Linear-Algebra-Done-Right · Unit 15 · Video 2 · Interactive Practice

Gram-Schmidt, Orthonormal Bases and Schur's Theorem

IKey Formulas

Formula	Name	Hypotheses
$f_1 = v_1$ , $\ f_k = v_k - \dfrac{\langle v_k, f_1 \rangle}{\\|f_1\\|^2} f_1 - \cdots - \dfrac{\langle v_k, f_{k-1} \rangle}{\\|f_{k-1}\\|^2} f_{k-1}$	Gram–Schmidt step	$v_1, \dots, v_m$ linearly independent in $V$ , $k = 2, \dots, m$
$e_k = \dfrac{f_k}{\\|f_k\\|}$ , $\ \operatorname{span}(v_1, \dots, v_k) = \operatorname{span}(e_1, \dots, e_k)$	Normalize; initial spans preserved	$f_k \neq 0$ because the $v$ 's are independent; $k = 1, \dots, m$
Every orthonormal list $e_1, \dots, e_m$ extends to an orthonormal basis of $V$	Existence and extension	$V$ a finite-dimensional inner product space
$\mathcal{M}\big(T, (e_1, \dots, e_n)\big)$ is upper triangular for some orthonormal basis	Schur's theorem	$V$ a finite-dimensional complex inner product space, $T \in \mathcal{L}(V)$

Key Insight: Gram–Schmidt keeps every initial span $\operatorname{span}(v_1, \dots, v_k)$ , so anything defined by those subspaces — such as their invariance under $T$ — passes from a basis to an orthonormal basis.

IIOne Gram–Schmidt Step in $\mathbf{R}^2$

Subtracting the component of $v_2$ along $f_1$ leaves a vector $f_2$ orthogonal to $f_1$ — unless $v_2 \in \operatorname{span}(v_1)$ .

IIIGram–Schmidt on $1, x, x^2$ in $\mathcal{P}_2(\mathbf{R})$

With $\langle p, q \rangle = \int_{-1}^{1} pq$ , every inner product in the procedure is a signed area.

Step 1

Step 1 — $f_1$ and its norm

f_1 = 1, \qquad \|f_1\|^2 = \int_{-1}^{1} 1\,dt = 2

IVWhy Gram–Schmidt Proves Schur's Theorem

On a two-dimensional space, $\mathcal{M}\big(T, (v_1, v_2)\big)$ is upper triangular exactly when $\operatorname{span}(v_1)$ is invariant, and Gram–Schmidt gives $\operatorname{span}(e_1) = \operatorname{span}(v_1)$ .

Operator

💡 Over $\mathbf{C}^2$ the rotation has eigenvalues $i$ and $-i$ , so its minimal polynomial $z^2 + 1 = (z - i)(z + i)$ splits and Schur's theorem applies.

VQuiz Questions

Problem 1 · One Gram–Schmidt Step

Given: In $\mathbf{R}^3$ with the Euclidean inner product, $v_1 = (1, 1, 0)$ and $v_2 = (1, 0, 1)$ . Find $f_2$ from the Gram–Schmidt procedure.

✅ Correct!

\langle v_2, f_1 \rangle = 1

and

\|f_1\|^2 = 2

, so

f_2 = v_2 - \tfrac{1}{2} f_1

, and indeed

\langle f_2, f_1 \rangle = \tfrac12 - \tfrac12 + 0 = 0

❌ Not quite. You subtracted

\langle v_2, f_1 \rangle f_1

without dividing by

\|f_1\|^2 = 2

; that formula is only right when

f_1

is a unit vector. Check:

\langle (0, -1, 1), (1, 1, 0) \rangle = -1 \neq 0

❌ Not quite. Nothing is subtracted only when

\langle v_2, f_1 \rangle = 0

, but here

\langle v_2, f_1 \rangle = 1

❌ Close, but… the coefficient divides by

\|f_1\|^2

, not

\|f_1\|

. The component of

v_2

along

f_1

\dfrac{\langle v_2, f_1 \rangle}{\|f_1\|^2} f_1

❌ Not quite. Use

f_2 = v_2 - \dfrac{\langle v_2, f_1 \rangle}{\|f_1\|^2} f_1

with

f_1 = v_1

Show solution

Set $f_1 = v_1 = (1, 1, 0)$ , so $\|f_1\|^2 = 1 + 1 + 0 = 2$ . Next $\langle v_2, f_1 \rangle = 1\cdot 1 + 0 \cdot 1 + 1 \cdot 0 = 1$ . Therefore

f_2 = v_2 - \frac{\langle v_2, f_1 \rangle}{\|f_1\|^2} f_1 = (1, 0, 1) - \tfrac{1}{2}(1, 1, 0) = \big(\tfrac{1}{2}, -\tfrac{1}{2}, 1\big).

Check: $\langle f_2, f_1 \rangle = \tfrac{1}{2} - \tfrac{1}{2} + 0 = 0$ . Normalizing, $\|f_2\|^2 = \tfrac14 + \tfrac14 + 1 = \tfrac32$ , so $e_1 = \tfrac{1}{\sqrt{2}}(1, 1, 0)$ and $e_2 = \tfrac{1}{\sqrt{6}}(1, -1, 2)$ .

Problem 2 · Where Linear Independence Is Used

Given: In $\mathbf{R}^3$ with the Euclidean inner product, run the Gram–Schmidt formulas on $v_1 = (1, 0, 1)$ , $v_2 = (2, 1, 2)$ , $v_3 = (0, 1, 0)$ . What happens at $k = 3$ ?

✅ Correct!

v_3 = v_2 - 2v_1

lies in

\operatorname{span}(v_1, v_2) = \operatorname{span}(f_1, f_2)

, so subtracting its components along

f_1

and

f_2

leaves

0

❌ Not quite. You subtracted only the component along

f_1

. The formula also subtracts the component along

f_2 = (0, 1, 0)

, and

\langle v_3, f_2 \rangle = 1

❌ Close, but…

f_3 = 0

is right, but

e_3 = f_3 / \|f_3\|

divides by

\|f_3\| = 0

, so it is undefined, not

0

. And a list containing

0

cannot be orthonormal: every vector in it must have norm

1

❌ Not quite.

\tfrac{1}{\sqrt 2}(1, 0, -1)

would extend

e_1, e_2

to an orthonormal basis of

\mathbf{R}^3

, but it is not what the formula produces from

v_3

: compute

f_3

first.

❌ Not quite. Compute

f_2

first, then

f_3 = v_3 - \frac{\langle v_3, f_1 \rangle}{\|f_1\|^2} f_1 - \frac{\langle v_3, f_2 \rangle}{\|f_2\|^2} f_2

Show solution

$f_1 = (1, 0, 1)$ , $\|f_1\|^2 = 2$ , $\langle v_2, f_1 \rangle = 4$ , so $f_2 = (2, 1, 2) - \tfrac{4}{2}(1, 0, 1) = (0, 1, 0)$ with $\|f_2\|^2 = 1$ .

Then $\langle v_3, f_1 \rangle = 0$ and $\langle v_3, f_2 \rangle = 1$ , so

f_3 = (0, 1, 0) - \tfrac{0}{2}(1, 0, 1) - \tfrac{1}{1}(0, 1, 0) = (0, 0, 0).

Dividing by $\|f_3\| = 0$ is impossible. This is exactly the case the proof rules out: $v_3 = v_2 - 2v_1 \in \operatorname{span}(v_1, v_2)$ , so the list is linearly dependent, and the first vector in the span of the earlier ones gives $f_k = 0$ .

Problem 3 · The Example Moved to $[0, 1]$

Given: $\mathcal{P}_2(\mathbf{R})$ with $\langle p, q \rangle = \int_{0}^{1} pq$ , and the basis $1, x, x^2$ . Find $f_2$ , $e_2$ and $f_3$ from the Gram–Schmidt procedure.

What is $f_2$ ?

