Problems for Chapter 6

(a) For all vectors $𝐱,𝐲$ in an inner product space, $|⟨𝐱,𝐲⟩|\le$$\Vert 𝐱\Vert \Vert 𝐲\Vert$, with equality if and only if one of $𝐱,𝐲$ is a scalar multiple of the other (Theorem 6.2.1).

(b) If $𝐲=\mathrm{𝟎}$, both sides are $0$ and $𝐲=0𝐱$. Let $𝐲\ne \mathrm{𝟎}$, $𝑐=⟨𝐱,𝐲⟩/{\Vert 𝐲\Vert }^2$, and $𝐰=𝐱-𝑐𝐲$. Then $⟨𝐰,𝐲⟩=$$⟨𝐱,𝐲⟩-$$𝑐{\Vert 𝐲\Vert }^2=$$0$, so by Proposition 6.1.29 the cross term in ${\Vert 𝐰+𝑐𝐲\Vert }^2$ vanishes:

Dropping ${\Vert 𝐰\Vert }^2\ge 0$ and multiplying by ${\Vert 𝐲\Vert }^2$ gives the inequality. Equality holds exactly when $𝐰=\mathrm{𝟎}$, that is, $𝐱=𝑐𝐲$; the converse is Lemma 6.2.3.

(c) Let $𝑎=\Vert 𝐱\Vert >0$, $𝑏=\Vert 𝐲\Vert >0$, and $𝑟=Re⟨𝐱,𝐲⟩$. By Proposition 6.1.29 with the real scalars $1/𝑎$ and $1/𝑏$,

and ${\Vert 𝐱-𝐲\Vert }^2=$${𝑎}^2-$$2𝑟+$${𝑏}^2=$${(𝑎-𝑏)}^2+$$2(𝑎𝑏-𝑟)$. Both sides of the claimed inequality are nonnegative, so it is equivalent to the inequality between their squares, and after multiplying by $𝑎𝑏{(𝑎+𝑏)}^2/2>0$, to

Since ${(𝑎+𝑏)}^2=$${(𝑎-𝑏)}^2+$$4𝑎𝑏$, the left side is $(𝑎𝑏-𝑟){(𝑎-𝑏)}^2+$$4𝑎𝑏(𝑎𝑏-𝑟)$, and the right side minus the left side is

By Cauchy–Schwarz, $-𝑟\le$$|⟨𝐱,𝐲⟩|\le$$𝑎𝑏$, so $𝑎𝑏+𝑟\ge 0$, which proves the inequality.

Equality holds if and only if ${(𝑎-𝑏)}^2(𝑎𝑏+𝑟)=$$0$, that is, $𝑎=𝑏$ or $𝑟=-𝑎𝑏$. If $𝑟=-𝑎𝑏$, then $Re⟨𝐱,-𝐲⟩=$$\Vert 𝐱\Vert \Vert -$$𝐲\Vert$. Since $Re𝑧\le |𝑧|$ and Cauchy–Schwarz bounds $|⟨𝐱,-𝐲⟩|$ by the same number, $⟨𝐱,-𝐲⟩=$$\Vert 𝐱\Vert \Vert -$$𝐲\Vert$, and by Theorem 6.2.11(b) one of $𝐱,-𝐲$ is a nonnegative real multiple of the other. Both are nonzero, so $𝐲=-𝑠𝐱$ with $𝑠>0$. Conversely, if $𝐲=-𝑠𝐱$ with $𝑠>0$, then $𝑟=-𝑠{𝑎}^2=-𝑎𝑏$.

The inequality says that sending nonzero vectors to the unit sphere along rays moves them apart by at most $2/(\Vert 𝐱\Vert +\Vert 𝐲\Vert )$ times their distance. Its proof used only $Re⟨𝐱,𝐲⟩\ge$$-\Vert 𝐱\Vert \Vert 𝐲\Vert$. In a general normed space the same inequality holds with $4$ in place of $2$ (Dunkl and Williams, 1964).

