Linear-Algebra-Done-Right Β· Unit 16 Β· Video 3 Β· Interactive Practice

Riesz Representation Without a Basis: v βŠ₯ null Ο†

IKey Formulas

FormulaNameWhat it says
Ο†=Ο†vβ‡’v∈(null⁑φ)βŠ₯\varphi = \varphi_v \Rightarrow v \in (\operatorname{null}\varphi)^\perpNecessary conditionu∈nullβ‘Ο†β‡’βŸ¨u,v⟩=Ο†(u)=0u \in \operatorname{null}\varphi \Rightarrow \langle u, v \rangle = \varphi(u) = 0
dim⁑(null⁑φ)βŠ₯=dim⁑Vβˆ’(dim⁑Vβˆ’1)=1\dim (\operatorname{null}\varphi)^\perp = \dim V - (\dim V - 1) = 1Dimension countHolds for every Ο†β‰ 0\varphi \ne 0 on finite-dimensional VV
v=Ο†(w)β€Ύβˆ₯wβˆ₯2 wv = \dfrac{\overline{\varphi(w)}}{\|w\|^2}\, wDefining equationΟ†β‰ 0\varphi \ne 0; ww any nonzero vector in (null⁑φ)βŠ₯(\operatorname{null}\varphi)^\perp
u=(uβˆ’Ο†(u)Ο†(v) v)+Ο†(u)βˆ₯vβˆ₯2 vu = \left(u - \dfrac{\varphi(u)}{\varphi(v)}\, v\right) + \dfrac{\varphi(u)}{\|v\|^2}\, vSplitting uuUses Ο†(v)=βˆ₯vβˆ₯2\varphi(v) = \|v\|^2; the first piece is in null⁑φ\operatorname{null}\varphi

Key Insight: For Ο†β‰ 0\varphi \ne 0, the representing vector is the nonzero vβŠ₯null⁑φv \perp \operatorname{null}\varphi with Ο†(v)=βˆ₯vβˆ₯2\varphi(v) = \|v\|^2 β€” every choice of ww gives the same vv, because the representing vector is unique.

IIVisualization 1 β€” Every ww in (null⁑φ)βŠ₯(\operatorname{null}\varphi)^\perp Gives the Same vv

For Ο†(x,y)=2xβˆ’y\varphi(x, y) = 2x - y, scaling any nonzero wβŠ₯null⁑φw \perp \operatorname{null}\varphi by Ο†(w)/βˆ₯wβˆ₯2\varphi(w)/\|w\|^2 lands on v=(2,βˆ’1)v = (2, -1).

IIIVisualization 2 β€” Why ⟨u,v⟩=Ο†(u)\langle u, v \rangle = \varphi(u)

Every uu splits into a piece in null⁑φ\operatorname{null}\varphi plus a multiple of vv, which forces ⟨u,v⟩=Ο†(u)\langle u, v \rangle = \varphi(u).

IVVisualization 3 β€” The Example on C2\mathbf{C}^2 and Its Conjugate

For Ο†(z1,z2)=iz1+2z2\varphi(z_1, z_2) = i z_1 + 2 z_2, every w=c (1,2i)w = c\,(1, 2i) with cβ‰ 0c \ne 0 yields v=(βˆ’i,2)v = (-i, 2) β€” with the conjugate; drop it and most choices of cc miss.

πŸ’‘ Challenge: find every cc for which dropping the bar still gives v=(βˆ’i,2)v = (-i, 2) β€” exactly those where Ο†(w)\varphi(w) is real.

VQuiz Questions

Problem 1 Β· The Defining Equation on RΒ³

Given: Ο†(x,y,z)=x+2yβˆ’2z\varphi(x, y, z) = x + 2y - 2z on R3\mathbf{R}^3 with the Euclidean inner product, and w=(βˆ’2,βˆ’4,4)w = (-2, -4, 4), a nonzero vector in (null⁑φ)βŠ₯(\operatorname{null}\varphi)^\perp. Find v=Ο†(w)β€Ύβˆ₯wβˆ₯2 wv = \dfrac{\overline{\varphi(w)}}{\|w\|^2}\, w.

