LinArg assignment 6

1. Subspaces of function spaces and ${\mathbb{R}}^{m\times m}$

a)

• Non-empty: all constant function $f\in U$
• closed under addition: $f,g\in U$, arbitrary $x\in [0,1]$
  • $(f+g)(x)=f(x)+g(x)=f(1-x)+g(1-x)=(f+g)(1-x)$
• closed under scalar multiplication: $f,g\in U$, $\alpha \in \mathbb{R}$, arbitrary $x\in [0,1]$
  • $(\alpha f)(x)=\alpha f(x)=\alpha f(1-x)=(\alpha f)(1-x)$ $⟹U$ is a subspace

b)

$\text{dim}({\mathcal{D}}_m)=m$ Proof: For $i\in [m]$ let $E_{ii}$ be the $m\times m$ matrix with 1 at $(i,i)$ and 0 elsewhere We try to prove that $\{E_{ii}|i\in [m]\}$ is the basis of ${\mathcal{D}}_m$

• Spanning: Any diagonal matrix can be expressed as ${\sum }_{i=1}^n{c}_i{E}_{ii}$ for $c_i\in \mathbb{R}$
• Linear independence: Consider the equation ${\sum }_{i=1}^n{d}_i{E}_{ii}=0=\left[\begin{array}{cccc}{d}_1& 0& \dots & 0\\ 0& d_2& \dots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & d_n\end{array}\right]$.
  • $d_i$ must be zero, which shows the linear independence of $⟹$ $\{E_{ii}|i\in [m]\}$ is the basis of ${\mathcal{D}}_m$ Since the dimension is the cardinality of a basis. $\text{dim}({\mathcal{D}}_m)=|\{E_{ii}|i\in [m]\}|=m$

exercise

2. Skew-symmetric matrices as a subspace

a)

we prove the claim by showing the following statements (using the definition of Subspaces)

• for all $A,B\in S_m$ we have $A+B\in S_m$
  • $A+B\stackrel{def}{=}-A^{\top }+(-B^{\top })$
  • $⟹A+B=-(A^{\top }+B^{\top })$
  • $⟹A+B=-(A+B{)}^{\top }$
  • $⟹A+B$ is a skew-symmetric matrix and thus $A+B\in S_m$
• for all $A\in S_m$ and $\lambda \in \mathbb{R}$ we have $\lambda A\in S_m$
  • $\lambda A=\lambda (-A^{\top })=-\lambda A^{\top }=-(\lambda A^{\top })=-(\lambda A{)}^{\top }$
  • $⟹\lambda A$ is a skew-symmetric matrix and thus $\lambda A\in S_m$ Hence, $S_m$ is a subspace of ${\mathbb{R}}^{m\times m}$

b)

The dimension of $S_m$ is $\frac{m^2-m}{2}$

We define:

$$A_{xy}=[a_{ij}{]}_{m\times m},a_{ij}=\{\begin{array}{l}1,\text{if }i=x\text{ and }j=y,\\ 0,\text{otherwise}\\ -1,\text{if }i=y\text{ and }j=x\end{array}$$

We prove the set $A=\{A_{xy}|x,y\in [m],y>x\}$ is a basis of $S_m$

• This set spans $S_m$
  • For any matrix $B$ in $S_m$ we can write $B$ as a linear combination of the matrices $A_{ij}$
  • $B={\sum }_{1\le i<j\le m}{b}_{ij}{A}_{ij}$
  • so the $(i,j)$-th entry is $b_{ij}$ ($i<j$), and the $(j,i)$-th entry is $-b_{ij}$, matching the structure of $B$
  • Therefore $A$ spans $S_m$
• This set $A$ is linear independent Assume

$$\sum _{1\le x<y\le m}{c}_{xy}{A}_{xy}=0$$

Consider the $i,j$-entry with $i<j$. Since only $A_{ij}$ has a nonzero element in that position, we obtain $c_{ij}=0$ This holds for all $i<j$ Hence all coefficient $c_{ij}$ are zero, and the set $A$ is linear independent. Conclusion Now we have a basis of $S_m$ The size of this basis is the number of elements on top of the diagonal of a matrix in ${\mathbb{R}}^{m\times m}$, which can be calculated by $\frac{m(m-1)}{2}=\frac{m^2-m}{2}$