LinArg assignment 3

exercise

1. Linear functional

a) $T$ is linear functional if $T({\lambda }_1\mathbf{u}+{\lambda }_2\mathbf{v})={\lambda }_{1}T(\mathbf{u})+{\lambda }_{2}T(\mathbf{v})$

$$T({\lambda }_1\mathbf{u}+{\lambda }_2\mathbf{v})=\sum _{k=1}^{n}k({\lambda }_1{u}_k+{\lambda }_2{v}_k)=\sum _{k=1}^{n}k{\lambda }_1{u}_k+\sum _{k=1}^{n}k{\lambda }_2{v}_k$$ $$={\lambda }_1\sum _{k=1}^{n}k{u}_k+{\lambda }_2\sum _{k=1}^{n}k{v}_k={\lambda }_{1}T(\mathbf{u})+{\lambda }_{2}T(\mathbf{v})$$

The statement is proved

b)

$$T({\lambda }_1\mathbf{u}+{\lambda }_2\mathbf{v})=\sum _{k=1}^n({\lambda }_1{u}_k+{\lambda }_2{v}_k{)}^k$$ $${\lambda }_{1}T(\mathbf{u})+{\lambda }_{2}T(\mathbf{v})={\lambda }_1\sum _{k=1}^n(u_k{)}^k+{\lambda }_2\sum _{k=1}^n(v_k{)}^k=\sum _{k=1}^n({\lambda }_1(u_k{)}^k+{\lambda }_2(v_k{)}^k)$$ $$\sum _{k=1}^n({\lambda }_1{u}_k+{\lambda }_2{v}_k{)}^k\ne \sum _{k=1}^n({\lambda }_1(u_k{)}^k+{\lambda }_2(v_k{)}^k)\text{when }n\ge 2$$

Thus, $T$ is not linear functional

2. Matrix powers

IH: $A^k=\left[\begin{array}{cc}1+k& k\\ -k& 1-k\end{array}\right]$ for any $k\ge 0$ Base: $k=0$ $A^0=\left[\begin{array}{cc}1+0& 0\\ -0& 1-0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$

IS: $k\to k+1$ $A^{k+1}=A{A}^k\stackrel{IH}{=}A\left[\begin{array}{cc}1+k& k\\ -k& 1-k\end{array}\right]=\left[\begin{array}{cc}2& 1\\ -1& 0\end{array}\right]\left[\begin{array}{cc}1+k& k\\ -k& 1-k\end{array}\right]$ $=\left[\begin{array}{cc}2(1+k)-k& 2k+(1-k)\\ -(1+k)& -k\end{array}\right]=\left[\begin{array}{cc}k+2& k+1\\ -k-1& -k\end{array}\right]=\left[\begin{array}{cc}1+(k+1)& k+1\\ -(k+1)& 1-(k+1)\end{array}\right]$

By the principle of mathematical induction, $A^k=\left[\begin{array}{cc}1+k& k\\ -k& 1-k\end{array}\right]$ is true for any $k\ge 0$

3. Reconstruct a linear transformation

a) $T\left(\begin{array}{c}\left(\begin{array}{c}x\\ y\end{array}\right)\end{array}\right)=\left(\begin{array}{c}x+y\\ x+2y\\ 2x\end{array}\right)$ b) $A=\left[\begin{array}{cc}1& 1\\ 1& 2\\ 2& 0\end{array}\right]$

4. Linear transformation

$T(\mathbf{x})=A\left(\begin{array}{c}\mathbf{x}\\ 1\end{array}\right)={\mathbf{v}}_{n+1}\cdot 1+{\sum }_{i=1}^n{\mathbf{v}}_i{x}_i={\mathbf{v}}_{n+1}+{\sum }_{i=1}^n{\mathbf{v}}_i{x}_i$ Necessity $T$ is linear when $T(0)=0$ $T(0)=A\left(\begin{array}{c}0\\ 1\end{array}\right)={\mathbf{v}}_{n+1}\cdot 1={\mathbf{v}}_{n+1}$ meaning that ${\mathbf{v}}_{n+1}=0$

