*Question 1*
For all $i,j\in\{ 1,\dots,n \},$ $\langle w_{i},w_{i}\rangle=1$ and $\langle w_{i},w_{j}\rangle=0$ for $i \neq j.$
By bilinearity, $\begin{align}
\langle v,w_{i}\rangle = \langle \lambda_{1}w_{1} +\dots+ \lambda_{n}v_{n},w_{i}\rangle = \lambda_{1} \langle w_{1},w_{i}\rangle + \lambda_{2} \langle w_{2},w_{i}\rangle +\dots+ \lambda_{n} \langle w_{n},\; w_{i}\rangle = \lambda_{i}.
\end{align}$
*Question 2*
(a) Step 1: $||v_{1}||=\sqrt{2 }$ so $w_{1} = \frac{1}{||v_{1}||} v_{1}=\frac{1}{\sqrt{2}} (1,1,0)^{T}.$
Step 2: $u_{2}:=v_{2}-(v_{2}\cdot w_{1})w_{1}=(2,0,1)^{T}-(1,1,0)^{T}=(1,-1,0)^{T}.$
$||u_{2}||=\sqrt{2}$ so $w_{2}=\frac{1}{||u_{2}||}=\frac{1}{\sqrt{2 }}(1,-1,0)^{T}.$
Step 3: $u_{3}:=v_{3}-(v_{3}\cdot w_{2})w_{2} -(v_{3}\cdot w_{1})w_{1}=(1,1,2)^{T}- 0_{V}-(1,1,0)^{T}=(0,0,2)^{T}$.
$||u_{3}||=2$ so $w_{3}=(0,0,1)$.
(b) Then $(\lambda_{1},\lambda_{2},\lambda_{3})=(0,0,1).$ Check: $0w_{1}+0w_{2}+w_{3}= (0,0,1)^{T}=v.$
*Question 3*
(a) Note that $\varphi(e_{1}) = \begin{pmatrix}
1 & 0
\end{pmatrix} \begin{pmatrix}
a & \frac{b}{2 } \\
\frac{b}{2} & c
\end{pmatrix} \begin{pmatrix}
1 \\
0
\end{pmatrix} = \begin{pmatrix}
a & \frac{b}{2}
\end{pmatrix} \begin{pmatrix}
1 \\
0
\end{pmatrix} = a \leq 0.$
(b) $\varphi (v_{x}) = \begin{pmatrix}
x & 1
\end{pmatrix} \begin{pmatrix}
a & \frac{b}{2 } \\
\frac{b}{2} & c
\end{pmatrix} \begin{pmatrix}
x \\
1
\end{pmatrix} = \begin{pmatrix}
ax+\frac{b}{2} & \frac{bx}{2} +c
\end{pmatrix} \begin{pmatrix}
x \\
1
\end{pmatrix} = ax^{2} + bx + c $
(c) Note that $\varphi(v_{x}) = ax^{2}+bx+c=a\left( x+\frac{b}{2a} \right)^{2}-\frac{b^{2}}{4a}+c.$ Also $\left( x+\frac{b}{2a} \right)^{2}\geq 0$ so $\varphi(v_{x})>0$ for all $x$ iff $-\frac{b^{2}}{4a}+c>0 \iff b^{2}-4ac<0.$
*Question 4*
(a) We have $\begin{align}
\langle w, v \rangle &= w^{T}Av \\
& =(v^{T}A^{T}w)^{T} & \text{since $(AB)^{T}=B^{T}A^{T}$ }\ \\
&=v^{T}A^{T}w & \text{since $(v^{T} A^{T} w)^{T }=v^{T}A^{T}w \;$ is a scalar} \\
&= v^{T}Aw & \text{since $A^{T}= A$} \\
&= \langle v, w \rangle
\end{align}$ (b) We have $\begin{align}
\langle \lambda_{1} v_{1} + \lambda_{2} v_{2}, w\rangle & = (\lambda_{1}v_{1}^{T}+ \lambda_{2}v_{2}^{T } ) A w \\
&= \lambda_{1} v_{1}^{T}Aw + \lambda_{2}v_{2}^T Aw & \text{ using distributivity of matrix multiplication} \\
&= \lambda_{1} \langle v_{1},w \rangle + \lambda_{2} \langle v_{2}, w \rangle
\end{align}$
(c) Write $v=(x,y)^{T}$ then $\langle v, v\rangle = a x^{2} + bxy + c^{2}y^{2}$which is clearly non-negative if
(d) All of the above since they are all symmetric.