(a) Orthogonal decomposition (Theorem 6.4.24): every $𝐯\in 𝑉$ can be written in exactly one way as $𝐯=𝐰+{𝐰}^{\prime }$ with $𝐰\in 𝑊$ and ${𝐰}^{\prime }\in {𝑊}^{⟂}$, namely with $𝐰={Pr}_{𝑊}(𝐯)$. Best approximation (Theorem 6.4.33): $\Vert 𝐯-$${Pr}_{𝑊}(𝐯)\Vert \le$$\Vert 𝐯-$$𝐰\Vert$ for every $𝐰\in 𝑊$, with equality only for $𝐰={Pr}_{𝑊}(𝐯)$.

(b) Existence. Let $𝐩=$${Pr}_{𝑊}(𝐯)=$${\sum }_{𝑖}⟨𝐯,{𝐮}_{𝑖}⟩{𝐮}_{𝑖}\in$$𝑊$. For each $𝑗$, since $⟨{𝐮}_{𝑖},{𝐮}_{𝑗}⟩=$${𝛿}_{𝑖𝑗}$,

A vector orthogonal to each ${𝐮}_{𝑗}$ is orthogonal to their span $𝑊$ (Proposition 6.3.9(c)), so $𝐯-𝐩\in {𝑊}^{⟂}$, and $𝐯=𝐩+(𝐯-𝐩)$ is a decomposition.

Uniqueness. If ${𝐰}_1+{𝐰}_1^{\prime }={𝐰}_2+{𝐰}_2^{\prime }$ with ${𝐰}_{𝑖}\in 𝑊$ and ${𝐰}_{𝑖}^{\prime }\in {𝑊}^{⟂}$, then $𝐳=$${𝐰}_1-$${𝐰}_2=$${𝐰}_2^{\prime }-$${𝐰}_1^{\prime }$ lies in $𝑊\cap {𝑊}^{⟂}$, so $⟨𝐳,𝐳⟩=0$ and $𝐳=\mathrm{𝟎}$.

Best approximation. Let $𝐰\in 𝑊$. Then $𝐯-$$𝐰=$$(𝐯-𝐩)+$$(𝐩-𝐰)$, where the first piece lies in ${𝑊}^{⟂}$ and the second in $𝑊$. By Theorem 6.3.3,

with equality if and only if $\Vert 𝐩-𝐰\Vert =0$, that is, $𝐰=𝐩$.

(c) The points of $𝑊$ are $𝑡{𝐞}_1=(𝑡,0)$, and ${\Vert {𝐞}_2-𝑡{𝐞}_1\Vert }_{\infty }=$${\Vert (-𝑡,1)\Vert }_{\infty }=$$max(|𝑡|,1)\ge$$1$, with equality if and only if $|𝑡|\le 1$. So the distance from ${𝐞}_2$ to $𝑊$ is $1$, and it is attained at every point $(𝑡,0)$ with $-1\le 𝑡\le 1$.

The step with no analogue is the Pythagorean identity, which made the closest point unique: it adds the positive number ${\Vert 𝐩-𝐰\Vert }^2$ for every $𝐰\ne 𝐩$. The norm ${\Vert \cdot \Vert }_{\infty }$ does not come from any inner product, since it violates the parallelogram law (Proposition 6.1.30): ${\Vert {𝐞}_1+{𝐞}_2\Vert }_{\infty }^2+$${\Vert {𝐞}_1-{𝐞}_2\Vert }_{\infty }^2=$$2$, while $2{\Vert {𝐞}_1\Vert }_{\infty }^2+$$2{\Vert {𝐞}_2\Vert }_{\infty }^2=$$4$. So there is no orthogonality to make the error perpendicular to $𝑊$ (compare Aside (Norms that do not come from inner products)). Closest points still exist here, but uniqueness, and the linear formula for ${Pr}_{𝑊}$, are features of inner products.

