Commit 56ebebc7 authored by Jim Hefferon's avatar Jim Hefferon

det1 edits

parent 41be78e0
This diff is collapsed.
......@@ -22019,7 +22019,7 @@ octave:6> gplot z
\end{ans}
\subsection{Subsection Four.I.3: The Permutation Expansion}
\begin{ans}{Four.I.3.14}
\begin{ans}{Four.I.3.15}
\begin{exparts}
\partsitem This matrix is singular.
\begin{align*}
......@@ -22062,7 +22062,7 @@ octave:6> gplot z
\end{exparts}
\end{ans}
\begin{ans}{Four.I.3.15}
\begin{ans}{Four.I.3.16}
\begin{exparts}
\partsitem Gauss' method gives this
\begin{equation*}
......@@ -22147,7 +22147,7 @@ octave:6> gplot z
\end{exparts}
\end{ans}
\begin{ans}{Four.I.3.16}
\begin{ans}{Four.I.3.17}
Following \nearbyexample{ex:SamplePermExp} gives this.
\begin{align*}
\begin{vmat}
......@@ -22178,7 +22178,7 @@ octave:6> gplot z
\end{align*}
\end{ans}
\begin{ans}{Four.I.3.17}
\begin{ans}{Four.I.3.18}
This is all of the permutations where $\phi(1)=1$
\begin{align*}
&\phi_1=\sequence{1,2,3,4}
......@@ -22237,7 +22237,7 @@ octave:6> gplot z
\end{align*}
\end{ans}
\begin{ans}{Four.I.3.18}
\begin{ans}{Four.I.3.19}
Each of these is easy to check.
\begin{exparts*}
\partsitem
......@@ -22257,7 +22257,7 @@ octave:6> gplot z
\end{exparts*}
\end{ans}
\begin{ans}{Four.I.3.19}
\begin{ans}{Four.I.3.20}
For the `if' half, the first condition of
\nearbydefinition{def:multilinear} follows from taking $k_1=k_2=1$
and the second condition follows from taking $k_2=0$.
......@@ -22278,11 +22278,11 @@ octave:6> gplot z
result.
\end{ans}
\begin{ans}{Four.I.3.20}
\begin{ans}{Four.I.3.21}
They would all double.
\end{ans}
\begin{ans}{Four.I.3.21}
\begin{ans}{Four.I.3.22}
For the second statement,
given a matrix, transpose it, swap rows, and transpose back.
The result is swapped columns, and the determinant changes by a factor
......@@ -22292,12 +22292,12 @@ octave:6> gplot z
rows, and then transpose back the resulting matrices.
\end{ans}
\begin{ans}{Four.I.3.22}
\begin{ans}{Four.I.3.23}
An \( \nbyn{n} \) matrix with a nonzero determinant has rank
\( n \) so its columns form a basis for \( \Re^n \).
\end{ans}
\begin{ans}{Four.I.3.23}
\begin{ans}{Four.I.3.24}
False.
\begin{equation*}
\begin{vmat}[r]
......@@ -22308,7 +22308,7 @@ octave:6> gplot z
\end{equation*}
\end{ans}
\begin{ans}{Four.I.3.24}
\begin{ans}{Four.I.3.25}
\begin{exparts}
\partsitem For the column index of the entry in the first row there are
five choices.
......@@ -22323,23 +22323,23 @@ octave:6> gplot z
\end{exparts}
\end{ans}
\begin{ans}{Four.I.3.25}
\begin{ans}{Four.I.3.26}
\( n\cdot(n-1)\cdots 2\cdot 1=n! \)
\end{ans}
\begin{ans}{Four.I.3.26}
\begin{ans}{Four.I.3.27}
In
\( \deter{A}=\deter{\trans{A}}=\deter{-A}=(-1)^n\deter{A} \)
the exponent $n$ must be even.
\end{ans}
\begin{ans}{Four.I.3.27}
\begin{ans}{Four.I.3.28}
Showing that no placement of three zeros suffices is routine.
Four zeroes does suffice; put them all in the same
row or column.
\end{ans}
\begin{ans}{Four.I.3.28}
\begin{ans}{Four.I.3.29}
The $n=3$ case shows what to do.
The row combination operations of
$-x_1\rho_2+\rho_3$ and $-x_1\rho_1+\rho_2$
......@@ -22376,7 +22376,7 @@ octave:6> gplot z
\end{equation*}
\end{ans}
\begin{ans}{Four.I.3.29}
\begin{ans}{Four.I.3.30}
Let \( T \) be \( \nbyn{n} \),
let \( J \) be \( \nbyn{p} \),
and let \( K \) be \( \nbyn{q} \).