βœ… Correct! The scalar is βˆ’18/36=βˆ’12-18/36 = -\tfrac12, and v=(1,2,βˆ’2)v = (1, 2, -2) is the coefficient vector of Ο†\varphi, as it must be.
❌ Not quite. That is ww itself. It has the right direction, but Ο†(w)=βˆ’18\varphi(w) = -18 while βˆ₯wβˆ₯2=36\|w\|^2 = 36, so Ο†(w)β‰ βˆ₯wβˆ₯2\varphi(w) \ne \|w\|^2 and ww does not represent Ο†\varphi.
❌ Close, but… you divided by βˆ₯wβˆ₯=6\|w\| = 6. The defining equation divides by βˆ₯wβˆ₯2=36\|w\|^2 = 36, giving the scalar βˆ’12-\tfrac12, not βˆ’3-3.
❌ Close, but… the bar is complex conjugation, not absolute value. Over R\mathbf{R} it does nothing, so the scalar keeps its sign: βˆ’18β€Ύ=βˆ’18\overline{-18} = -18, not 1818.
Show solution

Compute the two numbers in the defining equation:

Ο†(w)=βˆ’2+2(βˆ’4)βˆ’2(4)=βˆ’18,βˆ₯wβˆ₯2=4+16+16=36.\varphi(w) = -2 + 2(-4) - 2(4) = -18, \qquad \|w\|^2 = 4 + 16 + 16 = 36.

Everything is real, so Ο†(w)β€Ύ=βˆ’18\overline{\varphi(w)} = -18 and

v=βˆ’1836 (βˆ’2,βˆ’4,4)=βˆ’12 (βˆ’2,βˆ’4,4)=(1,2,βˆ’2).v = \frac{-18}{36}\,(-2, -4, 4) = -\tfrac12\,(-2, -4, 4) = (1, 2, -2).

Check: ⟨(x,y,z),(1,2,βˆ’2)⟩=x+2yβˆ’2z=Ο†(x,y,z)\langle (x, y, z), (1, 2, -2) \rangle = x + 2y - 2z = \varphi(x, y, z), and Ο†(v)=1+4+4=9=βˆ₯vβˆ₯2\varphi(v) = 1 + 4 + 4 = 9 = \|v\|^2.

Problem 2 Β· Where the Conjugate Matters

Given: Ο†(z1,z2)=3z1βˆ’iz2\varphi(z_1, z_2) = 3z_1 - i z_2 on C2\mathbf{C}^2 with the Euclidean inner product. Here null⁑φ=span⁑((i,3))\operatorname{null}\varphi = \operatorname{span}\big((i, 3)\big), and w=(3i,βˆ’1)w = (3i, -1) is a nonzero vector in (null⁑φ)βŠ₯(\operatorname{null}\varphi)^\perp. Find the representing vector vv.

βœ… Correct! The scalar is 10iβ€Ύ/10=βˆ’i\overline{10i}/10 = -i, and βˆ’i (3i,βˆ’1)=(3,i)-i\,(3i, -1) = (3, i) β€” the same vector formula (R) gives.
❌ Not quite. You used Ο†(w)/βˆ₯wβˆ₯2=i\varphi(w)/\|w\|^2 = i without the bar. The scalar is Ο†(w)β€Ύ/βˆ₯wβˆ₯2=10iβ€Ύ/10=βˆ’i\overline{\varphi(w)}/\|w\|^2 = \overline{10i}/10 = -i.
❌ Close, but… reading off the coefficients of Ο†\varphi forgets the conjugates: on C2\mathbf{C}^2, ⟨z,(a,b)⟩=z1aβ€Ύ+z2bβ€Ύ\langle z, (a, b) \rangle = z_1\overline{a} + z_2\overline{b}, so the second entry must be βˆ’iβ€Ύ=i\overline{-i} = i.
❌ Not quite. That is ww, unscaled. Even though βˆ£Ο†(w)∣=10=βˆ₯wβˆ₯2|\varphi(w)| = 10 = \|w\|^2, Ο†(w)=10iβ‰ 10\varphi(w) = 10i \ne 10, so Ο†(w)β‰ βˆ₯wβˆ₯2\varphi(w) \ne \|w\|^2; the phase still has to be corrected by the scalar βˆ’i-i.
Show solution

First confirm w∈(null⁑φ)βŠ₯w \in (\operatorname{null}\varphi)^\perp: ⟨(i,3),(3i,βˆ’1)⟩=iβ‹…3iβ€Ύ+3β‹…βˆ’1β€Ύ=i(βˆ’3i)βˆ’3=3βˆ’3=0\langle (i, 3), (3i, -1) \rangle = i\cdot\overline{3i} + 3\cdot\overline{-1} = i(-3i) - 3 = 3 - 3 = 0.