(a) Let $𝑇\in \mathrm{L}(𝑉,𝑊)$ have an adjoint. Then $ker{𝑇}^{\ast }={(im𝑇)}^{⟂}$ and $ker𝑇={(im{𝑇}^{\ast })}^{⟂}$. If $𝑉$ is finite-dimensional, also $im{𝑇}^{\ast }={(ker𝑇)}^{⟂}$ and $im𝑇={(ker{𝑇}^{\ast })}^{⟂}$, so that $𝑉=ker𝑇\oplus im{𝑇}^{\ast }$ and $𝑊=im𝑇\oplus ker{𝑇}^{\ast }$ (Theorem 6.6.21).

(b) Let $𝐰\in 𝑊$. A vector of $𝑉$ is $\mathrm{𝟎}$ if and only if it is orthogonal to every vector of $𝑉$ (Proposition 6.1.37). So

using Definition 6.6.8 in the middle. This proves $ker{𝑇}^{\ast }={(im𝑇)}^{⟂}$.

By Proposition 6.6.18, ${𝑇}^{\ast }$ has the adjoint $𝑇$. Applying what we just proved to ${𝑇}^{\ast }$ gives $ker𝑇={(im{𝑇}^{\ast })}^{⟂}$. If $𝑉$ is finite-dimensional, so is its subspace $im{𝑇}^{\ast }$, and Theorem 6.4.26 gives ${(ker𝑇)}^{⟂}=$${({(im{𝑇}^{\ast })}^{⟂})}^{⟂}=$$im{𝑇}^{\ast }$.

(c) With dot products, the map $𝐱\mapsto 𝐴𝐱$ from ${𝔽}^{𝑛}$ to ${𝔽}^{𝑚}$ has the adjoint $𝐲\mapsto {𝐴}^{\ast }𝐲$ (Example 6.6.10). Applying (b) to the adjoint, with the finite-dimensional space ${𝔽}^{𝑚}$ in the role of $𝑉$, gives $im𝐴={(ker{𝐴}^{\ast })}^{⟂}$. So $𝐴𝐱=𝐛$ has a solution if and only if ${𝐲}^{\ast }𝐛=⟨𝐛,𝐲⟩=0$ for every $𝐲\in ker{𝐴}^{\ast }$. The second alternative says precisely that this fails for some $𝐲$, so exactly one of the two holds.

For the example, $𝐴=\left(\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 1& 0& -1\end{array}\right)$ is real, so ${𝐴}^{\ast }={𝐴}^{\text{𝖳}}$. Subtracting the second equation from the first gives ${𝑥}_1-{𝑥}_3$, so $𝐲=(1,-1,-1)$ satisfies ${𝐴}^{\text{𝖳}}𝐲=\mathrm{𝟎}$, as one checks column by column. With $𝐛=(1,1,1)$, ${𝐲}^{\text{𝖳}}𝐛=1-1-1=-1\ne 0$, so the system has no solution. With ${𝐛}^{\prime }=(1,1,0)$, ${𝐲}^{\text{𝖳}}{𝐛}^{\prime }=0$. The first two rows of $𝐴$ are independent and the third is their difference, so $rank𝐴=2$, $rank{𝐴}^{\text{𝖳}}=2$ (Corollary 6.6.22), and $ker{𝐴}^{\text{𝖳}}=span(𝐲)$ by Theorem 4.2.11. So ${𝐛}^{\prime }⟂ker{𝐴}^{\text{𝖳}}$, and the system is solvable; indeed $𝐱=(1,0,1)$ works.

The vector $𝐲$ is a certificate of inconsistency that can be checked without solving anything. Row reduction finds it too (as the combination of rows that produces $0=-1$), but (b) explains why it must exist: the right-hand side has to be orthogonal to every relation among the rows.

