- #1
mathmari
Gold Member
MHB
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Hey!
I am looking at the proof of the following proposition:
Let $\displaystyle{\phi, \ldots , \phi_k\in V^{\star}}$ and $\displaystyle{\psi_i=\sum_{j=1}^ka_{ij}\phi_j, \ i=1, \ldots , k}$ with a matrix $\displaystyle{(a_{ij})\in M(k\times k, \mathbb{R})}$.
Then it holds that $\displaystyle{\psi_1\land \ldots \land \psi_k=\det (a_{ij})\cdot \phi_1\land \ldots \land \phi_k}$. The proof is the following:
We see that $\displaystyle{\det (a_{ij})=\sum_{\pi\in \text{Per}_k}\text{sign}(\pi)a_{1\pi(1)}a_{2\pi(2)}\cdot \ldots \cdot a_{k\pi (k)}}$, where $\text{Per}_k$ is the group of all permutations of $1, \ldots , k$.
It holds that \begin{align*}\psi_2\land \ldots \land \psi_k&=\left (\sum_{j_1=1}^ka_{1j_1}\phi_{j_1}\right )\land \ldots \land \left (\sum_{j_k=1}^ka_{kj_k}\phi_{j_k}\right ) \\ & = \sum_{j_1, \ldots j_k=1}^k\left (a_{1j_1}\ldots a_{kj_k}\right )\phi_{j_1}\land \ldots \land \phi_{j_k} \\ &= \sum_{\pi\in \text{Per}_k}^k\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\phi_{\pi (1)}\land \ldots \land \phi_{\pi (k)} \\ & = \sum_{\pi\in \text{Per}_k}^k\text{sign}(\pi)\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\phi_{1}\land \ldots \land \phi_{k} \\ & = \det (a_{ij})\phi_{1}\land \ldots \land \phi_{k}\end{align*}
q.e.d.
I want to understand that proof.
For the exterior product the following properties hold:
By induction we can show that it holds that $\displaystyle{\phi_1\land \ldots \land \left (\sum_{j=1}^n\lambda_j \phi_{i_j}\right )\land \ldots \land \phi_k=\sum_{j=1}^n\lambda_j \phi_1\land \ldots \land \phi_{i_j} \land \ldots \land \phi_k}$ for each $n\geq 2$.
Base case: For $n=2$ it holds by the first property.
Inductive hypothesis: We suppose that it holds for $n=m$ :
$\displaystyle{\phi_1\land \ldots \land \left (\sum_{j=1}^m\lambda_j \phi_{i_j}\right )\land \ldots \land \phi_k=\sum_{j=1}^m\lambda_j \phi_1\land \ldots \land \phi_{i_j} \land \ldots \land \phi_k}$.
Inductive step: We want to show that it holds also for $n=m+1$ :
\begin{align*}\phi_1\land \ldots \land \left (\sum_{j=1}^{m+1}\lambda_j \phi_{i_j}\right )\land \ldots \land \phi_k&=\phi_1\land \ldots \land \left (\sum_{j=1}^m\lambda_j \phi_{i_j}+\lambda_{m+1} \phi_{i_{m+1}}\right )\land \ldots \land \phi_k \\ & \overset{ \text{ Property } }{ = } \phi_1\land \ldots \land \left (\sum_{j=1}^m\lambda_j \phi_{i_j}\right )\land \ldots \land \phi_k +\phi_1\land \ldots \land \left (\lambda_{m+1} \phi_{i_{m+1}}\right )\land \ldots \land \phi_k \\ & = \phi_1\land \ldots \land \left (\sum_{j=1}^m\lambda_j \phi_{i_j}\right )\land \ldots \land \phi_k +\lambda_{m+1}\phi_1\land \ldots \land \phi_{i_{m+1}}\land \ldots \land \phi_k \\ & \overset{ \text{ In.Hyp. } }{ = } \sum_{j=1}^m\lambda_j \phi_1\land \ldots \land \phi_{i_j} \land \ldots \land \phi_k+\lambda_{m+1}\phi_1\land \ldots \land \phi_{i_{m+1}}\land \ldots \land \phi_k \\ & = \sum_{j=1}^{m+1}\lambda_j \phi_1\land \ldots \land \phi_{i_j} \land \ldots \land \phi_k\end{align*}
Is this correct? Is the under bound of $n$ (i.e. $n\geq 2$) right? (Wondering)
Therefore, at the first part of the proof we get \begin{align*}\psi_2\land \ldots \land \psi_k&=\left (\sum_{j_1=1}^ka_{1j_1}\phi_{j_1}\right )\land \ldots \land \left (\sum_{j_k=1}^ka_{kj_k}\phi_{j_k}\right ) \\ & = \sum_{j_1=1}^k\left (a_{1j_1}\right )\phi_{j_1}\land \ldots \land \left (\sum_{j_k=1}^ka_{kj_k}\phi_{j_k}\right ) \\ &= \sum_{j_1, \ldots , j_k=1}^k\left (a_{1j_1}\ldots a_{kj_k}\right )\phi_{j_1}\land \ldots \land \phi_{j_k}\end{align*}
