Solution. The inner product is \((x,y)=1\cdot 2+0\cdot 1+2\cdot 2=6\), and \[\|x\|=\sqrt {1+0+4}=\sqrt 5,\qquad \|y\|=\sqrt {4+1+4}=3,\] while \(x+y=(3,1,4)\) gives \(\|x+y\|=\sqrt {9+1+16}=\sqrt {26}\).

The angle \(\theta \) satisfies \[\cos \theta =\frac {(x,y)}{\|x\|\,\|y\|}=\frac {6}{3\sqrt 5}=\frac {2}{\sqrt 5} \approx 0.894,\] so \(\theta \approx 26.6^{\circ }\).

Cauchy–Schwarz

\(|(x,y)|=6\) and \(\|x\|\|y\|=3\sqrt 5\approx 6.708\), so \(6\leq 6.708\) holds. Equality would mean \(x\) and \(y\) are parallel; they are not.

Triangle inequality

\(\|x+y\|=\sqrt {26}\approx 5.099\) and \(\|x\|+\|y\|=\sqrt 5+3\approx 5.236\), so \(5.099\leq 5.236\) holds.

Solution. Expand both norms using \(\|v\|^2=(v,v)\) and bilinearity. In a real space \((x,y)=(y,x)\), so \[\|x+y\|^2=(x+y,x+y)=\|x\|^2+2(x,y)+\|y\|^2,\] \[\|x-y\|^2=(x-y,x-y)=\|x\|^2-2(x,y)+\|y\|^2.\] Subtracting the second from the first, the \(\|x\|^2\) and \(\|y\|^2\) terms cancel and the cross terms add: \[\|x+y\|^2-\|x-y\|^2=4(x,y).\] Adding them instead, the cross terms cancel and the rest doubles: \[\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2.\] The first identity says the inner product is recoverable from the norm alone; the second, the parallelogram law, is the test for whether a given norm comes from an inner product at all.

Solution. Apply the parallelogram law from the previous problem: \[\|x+y\|^2+\|x-y\|^2=2\left (\|x\|^2+\|y\|^2\right ).\] Substituting the given values, \[16+36=2(9+\|y\|^2)\implies 52=18+2\|y\|^2\implies \|y\|^2=17,\] so \(\|y\|=\sqrt {17}\). Note the data are not free: any norm satisfying the parallelogram law determines \(\|y\|\) once the other three are fixed.

Solution. If such an inner product existed, its norm would have to satisfy the parallelogram law. Test it on \(u=(1,0)\) and \(v=(0,1)\): \[\|u\|=1,\qquad \|v\|=1,\] \[u+v=(1,1)\implies \|u+v\|=2,\qquad u-v=(1,-1)\implies \|u-v\|=2.\] Then \[\|u+v\|^2+\|u-v\|^2=4+4=8,\] but \[2\left (\|u\|^2+\|v\|^2\right )=2(1+1)=4.\] Since \(8\neq 4\) the parallelogram law fails, so this norm cannot arise from any inner product.

Solution.

Orthogonality

\[(x,y)=1\cdot 3+3\cdot 1+2\cdot (-4)+c\cdot 1=3+3-8+c=c-2,\] which vanishes when \(c=2\).

The inner product

The form has matrix \(A=\begin {pmatrix}1&-3\\-3&c\end {pmatrix}\), since \((x,y)=x^tAy\). It is symmetric and bilinear whatever \(c\) is, so the only condition to impose is positive definiteness: \((x,x)>0\) for \(x\neq 0\). Completing the square, \[(x,x)=x_1^2-6x_1x_2+cx_2^2=(x_1-3x_2)^2+(c-9)x_2^2.\] If \(c>9\) this is a sum of two non-negative terms which vanish together only when \(x_2=0\) and \(x_1=3x_2=0\), so it is positive definite. If \(c\leq 9\), taking \(x=(3,1)\) gives \((x,x)=0+(c-9)\leq 0\) with \(x\neq 0\), so it fails.

Hence the form is an inner product exactly when \(\boldsymbol {c>9}\).

Solution. Compute the inner product of \(x'=x-cy\) with \(y\), using linearity in the first argument: \[(x',y)=(x-cy,\ y)=(x,y)-c(y,y)=(x,y)-c\|y\|^2.\] This is zero if and only if \(c\|y\|^2=(x,y)\), and since \(y\neq 0\) we have \(\|y\|^2\neq 0\), so we may divide: \[c=\frac {(x,y)}{\|y\|^2}.\] The calculation gives existence and uniqueness at once — the condition is a single linear equation in \(c\) with a non-zero coefficient.

The vector \(cy\) is the orthogonal projection of \(x\) onto the line spanned by \(y\), and this is exactly the step repeated at each stage of the Gram–Schmidt procedure.

Solution. By Theorem 7.2.8, \(V=U\oplus U^{\perp }\). For a direct sum the dimensions add: taking a basis of \(U\) and a basis of \(U^{\perp }\), their union spans \(V\) because every vector decomposes, and it is independent because a dependence would give a non-zero vector in \(U\cap U^{\perp }=\{0\}\). Hence \[\dim V=\dim U+\dim U^{\perp },\] which rearranges to the stated formula.

For the consequence: \(U^{\perp }=\{0\}\) means \(\dim U^{\perp }=0\), which by the formula holds exactly when \(\dim U=\dim V\); and a subspace of the same finite dimension as the whole space is the whole space. So \(U^{\perp }=\{0\}\) if and only if \(U=V\).

Solution. The procedure breaks down at the first vector that is a combination of its predecessors, and it does so in a visible way: that step produces the zero vector, which cannot then be normalised.

Concretely, at step \(k\) Gram–Schmidt forms \[w_k=v_k-\sum _{i<k}(v_k,e_i)e_i,\] the part of \(v_k\) orthogonal to the span of \(v_1,\dots ,v_{k-1}\). If \(v_k\) already lies in that span then subtracting its projection removes all of it and \(w_k=0\). The next instruction, \(e_k=w_k/\|w_k\|\), then divides by zero.

So the procedure is self-diagnosing: it is a test for independence as well as a method of orthogonalisation. The usual repair is to discard any \(v_k\) producing \(w_k=0\) and carry on, which yields an orthonormal basis for the span of the original list rather than one vector per input.