(a) Every point of $𝐴$ is $𝐜={𝐩}_0+𝐳$ with $𝐳\in$$𝑈=$$span({𝐯}_1,\dots ,{𝐯}_{𝑘})$. Since $𝐜-{𝐩}_{𝑖}=𝐳-{𝐯}_{𝑖}$, Proposition 6.1.29 gives

So $𝐜$ is equidistant from all the points if and only if

Write $𝐳={\sum }_{𝑗}{𝑡}_{𝑗}{𝐯}_{𝑗}$. Then $⟨𝐳,{𝐯}_{𝑖}⟩=$${\sum }_{𝑗}{𝑡}_{𝑗}⟨{𝐯}_{𝑗},{𝐯}_{𝑖}⟩=$${(𝐺𝐭)}_{𝑖}$, where $𝐺$ is the Gram matrix of ${𝐯}_1,\dots ,{𝐯}_{𝑘}$ (Definition 6.3.28). So the conditions say $𝐺𝐭=𝐝$ with ${𝑑}_{𝑖}={\Vert {𝐯}_{𝑖}\Vert }^2/2$. The ${𝐯}_{𝑖}$ are independent, so $𝐺$ is invertible (Proposition 6.3.30), and there is exactly one $𝐭$, hence (by independence again) exactly one $𝐳\in 𝑈$ and one $𝐜\in 𝐴$.

(b) The computation in (a) did not use $𝐳\in 𝑈$: for every $𝐪\in {ℝ}^{𝑛}$, $𝐪$ is equidistant from the points if and only if $𝐳=𝐪-{𝐩}_0$ satisfies these $𝑘$ conditions. Since $𝐜-{𝐩}_0$ satisfies it, subtracting shows that $𝐪$ is equidistant if and only if $⟨𝐪-𝐜,{𝐯}_{𝑖}⟩=0$ for all $𝑖$.

For such $𝐪$, the vector $𝐜-{𝐩}_0$ lies in $𝑈$, and $(𝐪-{𝐩}_0)-$$(𝐜-{𝐩}_0)=$$𝐪-$$𝐜$ is orthogonal to $𝑈$. By Theorem 6.4.6, $𝐜-$${𝐩}_0=$${Pr}_{𝑈}(𝐪-{𝐩}_0)$. For any $𝐚={𝐩}_0+𝐮\in 𝐴$, Theorem 6.4.33 gives

with equality only for $𝐚=𝐜$. If $𝑘=𝑛$, then $𝑈={ℝ}^{𝑛}$, so ${𝑈}^{⟂}=\{\mathrm{𝟎}\}$ and $𝐪=𝐜$.

(c) Here ${𝐯}_1=(2,0,0)$ and ${𝐯}_2=(1,2,2)$, so

So $𝐜=$$(9/32)(2,0,0)+$$(7/16)(1,2,2)=$$(1,7/8,7/8)$. Check: ${\Vert 𝐜\Vert }^2=$$1+$$49/32=$$81/32$, ${\Vert 𝐜-{𝐩}_1\Vert }^2=$$1+$$49/32=$$81/32$, and ${\Vert 𝐜-{𝐩}_2\Vert }^2=$$0+$$2\cdot$$81/64=$$81/32$. The common distance is $9/\sqrt{32}=9\sqrt{2}/8$.

By (b), the points of ${ℝ}^3$ equidistant from the three vertices form the line through $𝐜$ perpendicular to the plane of the triangle; the circumcenter is where that line meets the plane.

(a) The formula is linear in $𝑝$ and symmetric, since evaluation at a point is linear. Also $⟨𝑝,𝑝⟩=$${\sum }_{𝑘}{𝑝({𝑥}_{𝑘})}^2\ge$$0$, and if it is $0$, then every $𝑝({𝑥}_{𝑘})=0$ (Technique 6.1.9), so $𝑝$, which has degree at most $𝑛$ and $𝑛+1$ distinct roots, is $0$ by Lemma 3.4.23. So this is an inner product.

Since ${𝐿}_{𝑖}({𝑥}_{𝑘})={𝛿}_{𝑖𝑘}$,

So ${𝐿}_0,\dots ,{𝐿}_{𝑛}$ is orthonormal, and since $dim𝑉=𝑛+1$ it is an orthonormal basis (Corollary 6.3.14). By Proposition 6.3.18 the coordinates of $𝑝$ are $⟨𝑝,{𝐿}_{𝑗}⟩=$${\sum }_{𝑘}𝑝({𝑥}_{𝑘}){𝛿}_{𝑗𝑘}=$$𝑝({𝑥}_{𝑗})$, that is, $𝑝=$${\sum }_{𝑗}𝑝({𝑥}_{𝑗}){𝐿}_{𝑗}$. The Lagrange interpolation formula is the coordinate formula for this orthonormal basis.

