Linear-Algebra-Done-Right · Unit 30 · Video 5 · Interactive Practice

The Inner Product on V ⊗ W: Multiply the Factors

IKey Formulas

FormulaNameWhere it comes from
⟨v⊗w, u⊗x⟩=⟨v,u⟩⟨w,x⟩\langle v \otimes w,\ u \otimes x \rangle = \langle v, u \rangle \langle w, x \rangleMultiplication ruleDetermines a unique inner product on V⊗WV \otimes W
⟨∑j,kbj,k ej⊗fk, ∑j,kcj,k ej⊗fk⟩=∑j,kbj,k cj,k‾\Big\langle \sum_{j,k} b_{j,k}\, e_j \otimes f_k,\ \sum_{j,k} c_{j,k}\, e_j \otimes f_k \Big\rangle = \sum_{j,k} b_{j,k}\, \overline{c_{j,k}}Formula (∗∗)(\ast\ast)Existence, from orthonormal bases of VV and WW
∥v⊗w∥=∥v∥ ∥w∥\|v \otimes w\| = \|v\|\,\|w\|Norm of a simple tensorTake u=vu = v, x=wx = w in the rule
⟨ej⊗fk, eM⊗fN⟩=⟨ej,eM⟩⟨fk,fN⟩\langle e_j \otimes f_k,\ e_M \otimes f_N \rangle = \langle e_j, e_M \rangle \langle f_k, f_N \rangleOrthonormal basis of V⊗WV \otimes W11 exactly when j=Mj = M and k=Nk = N

Key Insight: In Fm⊗Fn\mathbf{F}^m \otimes \mathbf{F}^n with the standard inner products, identify each element with its m×nm \times n grid of coefficients; the inner product multiplies matching entries (conjugating the second) and adds.

IIThe Norm of v ⊗ w

The squared entries of the grid of v⊗wv \otimes w add up to ∥v∥2∥w∥2\|v\|^2 \|w\|^2.

IIIThe Multiplication Rule

Over R\mathbf{R}, with v=(1,2)v = (1, 2) and w=(3,4)w = (3, 4): matching entries of the two grids multiply and add to ⟨v,u⟩⟨w,x⟩\langle v, u \rangle \langle w, x \rangle.

💡 Challenge: make ⟨v⊗w,u⊗x⟩=0\langle v \otimes w, u \otimes x \rangle = 0 with u⊗x≠0u \otimes x \neq 0 in two different ways: u⊥vu \perp v, or x⊥wx \perp w.

IVQuiz Questions

Problem 1 · Norm of a Simple Tensor

Given: v=(1,0,2)∈R3v = (1, 0, 2) \in \mathbf{R}^3 and w=(2,−1)∈R2w = (2, -1) \in \mathbf{R}^2 with the standard inner products. Find ∥v⊗w∥\|v \otimes w\|.

✅ Correct! ∥v∥=5\|v\| = \sqrt{5} and ∥w∥=5\|w\| = \sqrt{5}, so ∥v⊗w∥=5⋅5=5\|v \otimes w\| = \sqrt{5} \cdot \sqrt{5} = 5.
❌ Not quite. 10=∥v∥2+∥w∥2\sqrt{10} = \sqrt{\|v\|^2 + \|w\|^2}; the norms multiply, they do not add in squares.
❌ Close, but… 25=∥v⊗w∥225 = \|v \otimes w\|^2; take the square root.
❌ Not quite. Use ∥v⊗w∥=∥v∥ ∥w∥\|v \otimes w\| = \|v\|\,\|w\|.
Show solution
∥v∥2=1+0+4=5,∥w∥2=4+1=5,\|v\|^2 = 1 + 0 + 4 = 5, \qquad \|w\|^2 = 4 + 1 = 5, ∥v⊗w∥=∥v∥ ∥w∥=5⋅5=5.\|v \otimes w\| = \|v\|\,\|w\| = \sqrt{5} \cdot \sqrt{5} = 5.

Check with the grid (2−1004−2)\begin{pmatrix} 2 & -1 \\ 0 & 0 \\ 4 & -2 \end{pmatrix}: the squares add to 4+1+16+4=254 + 1 + 16 + 4 = 25.

Problem 2 · Apply the Rule

Given: in R2\mathbf{R}^2, v=(1,2)v = (1, 2), u=(3,−1)u = (3, -1), w=(1,1)w = (1, 1), x=(2,5)x = (2, 5). Find ⟨v⊗w, u⊗x⟩\langle v \otimes w,\ u \otimes x \rangle.

