1}{3}(-18) = -6

, with no further division.

❌ This solves $Rx = b$ . Multiplying

QRx = b

on the left by

Q^*

changes the right side to

Q^*b

; back substitution must use

(-6, -3, 6)

❌ Sign slip in the top row. Row 1 reads

3x_1 + 3x_2 - 3x_3 = -6

; with

x_2 = -2

x_3 = 1

this is

3x_1 - 9 = -6

, so

x_1 = 1

❌ The right side was divided by $3$ twice. Back substitution with

(-2, -1, 2)

gives this; the right side is

Q^*b = (-6, -3, 6)

, three times larger, and so is the solution.

Show solution

$Q$ is real, so $Q^*$ is its transpose, whose rows are $e_1, e_2, e_3$ :

Q^*b = \Big(\tfrac{1}{3}(0 - 18 + 0),\; \tfrac{1}{3}(0 - 9 + 0),\; \tfrac{1}{3}(0 + 18 + 0)\Big) = (-6, -3, 6).

Back substitution on $Rx = Q^*b$ , from the last row up:

6x_3 = 6 \Rightarrow x_3 = 1, \qquad 3x_2 + 3x_3 = -3 \Rightarrow x_2 = -2, \qquad 3x_1 + 3x_2 - 3x_3 = -6 \Rightarrow 3x_1 - 9 = -6 \Rightarrow x_1 = 1.

Check: $A(1, -2, 1) = (1 - 6 + 5,\; 2 - 6 - 5,\; 2 - 0 - 2) = (0, -9, 0) = b$ .

Solved: 0 / 4

QR Factorization: Gram–Schmidt Recorded as A = QR

IKey Formulas

IIGram–Schmidt Fills In RRR

IIIWhy the Diagonal Must Be Positive

IVSolving Ax=bAx = bAx=b by Back Substitution

VQuiz Questions

IIGram–Schmidt Fills In $R$

IVSolving $Ax = b$ by Back Substitution