(b) The $(𝑖,𝑗)$ entry is ${\sum }_{𝑘}{𝑥}_{𝑘}^{𝑗}{𝑥}_{𝑘}^{𝑖}=$$⟨{𝑥}^{𝑗},{𝑥}^{𝑖}⟩$, so the matrix is the Gram matrix of $1,𝑥,\dots ,{𝑥}^{𝑛}$ (Definition 6.3.28). These polynomials are independent, so the Gram matrix is invertible by Proposition 6.3.30. Problem 5.9 computes its determinant as ${\prod }_{𝑖<𝑗}{({𝑥}_{𝑗}-{𝑥}_{𝑖})}^2$; here invertibility came from the positivity of an inner product instead.

(c) For every $𝑞\in {ℝ[𝑥]}_{\le 𝑑}$,

the sum of squared residuals of $𝑞$. By Theorem 6.4.33, this is smallest exactly for $𝑞={Pr}_{{ℝ[𝑥]}_{\le 𝑑}}(𝑝)$, so that projection is the least-squares polynomial of Proposition 6.5.21, and in particular it is unique.

(d) $𝑝=$$1+$$(3/2)𝑥+$$(1/2){𝑥}^2$ satisfies $𝑝(-1)=0$, $𝑝(0)=1$, $𝑝(1)=3$. For this inner product $⟨1,𝑥⟩=-1+0+1=0$, ${\Vert 1\Vert }^2=3$, and ${\Vert 𝑥\Vert }^2=2$, while $⟨𝑝,1⟩=0+1+3=4$ and $⟨𝑝,𝑥⟩=0+0+3=3$. By Corollary 6.4.10,

This agrees with Proposition 6.5.17: the mean of the nodes is $0$, the mean of the values is $4/3$, and the slope is ${\sum }_{𝑘}{𝑥}_{𝑘}{𝑦}_{𝑘}/{\sum }_{𝑘}{𝑥}_{𝑘}^2=$$3/2$.

The error. The last step cites Proposition 6.1.41, which is about complex inner product spaces. Over $ℝ$, $𝑆=𝑇-{𝑇}^{\ast }$ satisfies ${𝑆}^{\ast }=-𝑆$ (Proposition 6.6.18), and every such operator has $⟨𝑆𝐯,𝐯⟩=$$⟨𝐯,{𝑆}^{\ast }𝐯⟩=$$-⟨𝐯,𝑆𝐯⟩=$$-⟨𝑆𝐯,𝐯⟩$, hence $⟨𝑆𝐯,𝐯⟩=0$, whether or not $𝑆=0$. In fact over $ℝ$ the first display holds for every $𝑇$, so the argument never used the hypothesis, which is a sign that something is wrong.

The claim is false. Let $𝑇=\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$ on ${ℝ}^2$ with the dot product. Then

but ${𝑇}^{\ast }=$${𝑇}^{\text{𝖳}}=$$\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)\ne$$𝑇$. Here $𝑇={𝐼}_2+𝐽$ with $𝐽$ the rotation of Example 6.1.39, and $𝐽$ contributes nothing to $⟨𝑇𝐯,𝐯⟩$.

(a) Over $ℂ$ every step is correct. The conjugation step is the definition of the adjoint followed by conjugate symmetry, and the hypothesis is used exactly where it should be: a nonnegative number equals its conjugate. Then $⟨𝑆𝐯,𝐯⟩=0$ for every $𝐯$, and Proposition 6.1.41 gives $𝑆=0$. (This is one half of Proposition 6.6.30.)

