### finish leontief topics

parent 73d2b6e1
 ... ... @@ -3,6 +3,8 @@ TODO list for Linear Algebra http://joshua.smcvt.edu/linearalgebra * Look through bug reports ** Do I need lemma 2.4 where it is? Could I move it to after the defn of linear independence? ** In the index, make page references hyperlinks ... ...
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 ... ... @@ -1289,15 +1289,19 @@ If we have a non-$$\vec{0}$$ row written as a combination of the others $\rho_i=c_1\rho_1+\cdots+c_{i-1}\rho_{i-1}+ c_{i+1}\rho_{i+1}+\cdots+c_m\rho_m$ then we can rewrite the equation as then we can rewrite that as \begin{equation*} \vec{0}=c_1\rho_1+\cdots+c_{i-1}\rho_{i-1}+c_i\rho_i+ c_{i+1}\rho_{i+1}+\cdots+c_m\rho_m \tag{$*$} \end{equation*} where $c_i=-1$. We will use induction on the row index~$i$ to show that all of the coefficients~$c_i$ in that equation are~$0$. where not all the coefficients are zero; specifically, $c_i=-1$. The converse holds also:~given equation~($*$) where some $c_i\neq 0$ then we could express $\rho_i$ as a combination of the other rows by moving $c_i\rho_i$ to the left side and dividing by $c_i$. Therefore, we will have proved the theorem if we show that in~($*$) all of the coefficients are~$0$. For that, we use induction on the row index~$i$. The base case is the first row~$i=1$. Equation~($*$) defines an equation among the column $\ell_1$ entries of those ... ... @@ -1308,9 +1312,9 @@ the leading entry in row~$i$). 0=c_1r_{1,\ell_1}+c_2r_{2,\ell_1}+\cdots+c_mr_{m,\ell_1} \end{equation*} The matrix is in echelon form so every row after the first has a zero in that column, and thus every row after the first has a zero in that column $r_{2,\ell_1}=\cdots=r_{m,\ell_1}=0$. We conclude that $c_1=0$ because as the leading entry Thus $c_1=0$ because, as the leading entry in the row, $r_{1,\ell_1}\neq 0$. The inductive step is to prove the implication:~if ... ... @@ -1318,14 +1322,14 @@ for each row index $k\in\set{1,\ldots,i}$ the coefficient $c_k$ is $0$ then $c_{i+1}$ is also $0$. Consider the entries from column~$\ell_{i+1}$ in equation~($*$). \begin{equation*} 0=c_1r_{1,\ell_1}+\cdots+c_{i+1}r_{i+1,\ell_{i+1}}+\cdots+c_mr_{m,\ell_{i+1}} 0=c_1r_{1,\ell_{i+1}}+\cdots+c_{i+1}r_{i+1,\ell_{i+1}}+\cdots+c_mr_{m,\ell_{i+1}} \end{equation*} By the inductive hypothesis the coefficients $c_1$, \ldots $c_i$ are all $0$ so the equation reduces to $0=c_{i+1}r_{i+1,\ell_{i+1}}+\cdots+c_mr_{m,\ell_{i+1}}$. As in the base case we next note that the matrix is in echelon form so $r_{i+2,\ell_{i+1}}=\cdots=r_{m,\ell_{i+1}}=0$, and thus $c_{i+1}=0$ because $r_{i+1,\ell_{i+1}}\neq 0$ as it is the row's leading entry. The matrix is in echelon form so $r_{i+2,\ell_{i+1}}=\cdots=r_{m,\ell_{i+1}}=0$ and $r_{i+1,\ell_{i+1}}\neq 0$. Thus $c_{i+1}=0$. \end{proof} \begin{theorem} ... ...
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