Linear Algebra/Determinants Exist

''This subsection is optional. It consists of proofs of two results from the prior subsection. These proofs involve the properties of permutations, which will not be used later, except in the optional Jordan Canonical Form subsection.''

The prior subsection attacks the problem of showing that for any size there is a determinant function on the set of square matrices of that size by using multilinearity to develop the permutation expansion.


 * $$\begin{array}{rl}

\begin{vmatrix} t_{1,1} &t_{1,2}  &\ldots  &t_{1,n}  \\ t_{2,1} &t_{2,2}  &\ldots  &t_{2,n}  \\ &\vdots                     \\ t_{n,1} &t_{n,2}  &\ldots  &t_{n,n} \end{vmatrix} &= \begin{array}{l} t_{1,\phi_1(1)}t_{2,\phi_1(2)}\cdots t_{n,\phi_1(n)}\left|P_{\phi_1}\right|      \\ \quad+t_{1,\phi_2(1)}t_{2,\phi_2(2)}\cdots t_{n,\phi_2(n)}\left|P_{\phi_2}\right|      \\ \quad\vdots                             \\ \quad+t_{1,\phi_k(1)}t_{2,\phi_k(2)}\cdots t_{n,\phi_k(n)}\left|P_{\phi_k}\right| \end{array}                                                \\ &=\displaystyle\sum_{\text{permutations }\phi} t_{1,\phi(1)}t_{2,\phi(2)}\cdots t_{n,\phi(n)} \left|P_{\phi}\right| \end{array}$$

This reduces the problem to showing that there is a determinant function on the set of permutation matrices of that size.

Of course, a permutation matrix can be row-swapped to the identity matrix and to calculate its determinant we can keep track of the number of row swaps. However, the problem is still not solved. We still have not shown that the result is well-defined. For instance, the determinant of



P_{\phi}= \begin{pmatrix} 0 &1  &0  &0 \\ 1  &0  &0  &0 \\ 0  &0  &1  &0 \\ 0  &0  &0  &1 \end{pmatrix} $$

could be computed with one swap



P_{\phi} \xrightarrow[]{\rho_1\leftrightarrow\rho_2} \begin{pmatrix} 1 &0  &0  &0 \\ 0  &1  &0  &0 \\ 0  &0  &1  &0 \\ 0  &0  &0  &1 \end{pmatrix} $$

or with three.



P_{\phi} \xrightarrow[]{\rho_3\leftrightarrow\rho_1} \begin{pmatrix} 0 &0  &1  &0 \\ 1  &0  &0  &0 \\ 0  &1  &0  &0 \\ 0  &0  &0  &1 \end{pmatrix} \xrightarrow[]{\rho_2\leftrightarrow\rho_3} \begin{pmatrix} 0 &0  &1  &0 \\ 0  &1  &0  &0 \\ 1  &0  &0  &0 \\ 0  &0  &0  &1 \end{pmatrix} \xrightarrow[]{\rho_1\leftrightarrow\rho_3} \begin{pmatrix} 1 &0  &0  &0 \\ 0  &1  &0  &0 \\ 0  &0  &1  &0 \\ 0  &0  &0  &1 \end{pmatrix} $$

Both reductions have an odd number of swaps so we figure that $$ \left|P_{\phi}\right|=-1 $$ but how do we know that there isn't some way to do it with an even number of swaps? Corollary 4.6 below proves that there is no permutation matrix that can be row-swapped to an identity matrix in two ways, one with an even number of swaps and the other with an odd number of swaps.

We still have not shown that the permutation expansion is well-defined because we have not considered row operations on permutation matrices other than row swaps. We will finesse this problem: we will define a function $$ d:\mathcal{M}_{n \! \times \! n}\to \mathbb{R} $$ by altering the permutation expansion formula, replacing $$\left|P_\phi\right|$$ with $$\sgn(\phi)$$



d(T)= \sum_{\text{permutations }\phi}t_{1,\phi(1)}t_{2,\phi(2)}\dots t_{n,\phi(n)} \sgn(\phi) $$

(this gives the same value as the permutation expansion because the prior result shows that $$\det(P_\phi)=\sgn(\phi)$$). This formula's advantage is that the number of inversions is clearly well-defined &mdash; just count them. Therefore, we will show that a determinant function exists for all sizes by showing that $$ d $$ is it, that is, that $$d$$ satisfies the four conditions.

We have now shown that determinant functions exist for each size. We already know that for each size there is at most one determinant. Therefore, the permutation expansion computes the one and only determinant value of a square matrix.

We end this subsection by proving the other result remaining from the prior subsection, that the determinant of a matrix equals the determinant of its transpose.

Exercises
These summarize the notation used in this book for the $$2$$- and $$3$$- permutations. $$ \begin{array}{c|cc} i         &1      &2    \\ \hline \phi_1(i) &1      &2     \\ \phi_2(i) &2      &1 \end{array} \qquad \begin{array}{c|ccc} i         &1     &2   &3    \\ \hline \phi_1(i) &1     &2   &3    \\ \phi_2(i) &1     &3   &2    \\ \phi_3(i) &2     &1   &3    \\ \phi_4(i) &2     &3   &1    \\ \phi_5(i) &3     &1   &2    \\ \phi_6(i) &3     &2   &1 \end{array} $$

/Solutions/