✅ Correct! ⟨v,u⟩=3−2=1\langle v, u \rangle = 3 - 2 = 1 and ⟨w,x⟩=2+5=7\langle w, x \rangle = 2 + 5 = 7, so the product is 77.
❌ Not quite. 8=1+78 = 1 + 7: the two inner products multiply.
❌ Check the sign. ⟨v,u⟩=1⋅3+2⋅(−1)=1\langle v, u \rangle = 1 \cdot 3 + 2 \cdot (-1) = 1, which is positive.
❌ Not quite. Compute ⟨v,u⟩\langle v, u \rangle and ⟨w,x⟩\langle w, x \rangle separately, then multiply.
Show solution
⟨v⊗w, u⊗x⟩=⟨v,u⟩⟨w,x⟩=(3−2)(2+5)=1⋅7=7.\langle v \otimes w,\ u \otimes x \rangle = \langle v, u \rangle \langle w, x \rangle = (3 - 2)(2 + 5) = 1 \cdot 7 = 7.

Problem 3 · The Conjugate Slot

Given: C2⊗C2\mathbf{C}^2 \otimes \mathbf{C}^2 with the inner product from the standard inner products, and standard bases e1,e2e_1, e_2 and f1,f2f_1, f_2. Find ⟨e1⊗f1, (ie1)⊗f1⟩\langle e_1 \otimes f_1,\ (i e_1) \otimes f_1 \rangle.

✅ Correct! ⟨e1,ie1⟩⟨f1,f1⟩=iˉ⋅1=−i\langle e_1, i e_1 \rangle \langle f_1, f_1 \rangle = \bar{i} \cdot 1 = -i.
❌ Close, but… a scalar leaving the second slot of an inner product is conjugated: ⟨e1,ie1⟩=iˉ ⟨e1,e1⟩\langle e_1, i e_1 \rangle = \bar{i}\,\langle e_1, e_1 \rangle.
❌ Not quite. Apply the multiplication rule, then pull ii out of the second slot.
Show solution
⟨e1⊗f1, (ie1)⊗f1⟩=⟨e1,ie1⟩⟨f1,f1⟩=iˉ ⟨e1,e1⟩⋅1=−i.\langle e_1 \otimes f_1,\ (i e_1) \otimes f_1 \rangle = \langle e_1, i e_1 \rangle \langle f_1, f_1 \rangle = \bar{i}\,\langle e_1, e_1 \rangle \cdot 1 = -i.

Formula (∗∗)(\ast\ast) agrees: the coefficients are b1,1=1b_{1,1} = 1 and c1,1=ic_{1,1} = i, and b1,1 c1,1‾=−ib_{1,1}\,\overline{c_{1,1}} = -i.

Problem 4 · A Non-Standard Orthonormal Basis

Given: the orthonormal basis e1=12(1,1)e_1 = \tfrac{1}{\sqrt{2}}(1, 1), e2=12(1,−1)e_2 = \tfrac{1}{\sqrt{2}}(1, -1) of R2\mathbf{R}^2 and the standard basis f1,f2f_1, f_2 of R2\mathbf{R}^2. Find ∥e1⊗f2+2 e2⊗f1∥2\|e_1 \otimes f_2 + 2\, e_2 \otimes f_1\|^2.

✅ Correct! The tensors ej⊗fke_j \otimes f_k form an orthonormal basis, so the squared norm is 12+22=51^2 + 2^2 = 5.
❌ Not quite. The two terms are orthogonal: ⟨e1⊗f2, e2⊗f1⟩=⟨e1,e2⟩⟨f2,f1⟩=0\langle e_1 \otimes f_2,\ e_2 \otimes f_1 \rangle = \langle e_1, e_2 \rangle \langle f_2, f_1 \rangle = 0, so the squares add (Pythagoras).
❌ Close, but… 5\sqrt{5} is the norm; the question asks for its square.
Show solution

Since e1,e2e_1, e_2 and f1,f2f_1, f_2 are orthonormal bases, the four tensors ej⊗fke_j \otimes f_k form an orthonormal basis of R2⊗R2\mathbf{R}^2 \otimes \mathbf{R}^2. In an orthonormal basis the squared norm is the sum of the squared coefficients:

∥e1⊗f2+2 e2⊗f1∥2=12+22=5.\|e_1 \otimes f_2 + 2\, e_2 \otimes f_1\|^2 = 1^2 + 2^2 = 5.

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