#### Exercise 8.1.1

Let $V$ be a vector space and $(\ | \ )$ an inner product on $V$.

(a) Show that $(0|\beta)=0$ for all $\beta$ in $V$.

(b) Show that if $(\alpha|\beta)=0$ for all $\beta$ in $V$, then $\alpha=0$.

Solution:

(a) This comes from Part (b) of Definition on Page 271. Because $0=0 \cdot \beta$ (here the first $0$ is the zero vector), we have $$(0|\beta)=(0\cdot \beta|\beta)=0(\beta|\beta)=0.$$ (b) Because $(\alpha|\beta)=0$ for all $\beta$ in $V$, it holds particularly for $\beta=\alpha$. Hence we have $$(\alpha|\alpha)=0.$$It follows from Part (d) of Definition on Page 271 that $\alpha=0$ (otherwise $(\alpha|\alpha)>0$.

#### Exercise 8.1.9

let $V$ be a real or complex vector space with an inner product. Show that the quadratic form determined by the inner product satisfies the parallelogram law $$\|\alpha+\beta\|^2+\|\alpha-\beta\|^2=2\|\alpha\|^2+2\|\beta\|^2.$$ Solution: Because the quadratic form is determined by the inner product, we have $\|v\|=(v|v)$. Hence, we have\begin{align*}\|\alpha+\beta\|^2=&\ (\alpha+\beta|\alpha+\beta)\\=&\ (\alpha|\alpha)+(\alpha|\beta)+(\beta|\alpha)+(\beta|\beta)\end{align*} and \begin{align*}\|\alpha-\beta\|^2=&\ (\alpha-\beta|\alpha-\beta)\\=&\ (\alpha|\alpha)-(\alpha|\beta)-(\beta|\alpha)+(\beta|\beta).\end{align*}Therefore,\begin{align*}&\|\alpha+\beta\|^2+\|\alpha-\beta\|^2\\ =&\ (\alpha|\alpha)+(\alpha|\beta)+(\beta|\alpha)+(\beta|\beta)+(\alpha|\alpha)-(\alpha|\beta)-(\beta|\alpha)+(\beta|\beta)\\=&\ 2(\alpha|\alpha)+2(\beta|\beta)=2\|\alpha\|^2+2\|\beta\|^2.\end{align*}

#### Exercise 8.1.11

Show that the formula \begin{equation}\label{eq:8.1.11.1}\left(\sum_ja_jx^j|\sum_k b_kx^k\right)=\sum_{j,k}\frac{a_jb_k}{j+k+1}\end{equation}defines an inner product on the space $\mathbf R[x]$ of polynomial over the field $\mathbf R$. Let $W$ be the subspace of polynomials of degree less than or equal to $n$. Restrict the above inner product to $W$, and find the matrix of this inner product on $W$, relative to the ordered basis $\{1,x,x^2,\dots,x^n\}$. (*Hint*: To show that the formula defines an inner product, observe that $$(f|g)=\int_0^1 f(t)g(t)dt$$and work with the integral.)

Solution: We see similarly that $$(f|g)=\int_0^1 f(t)g(t)dt$$ defines an inner product on the vector space of all continuous real valued functions on the unit interval, $0\leqslant t\leqslant 1$. It is clear that $\mathbf R[x]$ is a subspace to this vector space and a restrict of inner product to subspace is an inner product on subspace. Hence the restriction $$(f|g)=\int_0^1 f(t)g(t)dt$$ defines an inner product on $\mathbf R[x]$. In particular, we can show that under this inner product (by integral), we have \begin{align*}\left(\sum_ja_jx^j|\sum_k b_kx^k\right)=&\ \int_0^1 \left(\sum_ja_jt^j\right)\left(\sum_k b_kt^k\right)dt\\ = &\ \sum_{j,k}\int_0^1 a_jb_kt^jt^kdt\\=&\ \sum_{j,k}\int_0^1 a_jb_kt^{j+k}dt \\ =&\ \sum_{j,k}\frac{a_jb_k}{j+k+1}.\end{align*}This is exactly the formula \eqref{eq:8.1.11.1}. Hence the formula \eqref{eq:8.1.11.1} defines an inner product on $\mathbf R[x]$.

The matrix of this inner product on $W$, relative to the ordered basis $\{1,x,x^2,\dots,x^n\}$, is given by $(\frac{1}{i+j-1})_{i,j=1}^{n+1}$. Indeed, the $(i,j)$ element of this matrix corresponds to the number $(x^{i-1}|x^{j-1})$, which is $\frac{1}{i+j-1}$.