(b) Over $ℝ$, $⟨{𝑇}^{\ast }𝐯,𝐯⟩=$$⟨𝐯,𝑇𝐯⟩=$$⟨𝑇𝐯,𝐯⟩$ by symmetry of the real inner product. So

The numbers $⟨𝑇𝐯,𝐯⟩$ see only the self-adjoint part $𝐻$ of a real operator, and adding any $𝐾$ with ${𝐾}^{\ast }=-𝐾$ to $𝑇$ does not change them. This is why Section 8.3 requires positive operators to be self-adjoint by definition.

(a) False. $𝐴=\left(\begin{array}{c}1\\ 0\end{array}\right)\in {ℝ}^{2\times 1}$ has one orthonormal column, but $𝐴{𝐴}^{\text{𝖳}}=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)\ne {𝐼}_2$. In general $𝐴{𝐴}^{\text{𝖳}}$ is the matrix of ${Pr}_{im𝐴}$ (Theorem 6.4.31), which is ${𝐼}_{𝑚}$ only when $im𝐴={ℝ}^{𝑚}$, that is, when $𝑚=𝑛$.

(b) True. By Theorem 5.4.10 and Exercise 6.3.57(a), $det({𝐴}^{\ast }𝐴)=$$det({𝐴}^{\ast })det𝐴=$$det𝐴det𝐴=$${|det𝐴|}^2\ge$$0$.

(c) True. Let $𝐯\in {ℝ}^{𝑛}$. Using the dot product, ${\Vert 𝑃𝐯\Vert }^2=$$⟨{𝑃}^{\text{𝖳}}𝑃𝐯,𝐯⟩=$$⟨𝑃{𝑃}^{\text{𝖳}}𝐯,𝐯⟩=$${\Vert {𝑃}^{\text{𝖳}}𝐯\Vert }^2$, so $ker𝑃=ker{𝑃}^{\text{𝖳}}$, and $ker{𝑃}^{\text{𝖳}}={(im𝑃)}^{⟂}$ by Lemma 6.5.4. Now $𝑃𝐯\in im𝑃$, and $𝑃(𝐯-𝑃𝐯)=$$𝑃𝐯-$${𝑃}^2𝐯=$$\mathrm{𝟎}$, so $𝐯-$$𝑃𝐯\in$$ker𝑃=$${(im𝑃)}^{⟂}$. By Theorem 6.4.6, $𝑃𝐯={Pr}_{im𝑃}(𝐯)$ for every $𝐯$.

(d) True. Let $𝑅:𝑈\to 𝑊$ be the restriction of ${Pr}_{𝑊}$ to $𝑈$; it maps into $𝑊$ and is linear. Its kernel is $𝑈\cap ker{Pr}_{𝑊}=𝑈\cap {𝑊}^{⟂}$ (Proposition 6.4.29(a)), and its image is a subspace of $𝑊$. By Theorem 4.2.11,

In words: if $𝑈$ is bigger than $𝑊$, then $𝑈$ contains $dim𝑈-dim𝑊$ independent directions perpendicular to all of $𝑊$.

(e) “Only if” is true. The columns of $𝐴$ are ${𝐚}_{𝑗}=𝐴{𝐞}_{𝑗}$, with $\Vert {𝐚}_{𝑗}\Vert \le \Vert {𝐞}_{𝑗}\Vert =1$. By Hadamard’s inequality (Exercise 6.4.59(c)), $|det𝐴|\le$$\Vert {𝐚}_1\Vert \cdots \Vert {𝐚}_{𝑛}\Vert \le$$1$. “If” is false: $𝐴=\left(\begin{array}{cc}2& 0\\ 0& 1/2\end{array}\right)$ has $det𝐴=1$ but $\Vert 𝐴{𝐞}_1\Vert =2>1$. The determinant measures the change of volume, and a map can stretch one direction while shrinking another.

(a) Taking $𝐲=\mathrm{𝟎}$ gives $\Vert 𝑓(𝐱)\Vert =$$\Vert 𝑓(𝐱)-$$𝑓(\mathrm{𝟎})\Vert =$$\Vert 𝐱\Vert$. By Proposition 6.1.29 over $ℝ$,

and the squared lengths agree, so $⟨𝑓(𝐱),𝑓(𝐲)⟩=$$⟨𝐱,𝐲⟩$.

