LinArg assignment 2

exercise

1. Rank of a matrix

a) $A_1=\left[\begin{array}{cc}2& 3\\ 3& 4\end{array}\right]$ $A_2=\left[\begin{array}{ccc}2& 3& 4\\ 3& 4& 5\\ 4& 5& 6\end{array}\right]$ $A_3=\left[\begin{array}{cccc}2& 3& 4& 5\\ 3& 4& 5& 6\\ 4& 5& 6& 7\\ 5& 6& 7& 8\end{array}\right]$

b) $rank(A_2)=2$ $\left(\begin{array}{c}2\\ 3\end{array}\right)\ne \lambda \left(\begin{array}{c}3\\ 4\end{array}\right)$ $rank(A_3)=2$ $\left(\begin{array}{c}4\\ 5\\ 6\end{array}\right)=2\left(\begin{array}{c}3\\ 4\\ 5\end{array}\right)-\left(\begin{array}{c}2\\ 3\\ 4\end{array}\right)$ $\left(\begin{array}{c}3\\ 4\\ 5\end{array}\right)\ne \lambda \left(\begin{array}{c}2\\ 3\\ 4\end{array}\right)$

$rank(A_4)=2$ for $3\le k\le m$: ${\mathbf{v}}_k-{\mathbf{v}}_1=\left(\begin{array}{c}1+k\\ ⋮\\ m+k\end{array}\right)-\left(\begin{array}{c}2\\ ⋮\\ m+1\end{array}\right)=(k-1)\left(\begin{array}{c}1\\ ⋮\\ 1\end{array}\right)=(k-1)({\mathbf{v}}_2-{\mathbf{v}}_1)$ ${\mathbf{v}}_k=(k-1)({\mathbf{v}}_2-{\mathbf{v}}_1)+{\mathbf{v}}_1=(2-k){\mathbf{v}}_1+(k-1){\mathbf{v}}_2$

2. Nullspace as a hyperplane

a) $rank(A)=1$ b) Let $\mathbf{\lambda }=\left(\begin{array}{c}{\lambda }_1\\ {\lambda }_2\\ ⋮\\ {\lambda }_n\end{array}\right)$ $A\mathbf{x}=0$ $\left[\begin{array}{cccc}{\lambda }_1\mathbf{v}& {\lambda }_2\mathbf{v}& \dots & {\lambda }_n\mathbf{v}\end{array}\right]\mathbf{x}=0$ ${\sum }_{i=1}^n{x}_i{\lambda }_i\mathbf{v}=0$ Since $\mathbf{v}\ne 0$, ${\sum }_{i=1}^n{x}_i{\lambda }_i=0$ This is the equation for the hyperplane. Hence the statement is proved.

3. Matrix transformations

a) rotate the input around y-axis by 45 degree (counter-clockwise)

b) $\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& -1\end{array}\right]$

4. Scalar product

a) $A(\lambda \mathbf{x}+\mu \mathbf{y})=\lambda A(\mathbf{x})+\mu A(\mathbf{y})=0\lambda +0\mu =0$ Let $\mathbf{v}=\lambda \mathbf{x}+\mu \mathbf{y}$ $⟹A\mathbf{v}=0$ $⟹\left(\begin{array}{c}{\mathbf{u}}_1^{\top }\mathbf{v}\\ {\mathbf{u}}_2^{\top }\mathbf{v}\\ ⋮\\ {\mathbf{u}}_m^{\top }\mathbf{v}\end{array}\right)=0$ $⟹\forall i\in [m],{\mathbf{u}}_i^{\top }\mathbf{v}=0$ $⟹\lambda \mathbf{x}+\mu \mathbf{y}$ is orthogonal to each of ${\mathbf{u}}_1,{\mathbf{u}}_2,\dots ,u_m$

5. Rank of matrices

a) $rank(A)=2$ b) $rank(A)=3$

6. Skew-symmetric matrices

a) $\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$ b) $A^{\top }=[a_{ji}{]}_{i=1,j=1}^{nm}=-A=[-a_{ij}{]}_{i=1,j=1}^{nm}$ $⟹\forall i\in [m]a_{ii}=-a_{ii}$ $⟹\forall i\in [m]a_{ii}=0$

c)

• $0$ $\forall i,j\in [m]a_{ij}=a_{ji}=-a_{ij}$ $⟹\forall i,j\in [m]a_{ij}=0$ $⟹A=0$

d) $A=\left[\begin{array}{ccc}0& a& b\\ -a& 0& c\\ -b& -c& 0\end{array}\right]$

$-\frac{c}{a}{\mathbf{a}}_1+\frac{b}{a}{\mathbf{a}}_2=-\frac{c}{a}\left(\begin{array}{c}0\\ -a\\ -b\end{array}\right)+\frac{b}{a}\left(\begin{array}{c}a\\ 0\\ -c\end{array}\right)=\left(\begin{array}{c}0\\ c\\ \frac{bc}{a}\end{array}\right)+\left(\begin{array}{c}b\\ 0\\ -\frac{bc}{a}\end{array}\right)=\left(\begin{array}{c}b\\ c\\ 0\end{array}\right)={\mathbf{a}}_3$ We see ${\mathbf{a}}_3$ is linear dependent on ${\mathbf{a}}_1$ and ${\mathbf{a}}_2$, so the rank muss be less or equal to 2

7. Embedding a line in ${\mathbb{R}}^m$

Let $A=[\mathbf{v}]$ $T_A(\mathbf{x})=A\mathbf{x}=[\mathbf{v}]\mathbf{x}=\left(\begin{array}{c}{v}_{1}x\\ v_{2}x\\ ⋮\\ v_{m}x\end{array}\right)=x\left(\begin{array}{c}{v}_1\\ v_2\\ ⋮\\ v_m\end{array}\right)=x\mathbf{v}$ Since $x\in \mathbb{R}$ $\{T_A(\mathbf{x}):\mathbf{x}\in {\mathbb{R}}^1\}=\{x\mathbf{v}:x\in \mathbb{R}\}=\{\lambda \mathbf{v}:\lambda \in \mathbb{R}\}=L$ The statement is proved.