- #1

Eclair_de_XII

- 1,063

- 89

- TL;DR Summary
- Let ##A## be an ##(n,n)## matrix. If ##A'## is the matrix obtained from adding one multiple of one row to another, then ##\det A=\det A'##.

Let ##S_n## denote ##\{1,\ldots,n\}##, where ##n\in\mathbb{N}##.

Recall that the ##\textrm{sgn}## function maps a permutation of ##S_n## to an element in ##\{1,0,-1\}##.

We want to rework the definition of ##\textrm{sgn}## because it is not sufficient for some proofs about determinants. For example, it is known that replacing one row of a given matrix with the sum of itself and a multiple of another row does not change the value of the determinant. The proof for this fact could start out in the following way.

Let ##A## be a matrix and let ##A'## be the matrix obtained from replacing the ##n\textrm{-th}## row with itself added to the first row multipled by ##\alpha##, where ##\alpha## is a constant. In theory, replacing any row with a itself added to a multiple off another would yield the same result, but these two rows are chosen specifically to economize on space. We proceed by computing ##\det A'##.

\begin{eqnarray*}

\det A'&=&\sum_{\textrm{n-lists}} \textrm{sgn}(j_1,\ldots,j_n) a_{1j_1}\cdot\ldots\cdot (\alpha\cdot a_{1j_1}+a_{nj_n})\\

&=&\alpha \sum_{\textrm{n-lists}} \textrm{sgn}(j_1,\ldots,j_n) a_{1j_1}\cdot\ldots\cdot a_{1j_1}\\

&+&\sum_{\textrm{n-lists}} \textrm{sgn}(j_1,\ldots,j_n) a_{1j_1}\cdot\ldots\cdot a_{nj_n}\\

&=&\alpha \sum_{\textrm{n-lists}} \textrm{sgn}(j_1,\ldots,j_n) a_{1j_1}\cdot\ldots\cdot a_{1j_1}+\det A

\end{eqnarray*}

The conclusion of the proof is that ##\det A'=\det A##, which means that the summation must be identically zero. This holds if each summand is zero, which is a fact cited by some linear-algebra texts in order to prove this theorem.

Assuming this is true, either there is an identically zero row in the matrix described by first determinant term, or there are two identical column indices ##j_i,j_l## in the permutation of ##S_n##. In the latter scenario, ##\textrm{sgn}## evaluates to zero. On the other hand, these elements ##j_i## are meant to represent distinct numbers in the set ##S_n##. But the notation does not reflect this. The revised definition of ##\textrm{sgn}## to follow aims to correct for this shortcoming.

It is slightly inaccurate to think of the elements in each permutation as integers; it would be more natural to think of them as integer-valued functions. We base this off the following idea: Each distinct row vector must be assigned a unique column index in each summand of the determinant expansion. By construction, if any of the rows are identical, they are assigned same column index, which means that the ##\textrm{sgn}## evaluates the list of said indices to zero.

Let ##V_n## be a list of vectors in ##\mathbb{F}^n##, not necessarily distinct, each of length ##n##, with ##|V_n|=n##.

Define ##f:S_n\longrightarrow V_n##. In other words, ##f## maps each integer in ##S_n## to some row vector of length ##n##. Generally speaking, we can interpret any matrix as an array of such mappings. Alternatively, ##f## can be interpreted as a function that determines the row vector in some row index of a given matrix.

Let ##T_n## denote a subset of ##S_n##, where ##T_n## has as many elements as there are distinct row vectors in the image of ##f##.

Now define a function ##g:V_n\longrightarrow T_n## as the function assigning each distinct row vector a unique column index. This defines a one-to-one function.

The composition ##g\circ f## applies to some element ##k\in S_n## as follows:

\begin{eqnarray*}

k\textrm{ is mapped to some row vector in a matrix via }f.\\

\textrm{This row vector is mapped to a column index }(g\circ f)(k).

\end{eqnarray*}

We redefine the argument of ##\textrm{sgn}## as a permutation of the function ##h## evaluated on ##S_n##, where ##h## is defined below.

\begin{equation}

h(S_n)=\left((g\circ f)(1),\ldots,(g\circ f)(n)\right)

\end{equation}

Going back to the proof from earlier, we have a matrix with two identical rows, one in the first row and the other in the last. Let ##a_1## denote the row vector in the first row of the matrix ##A'##.

\begin{eqnarray*}

f(1)=a_1=f(n)\\

g(f(1))=k=g(f(n))

\end{eqnarray*}

where ##k\in T_n##.

Let ##(i_1,\ldots,i_n)## be a permutation of ##h(S_n)##. It follows that at least two of the elements in the permutation must be identical, which means that ##\textrm{sgn}## evaluates it to zero, which means that each summand in the sum derived in the proof from earlier is identically zero.

