Understand Proof for Exterior Product Proposition

[mathmari]

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:

• 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 the first property we have that $\displaystyle{\phi_1\land \ldots \land \left (\lambda \phi_i'+\mu\phi_i''\right )\land \ldots \land \phi_k=\lambda \phi_1\land \ldots \land \phi_i' \land \ldots \land \phi_k+\mu \phi_1\land \ldots \land \phi_i''\land \ldots \land \phi_k}$.
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 like Serena]

Is this correct? Is the under bound of $n$ (i.e. $n\geq 2$) right? (Wondering)

Hey mathmari!

It looks all correct to me. (Smile)
The underbound of $n=2$ works, and for the record, so does $n=1$, and $n=0$.
The base case $n=0$ is trivially true - no properties needed.

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)

Yep. All correct.
Formally it needs to be proven by induction.
However, it is obvious that this will work out, so I believe it suffices to leave it as is, and perhaps just mention that it follows by induction. (Nerd)

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)

It aims to use the Leibniz formula of a determinant:
$$\det(A) = \sum_{\sigma \in S_n} \left( \operatorname{sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma_i}\right)$$

The summation $\sum_{j_1,...,j_k=1}^k$ sums over every k-tuple of values in 1,...,k, including duplicates.
And every element in $\text{Per}_k$ uniquely identifies every k-tuple of distinct values in 1,...,k.
We can pick the k-tuple $(\pi(1),...,\pi(k))$ for a specific $\pi \in \text{Per}_k$.
Btw, there shouldn't be a $k$ above the summation symbol in that case should there? (Wondering)
Anyway, what is left to prove for the step, is that if there is a duplicate value, that it doesn't contribute.

 

[mathmari]

It looks all correct to me. (Smile)
The underbound of $n=2$ works, and for the record, so does $n=1$, and $n=0$.
The base case $n=0$ is trivially true - no properties needed.

Ah ok! (Happy)

Yep. All correct.
Formally it needs to be proven by induction.
However, it is obvious that this will work out, so I believe it suffices to leave it as is, and perhaps just mention that it follows by induction. (Nerd)

Ah ok. If we want to apply the induction which is the general formula that we have to use? (Wondering)

The summation $\sum_{j_1,...,j_k=1}^k$ sums over every k-tuple of values in 1,...,k, including duplicates.
And every element in $\text{Per}_k$ uniquely identifies every k-tuple of distinct values in 1,...,k.
We can pick the k-tuple $(\pi(1),...,\pi(k))$ for a specific $\pi \in \text{Per}_k$.

We consider the permutation $\pi: (1, \ldots , k)\mapsto (j_1, \ldots , j_k)$ and we get that \begin{equation*}\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}\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\phi_{\pi (1)}\land \ldots \land \phi_{\pi (k)}\end{equation*}

The left hand side is the sum over all k-tuples. If two of the indices $j_i$'s are equal, say $j_s=j_t$ for $s\neq t$, then $\phi_s$ appears twice in the wedge product. It follows that this term of the sum is equal to $0$. So, in the sum only the terms apperear where the indices are distinct.

The right hand side is the sum over k-tuples where $j_1, \ldots , j_k$ are all distinct. They form therefore a permutation of $1, \ldots , k$. Therefore, we get the follwoing:
\begin{align*}\sum_{\pi\in \operatorname{Per}_k}\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\phi_{\pi (1)}\land \ldots \land \phi_{\pi (k)}&=\sum_{\pi\in \operatorname{Per}_k}\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\cdot \operatorname{sign}(\pi)\phi_{1}\land \ldots \land \phi_{k} \\ & =\sum_{\pi\in \operatorname{Per}_k}\left (\operatorname{sign}(\pi)\cdot \prod_{i=1}^ka_{i\pi(i)}\right )\cdot \phi_{1}\land \ldots \land \phi_{k} \\ & \overset{ \text{ Leibniz formula }}{ = }\det (a_{ij})\cdot \phi_{1}\land \ldots \land \phi_{k} \end{align*}

is everything correct? Could I improve something? (Wondering)

Btw, there shouldn't be a $k$ above the summation symbol in that case should there? (Wondering)

Oh yes, you're right! (Blush)