Ο†(w)=3(3i)βˆ’i(βˆ’1)=9i+i=10i,βˆ₯wβˆ₯2=∣3i∣2+βˆ£βˆ’1∣2=10.\varphi(w) = 3(3i) - i(-1) = 9i + i = 10i, \qquad \|w\|^2 = |3i|^2 + |-1|^2 = 10. v=10iβ€Ύ10 (3i,βˆ’1)=βˆ’i (3i,βˆ’1)=(βˆ’3i2,i)=(3,i).v = \frac{\overline{10i}}{10}\,(3i, -1) = -i\,(3i, -1) = (-3i^2, i) = (3, i).

Check: ⟨(z1,z2),(3,i)⟩=3z1+z2 iβ€Ύ=3z1βˆ’iz2=Ο†(z1,z2)\langle (z_1, z_2), (3, i) \rangle = 3z_1 + z_2\,\overline{i} = 3z_1 - i z_2 = \varphi(z_1, z_2).

Problem 3 Β· Splitting u in the Video's Example

Given: Ο†(x,y,z)=2xβˆ’5y+z\varphi(x, y, z) = 2x - 5y + z on R3\mathbf{R}^3, its representing vector v=(2,βˆ’5,1)v = (2, -5, 1), and u=(1,1,1)u = (1, 1, 1). Write u=(uβˆ’Ο†(u)Ο†(v) v)+Ο†(u)βˆ₯vβˆ₯2 vu = \left(u - \dfrac{\varphi(u)}{\varphi(v)}\, v\right) + \dfrac{\varphi(u)}{\|v\|^2}\, v.

What is the coefficient Ο†(u)/βˆ₯vβˆ₯2\varphi(u)/\|v\|^2 of vv in the second piece?

What is the first piece?

βœ… Correct! Ο†\varphi of the first piece is 3415βˆ’5015+1615=0\tfrac{34}{15} - \tfrac{50}{15} + \tfrac{16}{15} = 0, so it lies in null⁑φ\operatorname{null}\varphi and is orthogonal to vv.
❌ Not quite. βˆ’2-2 is Ο†(u)\varphi(u) itself; the coefficient divides it by βˆ₯vβˆ₯2=4+25+1=30\|v\|^2 = 4 + 25 + 1 = 30.
❌ Check the sign. Ο†(u)=2βˆ’5+1=βˆ’2\varphi(u) = 2 - 5 + 1 = -2 is negative, so the coefficient is βˆ’2/30-2/30.
❌ Close, but… that divides by βˆ₯vβˆ₯=30\|v\| = \sqrt{30}. The second piece divides by βˆ₯vβˆ₯2=30\|v\|^2 = 30, the same number as Ο†(v)\varphi(v).
❌ Check the sign. That is u+βˆ’230 vu + \tfrac{-2}{30}\,v; the first piece subtracts Ο†(u)Ο†(v) v=βˆ’115v\tfrac{\varphi(u)}{\varphi(v)}\,v = -\tfrac{1}{15}v, which adds 115(2,βˆ’5,1)\tfrac{1}{15}(2, -5, 1) to uu.
❌ Not quite. That is uβˆ’Ο†(u) vu - \varphi(u)\,v; the coefficient must be divided by Ο†(v)=30\varphi(v) = 30. Test it: Ο†(5,βˆ’9,3)=58β‰ 0\varphi(5, -9, 3) = 58 \ne 0, so it is not in null⁑φ\operatorname{null}\varphi.
Show solution
Ο†(u)=2βˆ’5+1=βˆ’2,Ο†(v)=βˆ₯vβˆ₯2=4+25+1=30.\varphi(u) = 2 - 5 + 1 = -2, \qquad \varphi(v) = \|v\|^2 = 4 + 25 + 1 = 30.

The coefficient of vv in the second piece is Ο†(u)/βˆ₯vβˆ₯2=βˆ’2/30=βˆ’115\varphi(u)/\|v\|^2 = -2/30 = -\tfrac{1}{15}, and the first piece is

uβˆ’Ο†(u)Ο†(v) v=(1,1,1)+115(2,βˆ’5,1)=(1715,1015,1615)=(1715,23,1615).u - \frac{\varphi(u)}{\varphi(v)}\, v = (1, 1, 1) + \tfrac{1}{15}(2, -5, 1) = \left(\tfrac{17}{15}, \tfrac{10}{15}, \tfrac{16}{15}\right) = \left(\tfrac{17}{15}, \tfrac{2}{3}, \tfrac{16}{15}\right).