What is $e_2$ ?

What is $f_3$ ?

✅ Correct! On

[0, 1]

nothing is odd, so every coefficient has to be computed:

f_2 = x - \tfrac12

\|f_2\|^2 = \tfrac{1}{12}

, and

f_3 = x^2 - x + \tfrac16

❌ Close, but…

\langle x, 1 \rangle = 0

came from odd symmetry on

[-1, 1]

. On

[0, 1]

\langle x, 1 \rangle = \int_0^1 t\,dt = \tfrac12

and

\|1\|^2 = 1

❌ Check $f_2$ .

f_2 = x - \dfrac{\langle x, 1 \rangle}{\|1\|^2}\cdot 1

with

\langle x, 1 \rangle = \int_0^1 t\,dt

and

\|1\|^2 = \int_0^1 1\,dt

❌ Not quite.

\sqrt{3/2}

normalizes

x

[-1, 1]

. Here

\|f_2\|^2 = \int_0^1 \big(t - \tfrac12\big)^2 dt = \tfrac{1}{12}

❌ Check $e_2$ . Divide

f_2

\|f_2\| = \sqrt{1/12}

— by the norm, not the norm squared — which multiplies by

\sqrt{12} = 2\sqrt{3}

❌ Not quite.

x^2 - \tfrac13

is the answer on

[-1, 1]

. On

[0, 1]

\langle x^2, f_2 \rangle = \tfrac{1}{12} \neq 0

, so a multiple of

f_2

must also be subtracted.

❌ Close, but… you subtracted

\langle x^2, f_2 \rangle f_2

without dividing by

\|f_2\|^2 = \tfrac{1}{12}

. The coefficient is

\tfrac{1/12}{1/12} = 1

❌ Check $f_3$ .

f_3 = x^2 - \dfrac{\langle x^2, 1 \rangle}{\|f_1\|^2}\cdot 1 - \dfrac{\langle x^2, f_2 \rangle}{\|f_2\|^2} f_2

; verify your answer integrates to

0

against

1

and against

x

Show solution

$f_1$ . $f_1 = 1$ , $\|f_1\|^2 = \int_0^1 1\,dt = 1$ .

$f_2$ . $\langle x, 1 \rangle = \int_0^1 t\,dt = \tfrac12$ , so $f_2 = x - \tfrac{1/2}{1}\cdot 1 = x - \tfrac12$ .

$e_2$ .

\|f_2\|^2 = \int_0^1 \Big(t - \tfrac12\Big)^2 dt = \tfrac13 - \tfrac12 + \tfrac14 = \tfrac{1}{12}, \qquad e_2 = \frac{f_2}{\|f_2\|} = \sqrt{12}\Big(x - \tfrac12\Big) = 2\sqrt{3}\Big(x - \tfrac12\Big).

$f_3$ . $\langle x^2, 1 \rangle = \tfrac13$ , and $\langle x^2, f_2 \rangle = \int_0^1 \big(t^3 - \tfrac12 t^2\big)dt = \tfrac14 - \tfrac16 = \tfrac{1}{12}$ , so

f_3 = x^2 - \frac{1/3}{1}\cdot 1 - \frac{1/12}{1/12}\Big(x - \tfrac12\Big) = x^2 - x + \tfrac16.

Check: $\int_0^1 \big(t^2 - t + \tfrac16\big)dt = \tfrac13 - \tfrac12 + \tfrac16 = 0$ and $\int_0^1 t\big(t^2 - t + \tfrac16\big)dt = \tfrac14 - \tfrac13 + \tfrac{1}{12} = 0$ .

Problem 4 · Schur's Proof by Hand

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ , $T(x, y) = (3x, x + 2y)$ , with the Euclidean inner product. Its matrix with respect to the standard basis is $\begin{pmatrix} 3 & 0 \\ 1 & 2 \end{pmatrix}$ , which is not upper triangular. Find an orthonormal basis $e_1, e_2$ that makes $\mathcal{M}(T)$ upper triangular, and that matrix.