......@@ -22412,7 +22412,7 @@ octave:6> gplot z
\end{equation*}
\end{ans}
\begin{ans}{Four.I.3.30}
\begin{ans}{Four.I.3.31}
The $n=3$ case shows what happens.
\begin{equation*}
\deter{T-rI}
......@@ -22435,7 +22435,7 @@ octave:6> gplot z
A polynomial of degree \( n \) has at most \( n \) roots.
\end{ans}
\begin{ans}{Four.I.3.31}
\begin{ans}{Four.I.3.32}
\answerasgiven
When two rows of a determinant are interchanged, the sign of the
determinant is changed.
......@@ -22446,7 +22446,7 @@ octave:6> gplot z
which sums to zero.
\end{ans}
\begin{ans}{Four.I.3.32}
\begin{ans}{Four.I.3.33}
\answerasgiven
When the elements of any column are subtracted from the elements of
each of the other two, the elements in two of the columns of the derived
......@@ -22471,7 +22471,7 @@ octave:6> gplot z
\end{equation*}
\end{ans}
\begin{ans}{Four.I.3.33}
\begin{ans}{Four.I.3.34}
\answerasgiven
Let
\begin{equation*}
......@@ -22515,7 +22515,7 @@ octave:6> gplot z
\end{equation*}
\end{ans}
\begin{ans}{Four.I.3.34}
\begin{ans}{Four.I.3.35}
\answerasgiven
Denote by \( D_n \) the determinant in question and by \( a_{i,j} \)
the element in the \( i \)-th row and \( j \)-th column.
......@@ -2280,7 +2280,7 @@ We can bring out the scalars.
0 &1 &0 \\
0 &0 &1
\end{vmat}
+(2)(\underline{0})(1)\begin{vmat}[r]
+(2)(\highlight{0})(1)\begin{vmat}[r]
1 &0 &0 \\
0 &0 &1 \\
0 &1 &0
......@@ -2290,7 +2290,7 @@ We can bring out the scalars.
1 &0 &0 \\
0 &0 &1
\end{vmat}
+(1)(\underline{0})(2)\begin{vmat}[r]
+(1)(\highlight{0})(2)\begin{vmat}[r]
0 &1 &0 \\
0 &0 &1 \\
1 &0 &0
......@@ -2308,7 +2308,7 @@ We can bring out the scalars.
\end{align*}
To finish, we evaluate those six determinants by row-swapping them
to the identity matrix,
keeping track of the resulting sign changes.
keeping track of the sign changes.
\begin{align*}
&=30\cdot (+1)+0\cdot (-1) \\
&\quad\hbox{}+20\cdot (-1)+0\cdot (+1) \\
......@@ -2316,31 +2316,39 @@ keeping track of the resulting sign changes.
\end{align*}
\end{example}
That example illustrates the key idea.
We've applied multilinearity to a $\nbyn{3}$ determinant to get
$3^3$ separate determinants, each with one distinguished entry per row.
We can drop most of these new determinants because the matrices are singular,
with one row a multiple of another.
That example captures the new calculation scheme.
We've apply multilinearity to a determinant to get
many separate determinants, each with one entry per row from the
original matrix.
Most of these matrices have
one row that is a multiple of another
so we can omit these determinants from the calculation.
We are left with the one-entry-per-row determinants
also having only one entry per column (one entry from the original determinant,
that is).
And, since we can factor scalars out, we can further reduce to
only considering determinants of
one-entry-per-row-and-column matrices where the entries are ones.
These are permutation matrices.
Thus, the determinant can be computed in this
three-step way
\textit{(Step~1)}~for each permutation matrix, multiply together
the entries from the original matrix
where that permutation matrix has ones,
\textit{(Step~2)}~multiply that by the determinant of the permutation matrix
and
\textit{(Step~3)}~do that for all permutation matrices
and sum the results together.
To state this as a formula, we introduce a notation for permutation matrices.
Let $\iota_j$ be the row vector that is all zeroes except for a one in its
also having only one entry per column from the original matrix.
And, since we can factor scalars out, we can further reduce the
determinants that we must compute to those
one-entry-per-row-and-column matrices where all the entries are $1$'s.
\begin{definition}
A square matrix whose entries are all $0$'s except for
one $1$ in each row and column is a
\definend{permutation matrix}.\index{permutation matrix}%
\index{matrix!permutation}
\end{definition}
% Thus, we can compute a determinant in this
% three-step way:
% \textit{(Step~1)}~for each correctly-sized permutation matrix, multiply together
% the entries from the original matrix
% where that permutation matrix has ones,
% \textit{(Step~2)}~multiply that by the determinant of the permutation matrix
% and
% \textit{(Step~3)}~do that for all permutation matrices
% and sum the results together.
To state the new calculation scheme as a formula
we introduce a notation for permutation matrices.
Let $\iota_j$ be the row vector that is all $0$'s except for a $1$ in its
$j$-th entry, so that the four-wide $\iota_2$ is $\rowvec{0 &1 &0 &0}$.
We can construct permutation matrices by
permuting \Dash that is, scrambling \Dash the numbers $1$, $2$, \ldots, $n$,
......@@ -2407,20 +2415,21 @@ The $3$-permutations are
\( \phi_4=\sequence{2,3,1} \),
\( \phi_5=\sequence{3,1,2} \), and
\( \phi_6=\sequence{3,2,1} \).
Here are two of the associated permutation matrices.
\begin{equation*}
P_{\phi_2}
=\begin{mat}
\iota_1 \\
\iota_3 \\
\iota_2
\end{mat}
=\begin{mat}[r]
1 &0 &0 \\
0 &0 &1 \\
0 &1 &0
\end{mat}
\qquad
Here is a sample permutation matrix; note that
the rows of $P_{\phi_5}$ are $\iota_{\phi_5(1)}=\iota_3$,
$\iota_{\phi_5(2)}=\iota_1$, and $\iota_{\phi_5(3)}=\iota_2$.\begin{equation*}
% P_{\phi_2}
% =\begin{mat}
% \iota_1 \\
% \iota_3 \\
% \iota_2
% \end{mat}
% =\begin{mat}[r]
% 1 &0 &0 \\
% 0 &0 &1 \\
% 0 &1 &0
% \end{mat}
% \qquad
P_{\phi_5}
=\begin{mat}
\iota_3 \\
......@@ -2433,12 +2442,8 @@ Here are two of the associated permutation matrices.
0 &1 &0
\end{mat}
\end{equation*}
For instance, the rows of $P_{\phi_5}$ are $\iota_{\phi_5(1)}=\iota_3$,
$\iota_{\phi_5(2)}=\iota_1$, and $\iota_{\phi_5(3)}=\iota_2$.
\end{example}
%Now we can restate the three-step procedure as a formula.
\begin{definition}
The \definend{permutation expansion}\index{determinant!permutation expansion}%
\index{permutation expansion}
......@@ -2476,15 +2481,10 @@ notation}\index{summation notation!for permutation expansion}
read aloud as
``the sum, over all permutations \( \phi \), of terms having the form
\( t_{1,\phi(1)}t_{2,\phi(2)}\cdots t_{n,\phi(n)} \deter{P_{\phi}} \)''.
This phrase is just a restating of the three-step process
\textit{(Step 1)}~for each permutation matrix, compute
\( t_{1,\phi(1)}t_{2,\phi(2)}\cdots t_{n,\phi(n)} \)
\textit{(Step 2)}~multiply that by \( \deter{P_{\phi}} \)
and \textit{(Step 3)}~sum all such terms together.
This just restates the computation scheme given above.
\begin{example}
The familiar formula for the determinant of a $\nbyn{2}$ matrix
can be derived in this way.
The familiar $\nbyn{2}$ determinant formula follows from the above
\begin{align*}
\begin{vmat}
t_{1,1} &t_{1,2} \\
......@@ -2508,8 +2508,7 @@ can be derived in this way.
\end{align*}
(the second permutation matrix takes one row swap to pass to the
identity).
Similarly, the formula for the determinant of a $\nbyn{3}$ matrix
is this.
So does the $\nbyn{3}$ determinant.
\begin{align*}
\begin{vmat}
t_{1,1} &t_{1,2} &t_{1,3} \\
......@@ -2541,11 +2540,10 @@ is this.
Computing a determinant by permutation expansion usually takes longer than
Gauss' method.
However, here we are not trying to do the computation efficiently,
we are instead
trying to give a determinant formula that we can prove to be well-defined.
While the permutation expansion is impractical for computations,
we will find it useful in the proofs below.
However,
While the permutation expansion is impractical for computations
we will find it useful in the proofs that
the determinant function is well-defined.
\begin{theorem}
\index{determinant!exists} \label{th:DetsExist}
......
This diff is collapsed.
Markdown is supported
0% or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment