Exercises — Inner product spaces — dot product generalization

Check the three axioms in order.

• Symmetry: $⟨\mathbf{u},\mathbf{v}⟩=u_1{v}_1+u_2{v}_2$ and $⟨\mathbf{v},\mathbf{u}⟩=v_1{u}_1+v_2{u}_2$. Numbers multiply in any order, so these are equal. ✓
• Linearity in slot 1: $⟨c\mathbf{u}+\mathbf{w},\mathbf{v}⟩=(c{u}_1+w_1)v_1+(c{u}_2+w_2)v_2=c(u_1{v}_1+u_2{v}_2)+(w_1{v}_1+w_2{v}_2)=c⟨\mathbf{u},\mathbf{v}⟩+⟨\mathbf{w},\mathbf{v}⟩$. ✓
• Positive-definite: $⟨\mathbf{v},\mathbf{v}⟩=v_1^2+v_2^2\ge 0$, and it equals $0$ only when $v_1=v_2=0$, i.e. $\mathbf{v}=0$. ✓

All three hold, so yes, it is an inner product.

Symmetry ✓ and linearity ✓ (same algebra as L1.1, just a minus sign). But test positivity with $\mathbf{v}=(0,1)$: $⟨(0,1),(0,1)⟩=0^2-1^2=-1<0.$ A "length-squared" came out negative. Positive-definiteness fails, so this is not an inner product.

Symmetry: swapping $\mathbf{u},\mathbf{v}$ gives $v_1{u}_2+v_2{u}_1$ — same thing, ✓. Linearity ✓. Positivity: $⟨\mathbf{v},\mathbf{v}⟩=v_1{v}_2+v_2{v}_1=2v_1{v}_2$. Take $\mathbf{v}=(1,-1)$: $2(1)(-1)=-2<0.$ Not an inner product — positivity fails again.

Norm from an inner product: $\Vert \mathbf{v}\Vert =\sqrt{⟨\mathbf{v},\mathbf{v}⟩}$ (see Norms and Distance). $⟨(1,1),(1,1)⟩=3(1)(1)+2(1)(1)=3+2=5,\Vert (1,1)\Vert =\sqrt{5}.$

Distance is the norm of the difference: $d(\mathbf{u},\mathbf{v})=\Vert \mathbf{u}-\mathbf{v}\Vert$. $\mathbf{u}-\mathbf{v}=(2-0,1-(-1))=(2,2).$ $⟨(2,2),(2,2)⟩=3(4)+2(4)=12+8=20,d=\sqrt{20}=2\sqrt{5}.$

$⟨x,x⟩={\int }_0^1{x}^{2}dx=[\frac{x^3}{3}{]}_0^1=\frac{1}{3},\Vert x\Vert =\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}.$

Angle formula: $\cos \theta =\frac{⟨\mathbf{u},\mathbf{v}⟩}{\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert }$. $⟨\mathbf{u},\mathbf{v}⟩=1\cdot 1+0\cdot 1=1,\Vert \mathbf{u}\Vert =1,\Vert \mathbf{v}\Vert =\sqrt{2}.$ $\cos \theta =\frac{1}{1\cdot \sqrt{2}}=\frac{1}{\sqrt{2}}\Rightarrow \theta =45^{\circ }.$ Figure s01 shows this: $\mathbf{u}=(1,0)$ is drawn flat along the x-axis (lavender), $\mathbf{v}=(1,1)$ points up-right at the diagonal (coral), and the mint arc between them is the $45^{\circ }$ opening. The arrow $\mathbf{v}$ rises one unit for every one unit right, so it splits the corner exactly in half — that half-of-a-right-angle is the $45^{\circ }$, and $\cos 45^{\circ }=1/\sqrt{2}$ is the horizontal shadow of the unit-scaled diagonal.

$⟨\mathbf{u},\mathbf{v}⟩=3(1)(1)+2(0)(1)=3.$ $\Vert \mathbf{u}\Vert =\sqrt{3(1)+2(0)}=\sqrt{3},\Vert \mathbf{v}\Vert =\sqrt{3(1)+2(1)}=\sqrt{5}.$ $\cos \theta =\frac{3}{\sqrt{3}\cdot \sqrt{5}}=\frac{3}{\sqrt{15}}=\sqrt{\frac{9}{15}}=\sqrt{\frac{3}{5}}\approx 0.7746\Rightarrow \theta \approx 39.2^{\circ }.$ Why different? The weights stretch the axes, so "angle" is measured with a different ruler. Same arrows on paper, different geometry. Figure s02 shows this: the identical arrows $\mathbf{u}=(1,0)$ (lavender) and $\mathbf{v}=(1,1)$ (coral) are drawn once, but two arcs sit in the corner — the mint arc labelled "dot: $45^{\circ }$" (the ordinary ruler from L3.1) and, tucked inside it, the smaller butter-coloured arc labelled "weighted: $39.2^{\circ }$". Because the weight on the $x$-part ($3$) is larger than on the $y$-part ($2$), the weighted ruler counts horizontal agreement more heavily, so the two arrows look "more aligned" and the angle shrinks from $45^{\circ }$ to about $39.2^{\circ }$.