Check: Ο†\varphi of the first piece is 3415βˆ’5015+1615=0\tfrac{34}{15} - \tfrac{50}{15} + \tfrac{16}{15} = 0. Taking the inner product of the splitting with vv: ⟨u,v⟩=0+(βˆ’115)β‹…30=βˆ’2=Ο†(u)\langle u, v \rangle = 0 + \left(-\tfrac{1}{15}\right)\cdot 30 = -2 = \varphi(u), matching ⟨(1,1,1),(2,βˆ’5,1)⟩=2βˆ’5+1\langle (1, 1, 1), (2, -5, 1) \rangle = 2 - 5 + 1.

Problem 4 Β· Evaluation at 0 on P₁(R)

Given: P1(R)\mathcal{P}_1(\mathbf{R}) with ⟨p,q⟩=∫01p(x) q(x) dx\langle p, q \rangle = \int_0^1 p(x)\,q(x)\,dx and Ο†(p)=p(0)\varphi(p) = p(0). Find the vv with Ο†(p)=⟨p,v⟩\varphi(p) = \langle p, v \rangle for every pp, without an orthonormal basis.

βœ… Correct! ⟨1,4βˆ’6x⟩=4βˆ’3=1\langle 1, 4 - 6x \rangle = 4 - 3 = 1 and ⟨x,4βˆ’6x⟩=2βˆ’2=0\langle x, 4 - 6x \rangle = 2 - 2 = 0, matching Ο†(1)=1\varphi(1) = 1 and Ο†(x)=0\varphi(x) = 0.
❌ Not quite. The constant 11 is not orthogonal to null⁑φ=span⁑(x)\operatorname{null}\varphi = \operatorname{span}(x): ⟨x,1⟩=12β‰ 0=Ο†(x)\langle x, 1 \rangle = \tfrac12 \ne 0 = \varphi(x).
❌ Close, but… 2βˆ’3x2 - 3x spans (null⁑φ)βŠ₯(\operatorname{null}\varphi)^\perp, but it still needs scaling: Ο†(2βˆ’3x)=2\varphi(2 - 3x) = 2 while βˆ₯2βˆ’3xβˆ₯2=1\|2 - 3x\|^2 = 1, so the scalar is 22.
❌ Close, but… the scalar is Ο†(w)β€Ύ/βˆ₯wβˆ₯2=2/1\overline{\varphi(w)}/\|w\|^2 = 2/1, not βˆ₯wβˆ₯2/Ο†(w)=12\|w\|^2/\varphi(w) = \tfrac12. Check: ⟨1,1βˆ’32x⟩=14β‰ 1\langle 1, 1 - \tfrac32 x \rangle = \tfrac14 \ne 1.
Show solution

Null space. Ο†(a+bx)=a\varphi(a + bx) = a, so null⁑φ=span⁑(x)\operatorname{null}\varphi = \operatorname{span}(x), of dimension dim⁑P1(R)βˆ’1=1\dim \mathcal{P}_1(\mathbf{R}) - 1 = 1.

Orthogonal complement. q=a+bxq = a + bx is orthogonal to xx when ∫01x(a+bx) dx=a2+b3=0\int_0^1 x(a + bx)\,dx = \tfrac{a}{2} + \tfrac{b}{3} = 0; take w=2βˆ’3xw = 2 - 3x.

Ο†(w)=2,βˆ₯wβˆ₯2=∫01(2βˆ’3x)2 dx=4βˆ’6+3=1.\varphi(w) = 2, \qquad \|w\|^2 = \int_0^1 (2 - 3x)^2\,dx = 4 - 6 + 3 = 1. v=Ο†(w)β€Ύβˆ₯wβˆ₯2 w=2(2βˆ’3x)=4βˆ’6x.v = \frac{\overline{\varphi(w)}}{\|w\|^2}\, w = 2(2 - 3x) = 4 - 6x.

Check: for p=a+bxp = a + bx, ⟨p,v⟩=a⟨1,v⟩+b⟨x,v⟩=aβ‹…1+bβ‹…0=p(0)\langle p, v \rangle = a\langle 1, v \rangle + b\langle x, v \rangle = a \cdot 1 + b \cdot 0 = p(0).

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