LinArg assignment 4

exercise

1. Matrix multiplication

a) ${\mathbf{v}}^{\top }\mathbf{w}=n^2+n$ b) $\mathbf{v}{\mathbf{w}}^{\top }=\left[\begin{array}{cccc}2& 2& 2& 2\\ 4& 4& 4& 4\\ 6& 6& 6& 6\\ 8& 8& 8& 8\end{array}\right]$ c) ${\mathbf{w}}^{\top }(\mathbf{v}{\mathbf{w}}^{\top })\mathbf{v}=({\mathbf{w}}^{\top }\mathbf{v}{)}^2=({\mathbf{v}}^{\top }\mathbf{w}{)}^2=(n^2+n{)}^2=n^4+2n^3+n^2$

2. Exercise 2.47

a) suppose $r=n$ $C=A$ $R^{\prime }=I$

b) suppose $r=0$ $C=[]$ $R^{\prime }=[]$

3. Matrix multiplication and invertibility

a) $BA=CA$ $⟹BA{A}^{-1}=CA{A}^{-1}$ $⟹BI=CI$ $⟹B=C$

b) It is false. Counterexample: $A=\left[\begin{array}{cc}1& 2\\ 0& 0\end{array}\right],B=I,C=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$ $BA=IA=A=\left[\begin{array}{cc}1& 2\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 0\end{array}\right]=CA$

$AB=\left[\begin{array}{cc}1& 2\\ 0& 0\end{array}\right]I=\left[\begin{array}{cc}1& 2\\ 0& 0\end{array}\right]$ $AC=\left[\begin{array}{cc}1& 2\\ 0& 0\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 3\\ 0& 0\end{array}\right]$ $⟹AB\ne AC$

c) $BA=CA$ $⟹BA-CA=0$ $⟹(B-C)A=0$ $⟹(B-C{)}^{-1}(B-C)A=(B-C{)}^{-1}0$ $⟹IA=0$ $⟹A=0$

4. Special matrix inverses

a) $A^{-1}=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]$ b) $D^{-1}=\left[\begin{array}{ccc}\frac{1}{2}& 0& 0\\ 0& \frac{1}{3}& 0\\ 0& 0& \frac{3}{2}\end{array}\right]$

c) $B^{-1}=\left[\begin{array}{ccc}0& \frac{1}{3}& 0\\ 0& 0& \frac{3}{2}\\ \frac{1}{2}& 0& 0\end{array}\right]$

5. Inverses of matrix powers

a) $(A^k{)}^{-1}=(A^{-1}{)}^k$ Proof: $A^k(A^{-1}{)}^k=\underset{k}{\underbrace{AA\dots A}}\underset{k}{\underbrace{A^{-1}{A}^{-1}\dots A^{-1}}}$ $=\underset{k-1}{\underbrace{AA\dots A}}(A{A}^{-1})\underset{k-1}{\underbrace{A^{-1}{A}^{-1}\dots A^{-1}}}$ $=I$

b) We prove the claim by showing a nilpotent matrix $A$ has an inverse leads to contradiction. $A^k=0$ $A^k(A^{-1}{)}^k=0(A^{-1}{)}^k$ $I=0$

c) $A{A}^3=A^4$ $AI=I$ $A=I$

d) $A=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$ e) $A=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$

6. Inverse of triangular matrices

a) $L^{-1}=\left[\begin{array}{cc}1& 0\\ -a& 1\end{array}\right]$

b) A matrix is invertible $⟺A\mathbf{x}=\mathbf{b}$ has an unique solution

$a_{11}{x}_1=b_1$ $a_{22}{x}_2+a_{21}{x}_1=b_2$ $⋮$ $a_{kk}{x}_k+{\sum }_{j=1}^{k-1}{a}_{kj}{x}_j=b_k$ Each step solves uniquely for $x_k$​ since $a_{kk}\ne 0$. Hence a unique solution exists for every $b$.

c)

d) Yes