This gives our desired result: ##\det A=\det A'##.

Recall that the ##\textrm{sgn}## function maps a permutation of ##S_n## to an element in ##\{1,0,-1\}##.

We want to rework the definition of ##\textrm{sgn}## because it is not sufficient for some proofs about determinants. For example, it is known that replacing one row of a given matrix with the sum of itself and a multiple of another row does not change the value of the determinant. The proof for this fact could start out in the following way.

Let ##A## be a matrix and let ##A'## be the matrix obtained from replacing the ##n\textrm{-th}## row with itself added to the first row multipled by ##\alpha##, where ##\alpha## is a constant. In theory, replacing any row with a itself added to a multiple off another would yield the same result, but these two rows are chosen specifically to economize on space. We proceed by computing ##\det A'##.

\begin{eqnarray*}

\det A'&=&\sum_{\textrm{n-lists}} \textrm{sgn}(j_1,\ldots,j_n) a_{1j_1}\cdot\ldots\cdot (\alpha\cdot a_{1j_1}+a_{nj_n})\\

&=&\alpha \sum_{\textrm{n-lists}} \textrm{sgn}(j_1,\ldots,j_n) a_{1j_1}\cdot\ldots\cdot a_{1j_1}\\

&+&\sum_{\textrm{n-lists}} \textrm{sgn}(j_1,\ldots,j_n) a_{1j_1}\cdot\ldots\cdot a_{nj_n}\\

&=&\alpha \sum_{\textrm{n-lists}} \textrm{sgn}(j_1,\ldots,j_n) a_{1j_1}\cdot\ldots\cdot a_{1j_1}+\det A

\end{eqnarray*}

The conclusion of the proof is that ##\det A'=\det A##, which means that the summation must be identically zero. This holds if each summand is zero, which is a fact cited by some linear-algebra texts in order to prove this theorem.

Assuming this is true, either there is an identically zero row in the matrix described by first determinant term, or there are two identical column indices ##j_i,j_l## in the permutation of ##S_n##. In the latter scenario, ##\textrm{sgn}## evaluates to zero. On the other hand, these elements ##j_i## are meant to represent distinct numbers in the set ##S_n##. But the notation does not reflect this. The revised definition of ##\textrm{sgn}## to follow aims to correct for this shortcoming.

It is slightly inaccurate to think of the elements in each permutation as integers; it would be more natural to think of them as integer-valued functions. We base this off the following idea: Each distinct row vector must be assigned a unique column index in each summand of the determinant expansion. By construction, if any of the rows are identical, they are assigned same column index, which means that the ##\textrm{sgn}## evaluates the list of said indices to zero.

Let ##V_n## be a list of vectors in ##\mathbb{F}^n##, not necessarily distinct, each of length ##n##, with ##|V_n|=n##.

Define ##f:S_n\longrightarrow V_n##. In other words, ##f## maps each integer in ##S_n## to some row vector of length ##n##. Generally speaking, we can interpret any matrix as an array of such mappings. Alternatively, ##f## can be interpreted as a function that determines the row vector in some row index of a given matrix.

Let ##T_n## denote a subset of ##S_n##, where ##T_n## has as many elements as there are distinct row vectors in the image of ##f##.

Now define a function ##g:V_n\longrightarrow T_n## as the function assigning each distinct row vector a unique column index. This defines a one-to-one function.

The composition ##g\circ f## applies to some element ##k\in S_n## as follows:

\begin{eqnarray*}

k\textrm{ is mapped to some row vector in a matrix via }f.\\

\textrm{This row vector is mapped to a column index }(g\circ f)(k).

\end{eqnarray*}

We redefine the argument of ##\textrm{sgn}## as a permutation of the function ##h## evaluated on ##S_n##, where ##h## is defined below.

\begin{equation}

h(S_n)=\left((g\circ f)(1),\ldots,(g\circ f)(n)\right)

\end{equation}

Going back to the proof from earlier, we have a matrix with two identical rows, one in the first row and the other in the last. Let ##a_1## denote the row vector in the first row of the matrix ##A'##.

\begin{eqnarray*}

f(1)=a_1=f(n)\\

g(f(1))=k=g(f(n))

\end{eqnarray*}

where ##k\in T_n##.

Let ##(i_1,\ldots,i_n)## be a permutation of ##h(S_n)##. It follows that at least two of the elements in the permutation must be identical, which means that ##\textrm{sgn}## evaluates it to zero, which means that each summand in the sum derived in the proof from earlier is identically zero.

This gives our desired result: ##\det A=\det A'##.

Last edited: