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# System prepended metadata

title: PSET 4

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# PSET 4
### Collaborators
* Gita Permatasari Sujatmiko
* Annisa Dian Nugrahani

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$$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$


### Soal Teori
2a) Buktikan bahwa jika ${Q}$ merupakan matriks orthogonal berukuran *m $\times$ m*, maka ${||Qx||_{2}=||x||_{2}}$ untuk setiap vektor **x** berukuran *n* $\times$ 1.
* ${||x||_{2}=\sqrt{\sum_{i}x_{i}^{2}}}$
* ${\langle\,x,y\rangle\,=x^{T}y=\sum\,x_{i}y_{i}}$
* ${||x||_{2}^{2}=\sum\,x_{i}^{2}=x^{T}x}$
* Note that ${||x||_{2}^{2}=x^{T}x}$ dan ${Q^{T}Q=I}$
* ${||Qx||_{2}^{2}=(Qx)^{T}Qx=x^{T}Q^{T}Qx=x^{T}(Q^{T}Q)x=x^{T}x=||x||_{2}^{2}}$
* sehingga ${||Qx||_{2}=||x||_{2}}$
$\therefore$ Terbukti

2b) Buktikan bahwa jika *Q1* dan *Q2* orthogonal, maka hasil kali *Q1Q2* juga orthogonal
* Misal: matriks ${A_{nxn}}$ dan ${B_{nxn}}$ orthogonal, sehingga ${A^{T}A = I_{n}}$ dan ${B^{T}B = I_{n}}$
* Ingin dibuktikan ${(AB)^{T}(AB)=I_{n}}$
* ${(AB)^{T}(AB)=(B^{T}A^{T})AB=B^{T}(A^{T}A)B=B^{T}B=I_{n}}$
$\therefore$ Terbukti

2c) Misalkan $v$ adalah vektor tak nol berukuran $m$ × 1 dan $I$ adalah matriks identitas berukuran $m \times m$. Buktikan bahwa matriks $H$ yang dibentuk dengan
$$ H = I - \frac{2}{v^{T}v}vv^{T} $$
merupakan matriks orthogonal. (Petunjuk: Agar tidak bingung, apa dimensi dari ${v^{T}v}$ dan $vv^{T}$?)

<!---
$I_{m} = \begin{bmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix}$, $v = \begin{bmatrix} v_{1} \\ v_{2} \\ v_{3} \\ \vdots \\ v_{m} \end{bmatrix}$, $v^T = \begin{bmatrix} v_{1} & v_{2} & v_{3} & \cdots & v_{m} \end{bmatrix}$

$$vv^T = \begin{bmatrix} v_{1} \\ v_{2} \\ v_{3} \\ \vdots \\ v_{m} \end{bmatrix} \begin{bmatrix} v_{1} & v_{2} & v_{3} & \cdots & v_{m} \end{bmatrix} = \begin{bmatrix} {v_1}^2 & 0 & 0 & \cdots & 0 \\ 0 & {v_2}^2 & 0 & \cdots & 0 \\ 0 & 0 & {v_3}^2 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & {v_m}^2 \end{bmatrix} $$

$$v^Tv = \begin{bmatrix} v_{1} & v_{2} & v_{3} & \cdots & v_{m} \end{bmatrix} \begin{bmatrix} v_{1} \\ v_{2} \\ v_{3} \\ \vdots \\ v_{m} \end{bmatrix} = \begin{bmatrix} {v_1}^2 + {v_2}^2 + {v_3}^2 + \cdots + {v_m}^2\end{bmatrix} $$

$$ H = I - \frac{2}{v^{T}v}vv^{T} = 
\begin{bmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix}-
2\cdot \begin{bmatrix} \frac{1}{{v_1}^2 + {v_2}^2 + \cdots + {v_m}^2} \end{bmatrix} \begin{bmatrix} {v_1}^2 & 0 & 0 & \cdots & 0 \\ 0 & {v_2}^2 & 0 & \cdots & 0 \\ 0 & 0 & {v_3}^2 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & {v_m}^2 \end{bmatrix} $$

$$ = \begin{bmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix} -
\begin{bmatrix} \frac{2{v_1}^2}{{v_1}^2 + {v_2}^2 + \cdots + {v_m}^2} & 0 & 0 & \cdots & 0 \\ 0 & \frac{2{v_2}^2}{{v_1}^2 + {v_2}^2 + \cdots + {v_m}^2} & 0 & \cdots & 0 \\ 0 & 0 & \frac{2{v_3}^2}{{v_1}^2 + {v_2}^2 + \cdots + {v_m}^2} & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \frac{2{v_m}^2}{{v_1}^2 + {v_2}^2 + \cdots + {v_m}^2} \end{bmatrix} $$
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Dimensi dari ${v^{T}v}$ adalah $1\times1$ dan dimensi dari $vv^{T}$ adalah $m\times m$