Is this correct? Or do we have to proof this part also by induction? (Wondering) I haven't really understood the next step of the proof. Do we consider the permutation $\pi: (1, \ldots , k)\mapsto (j_1, \ldots , j_k)$ and so we get that $\displaystyle{\sum_{j_1, \ldots , j_k=1}^k\left (a_{1j_1}\ldots a_{kj_k}\right )\phi_{j_1}\land \ldots \land \phi_{j_k}=\sum_{\pi\in \text{Per}_k}^k\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\phi_{\pi (1)}\land \ldots \land \phi_{\pi (k)}}$ ? But isn't $j_i$ the variable of the sum? (Wondering)
If we do that like that then we get that $\displaystyle{\sum_{\pi\in \text{Per}_k}^k\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\phi_{\pi (1)}\land \ldots \land \phi_{\pi (k)}=\sum_{\pi\in \text{Per}_k}^k\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\cdot \text{sign}(\pi)\phi_{1}\land \ldots \land \phi_{k}}$, right? (Wondering)
But why does it then hold that $\displaystyle{\sum_{\pi\in \text{Per}_k}^k\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\cdot \text{sign}(\pi)=\det (a_{ij})}$ ? (Wondering)
I am looking at the proof of the following proposition:
Let $\displaystyle{\phi, \ldots , \phi_k\in V^{\star}}$ and $\displaystyle{\psi_i=\sum_{j=1}^ka_{ij}\phi_j, \ i=1, \ldots , k}$ with a matrix $\displaystyle{(a_{ij})\in M(k\times k, \mathbb{R})}$.
Then it holds that $\displaystyle{\psi_1\land \ldots \land \psi_k=\det (a_{ij})\cdot \phi_1\land \ldots \land \phi_k}$. The proof is the following:
We see that $\displaystyle{\det (a_{ij})=\sum_{\pi\in \text{Per}_k}\text{sign}(\pi)a_{1\pi(1)}a_{2\pi(2)}\cdot \ldots \cdot a_{k\pi (k)}}$, where $\text{Per}_k$ is the group of all permutations of $1, \ldots , k$.
It holds that \begin{align*}\psi_2\land \ldots \land \psi_k&=\left (\sum_{j_1=1}^ka_{1j_1}\phi_{j_1}\right )\land \ldots \land \left (\sum_{j_k=1}^ka_{kj_k}\phi_{j_k}\right ) \\ & = \sum_{j_1, \ldots j_k=1}^k\left (a_{1j_1}\ldots a_{kj_k}\right )\phi_{j_1}\land \ldots \land \phi_{j_k} \\ &= \sum_{\pi\in \text{Per}_k}^k\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\phi_{\pi (1)}\land \ldots \land \phi_{\pi (k)} \\ & = \sum_{\pi\in \text{Per}_k}^k\text{sign}(\pi)\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\phi_{1}\land \ldots \land \phi_{k} \\ & = \det (a_{ij})\phi_{1}\land \ldots \land \phi_{k}\end{align*}
q.e.d.
I want to understand that proof.
For the exterior product the following properties hold:
- It is linear in each argument:
\begin{equation*}\phi_1\land \ldots \land \left (\lambda \phi_i'+\mu\phi_i''\right )\land \ldots \land \phi_k=\lambda \phi_1\land \ldots \land \phi_i' \land \ldots \land \phi_k+\mu \phi_1\land \ldots \land \phi_i''\land \ldots \land \phi_k\end{equation*} - It is alternating:
\begin{equation*}\phi_{\pi(1)}\land \ldots \land \phi_{\pi(k)}=\text{sign}(\pi)\phi_1\land \ldots \land \phi_k\end{equation*} for each permutation $\pi$ of $1, \ldots k$
By induction we can show that it holds that $\displaystyle{\phi_1\land \ldots \land \left (\sum_{j=1}^n\lambda_j \phi_{i_j}\right )\land \ldots \land \phi_k=\sum_{j=1}^n\lambda_j \phi_1\land \ldots \land \phi_{i_j} \land \ldots \land \phi_k}$ for each $n\geq 2$.
