Commit 56ebebc7 by Jim Hefferon

### det1 edits

parent 41be78e0
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 ... ... @@ -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} ... ...
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