# Linear Algebra hw2

## 1-4-11

### a.
for any $a_{ij}\in A,i\neq j, a_{ij}=c_{ij}+c_{ji}=c_{ji}+c_{ij}=a_{ji}$

therefore, $\forall a_{ij}\in A$, we have

$$\begin{equation}\left\{\begin{aligned}a_{ij}&=2c_{ij}&, i=j \\ a_{ij}&=c_{ij}+c_{ji}=a_{ji}&, i\neq j\end{aligned} \right . \end{equation}$$
$\Rightarrow A$ is symmetric.

### b.
for any $b_{ij}\in B,i\neq j, b_{ij}=c_{ij}-c_{ji}, b_{ji}=c_{ji}-c_{ij}$

$b_{ij}=b_{ji}\Leftrightarrow b_{ij} - b_{ji} = c_{ij}-c_{ji}-c_{ji}+c_{ij}=0$

$\Rightarrow b_{ij}=b_{ji}\Leftrightarrow c_{ij}=c{ji}$

since C is nonsymmetric, $c_{ij}\neq c{ji} \Rightarrow b_{ij}\neq b_{ji}\Rightarrow$B is nonsymmetric.

### c.
$\forall D_{ij}\in D, i\neq j,d_{ij}=\sum^n_{k=1}c_{ik}{c^{'}}_{kj}=\sum^n_{k=1}c_{ik}c_{jk}$

$d_{ji}=\sum^n_{k=1}c_{jk}{c^{'}}_{ki}=\sum^n_{k=1}c_{jk}c_{ik}$

$\because d_{ij}=\sum^n_{k=1}c_{ik}c_{jk}=\sum^n_{k=1}c_{jk}c_{ik}=d_{ji}$

$\therefore$ D is symmetric

### d.

according to the result from (c.), $C^TC$ is symmetric for any nonsymmetric matrix C
$\therefore CC^T=(C^T)^T(C^T)$ is symmetric
$\therefore$ any symmetric matrix - symmetric matrix is also symmetric.

### e.
$F=(I+C)(I+C^T)=I^2+CI+IC^T+CC^T=I+(C+C^T)+CC^T$
according to the result of (a.) and (d.), F is symmetric.

### f

$G=(I+C)(I-C^T)=I^2+CI-IC^T-CC^T=I+(C-C^T)+CC^T$

according to the result of (b.) and (d.), F is nonsymmetric.

## 1-5-14


$T=UR, \forall t_{ij}\in T, t_{ij}=\sum_{k=1}^nu_{ik}r_{kj}, u_{ik}\in U, r_{kj}\in R$

U and R are upper triangular matrices
$\Rightarrow u_{ij}=r_{ij}=0, i>j$

$$\begin{equation}t_{ij}=\sum^n_{k=1}u_{ik}r_{kj}\end{equation}$$

if $i>j$:

$$\begin{equation}\left\{\begin{aligned}u_{ik}&=0&, k<i \\ r_{kj}&=0&, k=i \\ r_{kj}&=0&,k>i>j\end{aligned}\right .\end{equation}$$

$\Rightarrow t_{ij}=0$

if $i=j$:

$$\begin{equation}\left\{\begin{aligned}u_{ik}&=0&, k<i \\ &u_{ik}r_{kj} &, k=i \\ r_{kj}&=0&,k>i>j\end{aligned}\right .\end{equation}$$

$\Rightarrow t_{ij}=u_{ij}r_{ij}, i=j$

Q.E.D.

## 1-5-24


### a.

$\because A$ is row equivalent to $B$ and $B$ is row equivalent to $C$

let $E=\{E_1 \ldots E_n\}$ be the elementary matrices of the series of row operation
we have
$$\begin{equation}B=E_{a_n}\ldots E_{a_1}A\end{equation}$$
and we have
$$\begin{equation}C=E_{b_n}\ldots E_{b_1}B\end{equation}$$

and we substitute equation (5) into equation (6), we get
$$\begin{equation}C=E_{b_n}\ldots E_{b_1}E_{a_n}\ldots E_{a_1}A\end{equation}$$
since $E_{b_n}\ldots E_{b_1}E_{a_n}\ldots E_{a_1}$ is also a series of row operation, we get A is row equivalent to C. Q.E.D.

### b.

since $A$ and $B$ are nonsingular, we can find a series of row operation $E_n \ldots E_2E_1$ is equal to $A^{-1}$ and $B^{-1}$
let $E_a = E_{a_n}\ldots E_{a_1}=A^{-1}$, $E_b = E_{b_n}\ldots E_{b_1}=B^{-1}$

$E_aA=A^{-1}A=I$, $E_bB=B^{-1}B=I$
since row operation is invertible, if A is row equivalent to C and B equivalent to C, A is row equivalent to B. Q.E.D.