Sufficiency if ${\mathbf{v}}_{n+1}=0$, then $T={\sum }_{i=1}^n{\mathbf{v}}_i{x}_i$ which is a matrix transformation and therefore a linear transformation

5. Matrix multiplication

$$\left[\begin{array}{ccc}1& 0& 3\\ 3& 1& 9\\ 0& 3& 1\end{array}\right]+x\left[\begin{array}{ccc}0& 1& 0\\ 0& 3& 1\\ 1& 0& 3\end{array}\right]+y\left[\begin{array}{ccc}0& 0& 1\\ 1& 0& 3\\ 0& 1& 0\end{array}\right]+z\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$$

Observing numbers at position $(1,1)$ $1+z=0$ $z=-1$ Observing numbers at position $(1,2)$ $x=0$ Observing numbers at position $(1,3)$ $3+y=0$ $y=-3$

Verification:

$$\left[\begin{array}{ccc}1& 0& 3\\ 3& 1& 9\\ 0& 3& 1\end{array}\right]-3\left[\begin{array}{ccc}0& 0& 1\\ 1& 0& 3\\ 0& 1& 0\end{array}\right]-\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$$ $$\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$$

The verification holds. The solutions are $x=0,y=-3,z=-1$

b) $(AB{)}^k=\underset{\text{k times}}{\underbrace{(AB)(AB)\dots (AB)}}=\underset{\text{k times}}{\underbrace{AA\dots A}}\underset{\text{k times}}{\underbrace{B\dots BB}}=A^k{B}^k$ c) $(AB{)}^k=A^k{B}^k=0B^k=0$ $AB$ is nilpotent, the nilpotent degree of is less or equal than $0$

d) $(I-A)(I+A+\cdots +A^{k-1})$ $=I(I+A+\cdots +A^{k-1})-A(I+A+\cdots +A^{k-1})$ $=(I+A+\cdots +A^{k-1})-(A+A^2+\cdots +A^k)$ $=I-A^k=I-0=I$ The equation is proved.

e)

6. Rotation matrices

a) Let $\varphi =\frac{\pi }{2}$ $A=Q\left(\frac{\pi }{2}\right)=\left[\begin{array}{cc}\cos \left(\frac{\pi }{2}\right)& \sin \left(\frac{\pi }{2}\right)\\ \sin \left(\frac{\pi }{2}\right)& \cos \left(\frac{\pi }{2}\right)\end{array}\right]=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$ b) $Q({\varphi }_3)=Q({\varphi }_1)Q({\varphi }_2)=\left[\begin{array}{cc}\cos ({\varphi }_1)& -\sin ({\varphi }_1)\\ \sin ({\varphi }_1)& \cos ({\varphi }_1)\end{array}\right]\left[\begin{array}{cc}\cos ({\varphi }_2)& -\sin ({\varphi }_2)\\ \sin ({\varphi }_2)& \cos ({\varphi }_2)\end{array}\right]$ $=\left[\begin{array}{cc}\cos ({\varphi }_1)\cos ({\varphi }_2)-\sin ({\varphi }_1)\sin ({\varphi }_2)& -\cos ({\varphi }_1)\sin ({\varphi }_2)-\sin ({\varphi }_1)\cos ({\varphi }_2)\\ \sin ({\varphi }_1)\cos ({\varphi }_2)+\cos ({\varphi }_1)\sin ({\varphi }_2)& \cos ({\varphi }_1)\cos ({\varphi }_2)-\sin ({\varphi }_1)\sin ({\varphi }_2)\end{array}\right]$ $=\left[\begin{array}{cc}\cos ({\varphi }_1+{\varphi }_2)& -\sin ({\varphi }_1+{\varphi }_2)\\ \sin ({\varphi }_1+{\varphi }_2)& \cos ({\varphi }_1+{\varphi }_2)\end{array}\right]$ $=Q({\varphi }_1+{\varphi }_2)$

c) Let $A=Q(\varphi )$ and let $B=Q(2\pi -\varphi )$ $AB=Q(\varphi )Q(2\pi -\varphi )=Q(\varphi +2\pi -\varphi )=Q(2\pi )=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$ $BA=Q(2\pi -\varphi )Q(\varphi )=Q(2\pi -\varphi +\varphi )=Q(2\pi )=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$