First the cross term: $⟨p,q⟩=p(0)q(0)+p(1)q(1)=(0)(1)+(1)(0)=0.$ Since $⟨p,q⟩=0$, they are orthogonal under this inner product (see Orthogonality and Gram-Schmidt). Now the norms, so the division in $\cos \theta$ is legal. Evaluate each polynomial at $x=0$ and $x=1$: $⟨p,p⟩=p(0{)}^2+p(1{)}^2=0^2+1^2=1\Rightarrow \Vert p\Vert =1,$ $⟨q,q⟩=q(0{)}^2+q(1{)}^2=1^2+0^2=1\Rightarrow \Vert q\Vert =1.$ Both norms are nonzero, so $\cos \theta =\frac{⟨p,q⟩}{\Vert p\Vert \Vert q\Vert }=\frac{0}{1\cdot 1}=0,$ which sits safely inside $[-1,1]$, so Cauchy–Schwarz holds ($\theta =90^{\circ }$ in this inner product).

Start from $\Vert \mathbf{w}{\Vert }^2=⟨\mathbf{w},\mathbf{w}⟩$ and expand by linearity in both slots (FOIL) — showing every cross term explicitly: $\Vert \mathbf{u}+\mathbf{v}{\Vert }^2=⟨\mathbf{u}+\mathbf{v},\mathbf{u}+\mathbf{v}⟩=⟨\mathbf{u},\mathbf{u}⟩+⟨\mathbf{u},\mathbf{v}⟩+⟨\mathbf{v},\mathbf{u}⟩+⟨\mathbf{v},\mathbf{v}⟩.$ Now apply symmetry ($⟨\mathbf{v},\mathbf{u}⟩=⟨\mathbf{u},\mathbf{v}⟩$): the two middle terms are equal, so they combine into $2⟨\mathbf{u},\mathbf{v}⟩$: $\Vert \mathbf{u}+\mathbf{v}{\Vert }^2=\Vert \mathbf{u}{\Vert }^2+2⟨\mathbf{u},\mathbf{v}⟩+\Vert \mathbf{v}{\Vert }^2.$ Identically, for the difference (the two cross terms are $⟨\mathbf{u},-\mathbf{v}⟩$ and $⟨-\mathbf{v},\mathbf{u}⟩$, each $=-⟨\mathbf{u},\mathbf{v}⟩$): $\Vert \mathbf{u}-\mathbf{v}{\Vert }^2=\Vert \mathbf{u}{\Vert }^2-2⟨\mathbf{u},\mathbf{v}⟩+\Vert \mathbf{v}{\Vert }^2.$ Add them: the $\pm 2⟨\mathbf{u},\mathbf{v}⟩$ cross terms cancel, leaving $2\Vert \mathbf{u}{\Vert }^2+2\Vert \mathbf{v}{\Vert }^2.■$ WHAT this means: the two diagonals of a parallelogram (sum $\mathbf{u}+\mathbf{v}$ and difference $\mathbf{u}-\mathbf{v}$) together carry exactly the energy of its four sides.

Use the two FOIL expansions established in L4.1, where each "$2⟨\mathbf{u},\mathbf{v}⟩$" came from the two equal cross terms $⟨\mathbf{u},\mathbf{v}⟩+⟨\mathbf{v},\mathbf{u}⟩$ collapsing by symmetry. Subtract them: $\Vert \mathbf{u}+\mathbf{v}{\Vert }^2-\Vert \mathbf{u}-\mathbf{v}{\Vert }^2=(\Vert \mathbf{u}{\Vert }^2+2⟨\mathbf{u},\mathbf{v}⟩+\Vert \mathbf{v}{\Vert }^2)-(\Vert \mathbf{u}{\Vert }^2-2⟨\mathbf{u},\mathbf{v}⟩+\Vert \mathbf{v}{\Vert }^2).$ The $\Vert \mathbf{u}{\Vert }^2$ and $\Vert \mathbf{v}{\Vert }^2$ terms cancel, and $2⟨\mathbf{u},\mathbf{v}⟩-(-2⟨\mathbf{u},\mathbf{v}⟩)=4⟨\mathbf{u},\mathbf{v}⟩$: $\Vert \mathbf{u}+\mathbf{v}{\Vert }^2-\Vert \mathbf{u}-\mathbf{v}{\Vert }^2=4⟨\mathbf{u},\mathbf{v}⟩.$ Divide by $4$. $■$ WHY it's remarkable: the inner product (angles!) is completely reconstructible from the norm (lengths!) alone. Length secretly knows angle.

