471-Homework

Q1) Parts 1 and 2, $L(V,V;F)=L^{sym}(V,V;F)\oplus L^{asym}(V,V;F)$

1. In class we found bases for Lsym(V, V ; F ) and Lasym(V, V ; F ). Prove that their union is disjoint and is a basis of L(V, V ; F ).
2. Consider the Anti-Symmetrizer map that is, a linear map $S:L(V,V;F)\to L(V,V;F)$ defined by $\Lambda F(\alpha ,\beta )=\frac{1}{2}(F(\alpha ,\beta )-F(\beta ,\alpha ))$ Prove it satisfies ${\Lambda }^2=\Lambda$. Prove its image is $L^{asym}(V,V;F)$ and its kernel is $L^{sym}(V,V;F)$. (Hint: we proved in class the condition ${\Lambda }^2=\Lambda$ implies S is a projection onto its image along its kernel.)

Check

1. ${\Lambda }^2=\Lambda$

$\begin{array}{rl}\Lambda (\Lambda F(\alpha ,\beta ))& =\Lambda (\frac{1}{2}(F(\alpha ,\beta )-F(\beta ,\alpha )))\\ & =\frac{1}{2}(\frac{1}{2}(F(\alpha ,\beta )-F(\beta ,\alpha ))-\frac{1}{2}(F(\beta ,\alpha )-F(\alpha ,\beta )))\\ & =\frac{1}{2}(F(\alpha ,\beta )-F(\beta ,\alpha ))\\ & =\Lambda F(\alpha ,\beta )\end{array}$

2. $Im\Lambda =L^{asym}(V,V;F)$

Pick an arbitrary bilinear map $F(\alpha ,\beta )$, then $\begin{array}{rl}\Lambda F(\alpha ,\beta )& =\frac{1}{2}(F(\alpha ,\beta )-F(\beta ,\alpha ))\\ & =-\frac{1}{2}(F(\beta ,\alpha )-F(\alpha ,\beta ))=-\Lambda F(\beta ,\alpha )\end{array}$ Therefore, we can clearly see that $\Lambda F(\alpha ,\beta )=-\Lambda F(\beta ,\alpha )$, hence $Im\Lambda =L^{asym}(V,V;F)$

3. $Ker\Lambda =L^{sym}(V,V;F)$ Suppose $\Lambda F(\alpha ,\beta )=0$, then $\frac{1}{2}(F(\alpha ,\beta )-F(\beta ,\alpha ))=0⟺F(\alpha ,\beta )=F(\beta ,\alpha )\Rightarrow F\in L^{sym}(V,V;F)$, therefore, $Ker\Lambda =L^{sym}(V,V;F)$

Q6) Anti-Symmetrizer Generalization

We can define the following group action below as :$S_k\times L(V^{\times k};F)\to L(V^{\times k};F)$ where $\sigma (x_{1,i_1}\cdot ...\cdot x_{k,i_k})=x_{\sigma (1),i_1}\cdot ...\cdot x_{\sigma (k),i_k}$

1. Consider its Anti-Symmetrizer : $\Lambda :L(V^{\times k};F)\to L(V^{\times k};F)$ $\Lambda f=\frac{1}{k!}{\sum }_{\sigma \in S_k}sgn(\sigma )\sigma f$ Prove ${\Lambda }^2=\Lambda$

$\begin{array}{rl}\Lambda (\Lambda f)=\Lambda (\frac{1}{k!}\sum _{\sigma \in S_k}sgn(\sigma )\sigma f)& =\frac{1}{k!}\sum _{\tau \in S_k}sgn(\tau )\tau \left(\frac{1}{k!}\sum _{\sigma \in S_k}sgn(\sigma )(\sigma f)\right)\\ & =\frac{1}{k!}\sum _{\tau \in S_k}\sum _{\sigma \in S_k}\frac{1}{k!}sgn(\tau )\cdot sgn(\sigma )(\tau \sigma f)\\ & =\frac{1}{k!}\sum _{\tau \sigma \in S_k}\frac{1}{k!}\left(\genfrac{}{}{0ex}{}{k}{1}\right)sgn(\tau \sigma )(\tau \sigma f)\\ & =\Lambda (f)\end{array}$

2. Prove $\Lambda$ is a projection of $L(V^{\times k};F)$ onto $L^{asym}(V^{\times k};F)$

1. $Im\Lambda =L^{asym}(V^{\times k};F)$

$\Lambda f({\alpha }_1,...,{\alpha }_k)=\frac{1}{k!}{\sum }_{\sigma \in S_k}sgn(\sigma )\sigma f=-\Lambda f(a_{\tau (1)},...,a_{\tau (k)})$

1. $ker\Lambda =L^{sym}(V^{\times k};F)$

$\Lambda f=0$, then $\frac{1}{k!}{\sum }_{\sigma \in S_k}sgn(\sigma )\sigma f=0⟺{\sum }_{\sigma \in S_k,sgn(\sigma )=1}\sigma f={\sum }_{\sigma \in S_k,sgn(\sigma )=-1}\sigma f$ Since the sum of every even permutation is equal to the sum of the odd permutation, we can represent any map f as a sequence of even and odd permutations which will always be positive!, therefore $f$ must be in $L^{sym}(V^{\times k};F)$

Q7) Determinant Formula

1. Write down all bijection of $S_3$

\Ematrix{e & \Ematrix{1 \mapsto 1 \\ 2 \mapsto 2\\ 3 \mapsto 3} & (12) & \Ematrix{1 \mapsto 2 \\ 2 \mapsto 1\\ 3 \mapsto 3} & (13) & \Ematrix{1 \mapsto 3 \\ 2 \mapsto 1\\ 3 \mapsto 1} \\\hline (23) & \Ematrix{1 \mapsto 1 \\ 2 \mapsto 3\\ 3 \mapsto 2} & (123) & \Ematrix{1 \mapsto 2 \\ 2 \mapsto 3\\ 3 \mapsto 1} & (132) & \Ematrix{1 \mapsto 3 \\ 2 \mapsto 1\\ 3 \mapsto 2}}

2. Use the formula for the determinant function $\det (A)={\sum }_{\sigma \in S_n}sgn(\sigma )a_{\sigma (1),1}\cdot ...\cdot a_{\sigma (n),n}$ and the result of point (1) – and nothing else, no expansions by minors of a row or a column are allowed – to write down an explicit formula for the determinant of a 3 by 3 matrix, all 6 terms with appropriate signs, no sigma-notation is allowed.

$\begin{array}{rl}& \det (A)\\ & =sgn(e)a_{e(1),1}\cdot a_{e(2),2}\cdot a_{e(3),3}+sgn((12))a_{(12)(1),1}\cdot a_{(12)(2),2}\cdot a_{(12)(3),3}\\ & +sgn((23))a_{(23)(1),1}\cdot a_{(23)(2),2}\cdot a_{(23)(3),3}+sgn((13))a_{(13)(1),1}\cdot a_{(13)(2),2}\cdot a_{(13)(3),3}\\ & +sgn((123))a_{(23)(1),1}\cdot a_{(123)(2),2}\cdot a_{(123)(3),3}\\ & +sgn((132))a_{(132)(1),1}\cdot a_{(132)(2),2}\cdot a_{(132)(3),3}\end{array}$ $\begin{array}{rl}\det (A)& =(1)a_{1,1}\cdot a_{2,2}\cdot a_{3,3}+(-1)a_{2,1}\cdot a_{1,2}\cdot a_{3,3}+(-1)a_{1,1}\cdot a_{3,2}\cdot a_{2,3}\\ & +(-1)a_{3,1}\cdot a_{2,2}\cdot a_{1,3}+(1)a_{2,1}\cdot a_{3,2}\cdot a_{1,3}+(1)a_{3,1}\cdot a_{1,2}\cdot a_{2,3}\end{array}$

8c)

Base Case : $B_1=\text{Cmatrix}2,\det B_0=2$ B_2 = \Cmatrix{}{2 & 1 \\ 1 & 2}, $\det (B_1)=3$ B_3 = \Cmatrix{c|cc}{\color{blue}2 & \color{blue}1 & \color{blue}0\\\hline 1 & \color{red}2 & \color{red}1 \\ 0 & \color{red}1 & \color{red}2} \rightarrow \Cmatrix{}{2 & 1 \\ 1 & 3}, $\det (B_2)=2\cdot 3-2=4$ B_4 = \Cmatrix{cccc}{\color{blue}2 & \color{blue}1 & \color{blue}0 & \color{blue}0\\ 1 & 2 & 1 & 0\\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 2}, \det(B_1) = \Cmatrix{c|ccc}{\color{blue}2 & 1 & 0 & 0\\\hline 1 & \color{red}2 & \color{red}1 & \color{red}0\\ 0 & \color{red}1 & \color{red}2 & \color{red}1 \\ 0 & \color{red}0 & \color{red}1 & \color{red}2} + (-1)\Cmatrix{c|c|cc}{2 & \color{blue}1 & 0 & 0\\\hline \color{red}1 & 2 & \color{red}1 & \color{red}0\\ \color{red}0 & 1 & \color{red}2 & \color{red}1 \\ \color{red}0 & 0 & \color{red}1 & \color{red}2}, $\det (B_4)=2\cdot 4-3$ Reccurance Relation : $\det B_n=2\cdot \det B_{n-1}-\det B_{n-2}$

Q9)

2. Prove that the dimension of the column space of a matrix equals the size of the maximal nonzero ‘subdeterminant’ of this matrix, that is to say, the maximal number m such that the determinant of a submatrix sitting at the intersection of rows numbers $i_1<i_2<\cdot \cdot \cdot <i_m$ and columns numbers $j_1<j_2<\cdot \cdot \cdot <j_m$ is nonzero.

$m$ be the dimension of the column space of the matrix. Then we clearly have m linearly independent columns, lets denote them as $i_1,i_2,\cdot \cdot \cdot ,i_m$. Since the column space is equal to m, then the row space is also equal to m, hence we can also find the m linearly independent row, lets label them $j_1,...,j_m$ using this intersection, we can create an $m\times m$ matrix using the intersection of the $i_{\ell }$ and $j_{\ell }$ element. Since it is a square matrix, we can find its determinant. Since every column/row is linearly independent, the determinant will always be non-zero! Hence the maximal non-zero sub determinant matrix must consist a sequence of linearly independent columns/rows.

Let