Base case: For $n=2$ it holds by the first property.
Inductive hypothesis: We suppose that it holds for $n=m$ :
$\displaystyle{\phi_1\land \ldots \land \left (\sum_{j=1}^m\lambda_j \phi_{i_j}\right )\land \ldots \land \phi_k=\sum_{j=1}^m\lambda_j \phi_1\land \ldots \land \phi_{i_j} \land \ldots \land \phi_k}$.
Inductive step: We want to show that it holds also for $n=m+1$ :
\begin{align*}\phi_1\land \ldots \land \left (\sum_{j=1}^{m+1}\lambda_j \phi_{i_j}\right )\land \ldots \land \phi_k&=\phi_1\land \ldots \land \left (\sum_{j=1}^m\lambda_j \phi_{i_j}+\lambda_{m+1} \phi_{i_{m+1}}\right )\land \ldots \land \phi_k \\ & \overset{ \text{ Property } }{ = } \phi_1\land \ldots \land \left (\sum_{j=1}^m\lambda_j \phi_{i_j}\right )\land \ldots \land \phi_k +\phi_1\land \ldots \land \left (\lambda_{m+1} \phi_{i_{m+1}}\right )\land \ldots \land \phi_k \\ & = \phi_1\land \ldots \land \left (\sum_{j=1}^m\lambda_j \phi_{i_j}\right )\land \ldots \land \phi_k +\lambda_{m+1}\phi_1\land \ldots \land \phi_{i_{m+1}}\land \ldots \land \phi_k \\ & \overset{ \text{ In.Hyp. } }{ = } \sum_{j=1}^m\lambda_j \phi_1\land \ldots \land \phi_{i_j} \land \ldots \land \phi_k+\lambda_{m+1}\phi_1\land \ldots \land \phi_{i_{m+1}}\land \ldots \land \phi_k \\ & = \sum_{j=1}^{m+1}\lambda_j \phi_1\land \ldots \land \phi_{i_j} \land \ldots \land \phi_k\end{align*}
Is this correct? Is the under bound of $n$ (i.e. $n\geq 2$) right? (Wondering)
Therefore, at the first part of the proof we get \begin{align*}\psi_2\land \ldots \land \psi_k&=\left (\sum_{j_1=1}^ka_{1j_1}\phi_{j_1}\right )\land \ldots \land \left (\sum_{j_k=1}^ka_{kj_k}\phi_{j_k}\right ) \\ & = \sum_{j_1=1}^k\left (a_{1j_1}\right )\phi_{j_1}\land \ldots \land \left (\sum_{j_k=1}^ka_{kj_k}\phi_{j_k}\right ) \\ &= \sum_{j_1, \ldots , j_k=1}^k\left (a_{1j_1}\ldots a_{kj_k}\right )\phi_{j_1}\land \ldots \land \phi_{j_k}\end{align*}
Is this correct? Or do we have to proof this part also by induction? (Wondering) I haven't really understood the next step of the proof. Do we consider the permutation $\pi: (1, \ldots , k)\mapsto (j_1, \ldots , j_k)$ and so we get that $\displaystyle{\sum_{j_1, \ldots , j_k=1}^k\left (a_{1j_1}\ldots a_{kj_k}\right )\phi_{j_1}\land \ldots \land \phi_{j_k}=\sum_{\pi\in \text{Per}_k}^k\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\phi_{\pi (1)}\land \ldots \land \phi_{\pi (k)}}$ ? But isn't $j_i$ the variable of the sum? (Wondering)
If we do that like that then we get that $\displaystyle{\sum_{\pi\in \text{Per}_k}^k\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\phi_{\pi (1)}\land \ldots \land \phi_{\pi (k)}=\sum_{\pi\in \text{Per}_k}^k\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\cdot \text{sign}(\pi)\phi_{1}\land \ldots \land \phi_{k}}$, right? (Wondering)
But why does it then hold that $\displaystyle{\sum_{\pi\in \text{Per}_k}^k\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\cdot \text{sign}(\pi)=\det (a_{ij})}$ ? (Wondering)
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