The transpose of a permutation matrix is its inverse.

Question

This is a question from the free Harvard online abstract algebra lectures. I'm posting my solutions here to get some feedback on them. For a fuller explanation, see this post.

This problem is from assignment 4.

Prove that the transpose of a permutation matrix $P$ is its inverse.

A permutation matrix $P$ has a single 1 in each row and a single 1 in each column, all other entries being 0. So column $j$ has a single 1 at position $e_{i_jj}$. $P$ acts by moving row $j$ to row $i_j$ for each column $j$. Taking the transpose of $P$ moves each 1 entry from $e_{i_jj}$ to $e_{ji_j}$. Then $P^t$ acts by moving row $i_j$ to row $j$ for each row $i_j$. Since this is the inverse operation, $P^t=P^{-1}$.

Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem.

Thanks.

Answer 1

A direct computation is also fine: $$(PP^T)_{ij} = \sum_{k=1}^n P_{ik} P^T_{kj} = \sum_{k=1}^n P_{ik} P_{jk}$$ but $P_{ik}$ is usually 0, and so $P_{ik} P_{jk}$ is usually 0. The only time $P_{ik}$ is nonzero is when it is 1, but then there are no other $i' \neq i$ such that $P_{i'k}$ is nonzero ($i$ is the only row with a 1 in column $k$). In other words, $$\sum_{k=1}^n P_{ik} P_{jk} = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{otherwise} \end{cases}$$ and this is exactly the formula for the entries of the identity matrix, so $$PP^T = I$$

Answer 2

Another way to prove it is to realize that any permutation matrix is the product of elementary permutations, where by elementary I mean a permutation that swaps two entries. Since in an identity matrix swapping $i$ with $j$ in a row is the same as swapping $j$ with $i$ in a column, such matrix is symmetric and it coincides with its inverse. Then, assuming $P=P_1\cdots P_k$, with $P_1,\ldots,P_k$ elementary, we have

$$ P^{-1} = (P_1\cdots P_k)^{-1}=P_k^{-1}\cdots P_1^{-1}=P_k\cdots P_1=P_k^t\cdots P_1^t = (P_1\cdots P_k)^t=P^t $$

Answer 3

Let $π$ be a permutation on $n$ objects and

\begin{equation} \pi=\left(\begin{matrix} 1 & 2 &\ldots& n \\ \pi(1) & \pi(2) &\ldots& \pi(n) \end{matrix} \right) \end{equation}

Assume that $P_π$ be a permutation matrix. We need to prove that $P_π^T P_π=I$.

Note that, $π$ sends the $i$th row of the identity matrix to the $π(i)$th row, i.e.,

\begin{eqnarray*} P_\pi=[P_{ij}]=\left\{ \begin{array}{ll} 1; & i=\pi(j)\\ 0; & i \ne \pi(j). \end{array} \right. \end{eqnarray*}

The $ij$th component of $P_\pi^TP_\pi$ is

\begin{eqnarray} (P_\pi^TP_\pi)_{ij}&=&\sum_{k=1}^n P^T_{ik}P_{kj}\\ &=&\sum_{k=1}^n P_{ki}P_{kj}\\ &=& P_{\pi(j)i}P_{\pi(j)j}\\ &=& P_{\pi(j)i}=\left\{ \begin{array}{ll} 1; & i=j\\ 0; & i \ne j. \end{array} \right. \end{eqnarray}

Therefore, $P^T_\pi P_\pi=I$.

Answer 4

Using a little knowledge about orthogonal matrices the following proof is pretty simple:

Since $v^tw=\sum_{k=0}^nv_iw_i$ if $v=(v_1,...,v_n),w=(w_1,...,w_n)$ we have $v^tv=1$ whenever v is a column of $P$. On the other hand $v^tw=0$ if $v$ and $w$ are two distinct columns of $P$. Therefore we can conclude that $(P^tP)_{i,j}=\delta_{i,j}$ and so $P^t=P^{-1}$.

Answer 5

Less sophisticated, you could just crunch it out.

First, a lemma:

The inverse of a matrix, if it exists, is unique.

Proof: If both $B$ and $C$ are inverse to $A$, then we have $B = BI = B(AC) = (BA)C = IC = C$ so $B = C$. (Here, $I$ denotes the identity matrix).

Using this, it follows in our specific case that in order to show $A^T = A^{-1}$, we need only show $A^TA = AA^T = I$.

Assume $i\neq j$. Then $(AA^T)_{ij} = \sum_k A_{ik}A^T_{kj} = \sum_k A_{ik}A_{jk}$. But for each $k$, $A_{ik}A_{jk} = 0$ since there is only one nonzero entry in the $k$th row and $i\neq j$ (so $A_{ik}$ and $A_{jk}$ can't both be the nonzero entry). So, $(AA^T)_{ij} = 0$ when $i\neq j$.

The argument that $(A^TA)_{ij} = 0$ when $i\neq j$ is almost identical, but uses the fact that the columns of $A$ contain only one nonzero entry.

Can you see what happens when, instead, $i = j$?

Answer 6

If we think of a matrix as a description of some linear map with respect to some bases, then the transpose matrix is a description of the dual map with respect to the dual bases. A permutation $\sigma$ of a basis induces the inverse permutation $\sigma^{-1}$ on the dual basis. This gives a conceptual explanation of the fact that the transpose of a permutation matrix is its inverse.

For a vector space $V$ over a field $k$, let $V^\ast$ denote the dual space — the space of linear maps from $V$ to $k$.

For a linear map $L:V\rightarrow W$, there is a dual map $L^\ast:W^\ast\rightarrow V^\ast$ defined in the following way: $$L^\ast(f)=f\circ L$$

Let $V$ be a finite-dimensional vector space with basis $\mathcal{B}$. For each $v_i\in\mathcal{B}$, define $v_i^\ast\in V^\ast$ to be the linear functional such that: $$v_i^\ast(v_j)=\delta_{ij}$$ Let $\mathcal{B}^\ast=\{v_i^\ast\}_{1\leq i\leq n}$. Then $\mathcal{B}^\ast$ forms a basis for the dual space $V^\ast$.

Let $L:V\rightarrow W$ be a linear map. Let $\mathcal{A}$ and $\mathcal{B}$ be bases for $V$ and $W$ respectively, and let $L_{\mathcal{AB}}$ be the matrix of $L$ with respect to $\mathcal{A}$ and $\mathcal{B}$. You may check that: $$L_{\mathcal{B^\ast A^\ast}}=L_{\mathcal{AB}}^T$$

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Once again, let $\mathcal{B}=\{v_i\}_{1\leq i\leq n}$ be a basis for $V$, and let $\mathcal{B}^\ast$ be the dual basis. Let $\sigma:\{1,2,\dots, n\}\rightarrow\{1,2,\dots, n\}$ be a permutation, and let $P_\sigma:V\rightarrow V$ be the linear map such that $P_\sigma(v_i)=v_{\sigma(i)}$. The matrix of $P_\sigma$ with respect to $\mathcal{B}$ is the permutation matrix representing $\sigma$.

It is easily checked that: $$P_{\sigma}^\ast(v_i^\ast)=v_{\sigma^{-1}(i)}^\ast$$ Thus, the transpose of the permutation matrix representing $\sigma$ is its inverse.

Answer 7

Let's P be an arbitrary permutation matrix. Some matrix is unitary iff their columns form a orthonormal base. Since the columns of a permutation matrix are distinct vectors of standard basis, it follows that P is unitary matrix.