(b) First proof. Let $𝐱,𝐲\in 𝑉$ and $𝐳=$$𝑓(𝐱+𝐲)-$$𝑓(𝐱)-$$𝑓(𝐲)$. Expanding,

By (a), each term equals the same term with $𝑓$ removed, and the result is the expansion of ${\Vert (𝐱+𝐲)-𝐱-𝐲\Vert }^2=$$0$. So $𝐳=\mathrm{𝟎}$ (Proposition 6.1.27). Similarly, for $𝑡\in ℝ$,

So $𝑓$ is additive and homogeneous, that is, linear. This proof works verbatim for $𝑓:𝑉\to 𝑊$ between any two real inner product spaces.

Second proof, for $dim𝑉=𝑛<\infty$. Let ${𝐞}_1,\dots ,{𝐞}_{𝑛}$ be an orthonormal basis. By (a), $⟨𝑓({𝐞}_{𝑖}),𝑓({𝐞}_{𝑗})⟩=$${𝛿}_{𝑖𝑗}$, so $𝑓({𝐞}_1),$$\dots ,$$𝑓({𝐞}_{𝑛})$ is an orthonormal list of length $𝑛$ in $𝑉$, hence an orthonormal basis (Corollary 6.3.14). By Proposition 6.3.18 and (a),

and each $𝐱\mapsto ⟨𝐱,{𝐞}_{𝑖}⟩$ is linear. This proof needs the values $𝑓({𝐞}_{𝑖})$ to span the target, which is why it asks for $𝑓:𝑉\to 𝑉$ with $𝑉$ finite-dimensional; the first proof needs nothing of the kind.

(c) Let $𝑓(𝐱)=$$𝑔(𝐱)-$$𝑔(\mathrm{𝟎})$. Then $𝑓(\mathrm{𝟎})=\mathrm{𝟎}$ and $\Vert 𝑓(𝐱)-$$𝑓(𝐲)\Vert =$$\Vert 𝑔(𝐱)-$$𝑔(𝐲)\Vert =$$\Vert 𝐱-$$𝐲\Vert$, so $𝑓$ is linear by (b), say $𝑓(𝐱)=𝑄𝐱$ (Proposition 2.3.17). By (a), $(𝑄{𝐞}_{𝑗})\cdot$$(𝑄{𝐞}_{𝑖})=$${𝛿}_{𝑖𝑗}$, so the columns of $𝑄$ are orthonormal and ${𝑄}^{\text{𝖳}}𝑄={𝐼}_{𝑛}$ (Theorem 6.3.29). With $𝐛=𝑔(\mathrm{𝟎})$, $𝑔(𝐱)=𝑄𝐱+𝐛$.

(d) Over $ℂ$, Proposition 6.1.29 contains $2Re⟨\cdot ,\cdot ⟩$ instead of $2⟨\cdot ,\cdot ⟩$, so (a) becomes $Re⟨𝑓(𝐱),𝑓(𝐲)⟩=$$Re⟨𝐱,𝐲⟩$, and the expansions in the first proof of (b) involve only lengths and real parts of inner products (for the second one, with $𝑡$ real). So they go through unchanged and show that $𝑓$ is additive and $𝑓(𝑡𝐱)=𝑡𝑓(𝐱)$ for $𝑡\in ℝ$. But $𝑓$ need not be $ℂ$-linear: on ${ℂ}^{𝑛}$ with the dot product, let $𝑓$ conjugate every entry. Then $𝑓(\mathrm{𝟎})=\mathrm{𝟎}$ and $\Vert 𝑓(𝐱)-$$𝑓(𝐲)\Vert =$$\Vert 𝐱-𝐲\Vert =$$\Vert 𝐱-$$𝐲\Vert$, while $𝑓(𝑖{𝐞}_1)=$$-𝑖{𝐞}_1\ne$$𝑖𝑓({𝐞}_1)$.

For surjective distance-preserving maps between real normed spaces, the conclusion of (b) still holds (the Mazur–Ulam theorem), but the proof is much harder: without an inner product there is no polarization.