From the FOIL expansion $\Vert \mathbf{u}+\mathbf{v}{\Vert }^2=\Vert \mathbf{u}{\Vert }^2+2⟨\mathbf{u},\mathbf{v}⟩+\Vert \mathbf{v}{\Vert }^2$, the middle term is $2\cdot 0=0$, so $\Vert \mathbf{u}+\mathbf{v}{\Vert }^2=\Vert \mathbf{u}{\Vert }^2+\Vert \mathbf{v}{\Vert }^2.■$ This is the abstract heart of every right-triangle picture.

Symmetry and linearity hold for every $c$. The only obstacle is positive-definiteness: $⟨\mathbf{v},\mathbf{v}⟩=v_1^2+c{v}_2^2.$ For this to be $\ge 0$ for all $\mathbf{v}$ and zero only at $\mathbf{v}=0$, take $\mathbf{v}=(0,1)$: we need $c>0$.

• If $c>0$: $v_1^2+c{v}_2^2\ge 0$, zero forces $v_1=0$ and $v_2=0$. ✓
• If $c=0$: $\mathbf{v}=(0,1)$ gives $⟨\mathbf{v},\mathbf{v}⟩=0$ but $\mathbf{v}\ne 0$ — positivity fails.
• If $c<0$: $\mathbf{v}=(0,1)$ gives a negative value — fails.

Answer: $\boxed{c>0}$.

$⟨1,\cos (2\pi x)⟩={\int }_0^1\cos (2\pi x)dx=[\frac{\sin (2\pi x)}{2\pi }{]}_0^1=\frac{\sin 2\pi -\sin 0}{2\pi }=0.$ Orthogonal ✓. Integrating a full-period cosine over $[0,1]$ washes it out to zero — the constant and the wave don't overlap.

The projection of $f$ onto $g$ is ${\mathrm{proj}}_{g}f=\frac{⟨f,g⟩}{⟨g,g⟩}g$. $⟨f,g⟩={\int }_0^{1}x\cdot 1dx=\frac{1}{2},⟨g,g⟩={\int }_0^{1}1dx=1.$ ${\mathrm{proj}}_{g}f=\frac{1/2}{1}\cdot 1=\frac{1}{2}.$ The best constant approximation to $x$ on $[0,1]$ is the constant $\frac{1}{2}$ — literally its average value. Figure s03 shows this: the coral line is $f(x)=x$ climbing from $(0,0)$ to $(1,1)$; the flat lavender line at height $\frac{1}{2}$ is the projection. The short mint segments are the vertical gaps $x-\frac{1}{2}$ (the residual) — they are positive on the right half and negative on the left half, and they cancel out over the interval, which is exactly why $\frac{1}{2}$ is the balanced (average) height.

Compute the three pieces as integrals. Cross term: $⟨f,g⟩={\int }_0^{1}x\cdot x^{2}dx={\int }_0^1{x}^{3}dx=[\frac{x^4}{4}{]}_0^1=\frac{1}{4}=\mathrm{0.25.}$ Norms: $\Vert f{\Vert }^2={\int }_0^1{x}^{2}dx=\frac{1}{3},\Vert g{\Vert }^2={\int }_0^1{x}^{4}dx=\frac{1}{5},$ $\Vert f\Vert \Vert g\Vert =\sqrt{\frac{1}{3}\cdot \frac{1}{5}}=\sqrt{\frac{1}{15}}\approx \mathrm{0.2582.}$ Compare: $|⟨f,g⟩|=0.25\le 0.2582=\Vert f\Vert \Vert g\Vert .✓$ Cauchy–Schwarz holds, and the inequality is strict ($0.25<0.2582$) because $x$ and $x^2$ are not scalar multiples of one another — equality in Cauchy–Schwarz happens only when the two vectors are parallel.

L1 test the axioms → L2 use the norm and distance formulas → L3 use angle and orthogonality → L4 prove identities from the axioms → L5 build new inner products and project functions. Each rung reuses only what the rung below earned.