Which unit vector $e_1$ has $\operatorname{span}(e_1)$ invariant under $T$ ?

With $e_2 = \tfrac{1}{\sqrt{2}}(1, -1)$ , what is $\mathcal{M}\big(T, (e_1, e_2)\big)$ ?

✅ Correct!

e_1

is an eigenvector (

Te_1 = 3e_1

), so

\operatorname{span}(e_1)

is invariant and the matrix is upper triangular, with the eigenvalues

3, 2

on the diagonal.

❌ Not quite.

T(1, 0) = (3, 1)

is not a multiple of

(1, 0)

, which is why the standard matrix has the entry

1

below the diagonal.

❌ Check $e_1$ .

\operatorname{span}(e_1)

is invariant exactly when

Te_1

is a multiple of

e_1

; apply

T

to each candidate.

❌ Not quite. That is the matrix with respect to the standard basis. The entries now are

\langle Te_j, e_i \rangle

in the new basis.

❌ Close, but… the first diagonal entry is the eigenvalue of

e_1

Te_1 = 3e_1

, so it is

3

, not

2

❌ Close, but… check the sign of

\langle Te_2, e_1 \rangle

Te_2 = \tfrac{1}{\sqrt2}(3, -1)

and

e_1 = \tfrac{1}{\sqrt2}(1, 1)

give

\tfrac{3 - 1}{2} = 1

. The value

-1

belongs to

e_2 = \tfrac{1}{\sqrt2}(-1, 1)

❌ Check the matrix. Column

j

holds the coordinates of

Te_j

; with an orthonormal basis, entry

(i, j)

\langle Te_j, e_i \rangle

Show solution

Choosing $e_1$ . Apply $T$ to each candidate: $T(1, 0) = (3, 1)$ , $T(1, -1) = (3, -1)$ , $T(1, 2) = (3, 5)$ , none a multiple of the input; but $T(1, 1) = (3, 3) = 3(1, 1)$ . So $(1, 1)$ is an eigenvector with eigenvalue $3$ and $e_1 = \tfrac{1}{\sqrt2}(1, 1)$ .

Following Schur's proof. The basis $v_1 = (1, 1)$ , $v_2 = (1, 0)$ makes $\mathcal{M}(T)$ upper triangular because $\operatorname{span}(v_1)$ is invariant. Gram–Schmidt gives $f_2 = (1, 0) - \tfrac{1}{2}(1, 1) = \tfrac12(1, -1)$ , so $e_1 = \tfrac{1}{\sqrt2}(1, 1)$ , $e_2 = \tfrac{1}{\sqrt2}(1, -1)$ , and $\operatorname{span}(e_1) = \operatorname{span}(v_1)$ is still invariant.

The matrix. $Te_1 = 3e_1$ and $Te_2 = \tfrac{1}{\sqrt2}(3, -1)$ , so

\langle Te_1, e_1 \rangle = 3,\quad \langle Te_1, e_2 \rangle = 0,\quad \langle Te_2, e_1 \rangle = \tfrac{3 - 1}{2} = 1,\quad \langle Te_2, e_2 \rangle = \tfrac{3 + 1}{2} = 2,

\mathcal{M}\big(T, (e_1, e_2)\big) = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}.

Indeed $Te_2 = e_1 + 2e_2$ . The diagonal holds the eigenvalues $3$ and $2$ .

Solved: 0 / 4

Gram-Schmidt, Orthonormal Bases and Schur's Theorem

IKey Formulas

IIOne Gram–Schmidt Step in R2\mathbf{R}^2R2

IIIGram–Schmidt on 1,x,x21, x, x^21,x,x2 in P2(R)\mathcal{P}_2(\mathbf{R})P2​(R)

IVWhy Gram–Schmidt Proves Schur's Theorem

VQuiz Questions

IIOne Gram–Schmidt Step in $\mathbf{R}^2$

IIIGram–Schmidt on $1, x, x^2$ in $\mathcal{P}_2(\mathbf{R})$