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Alan L.
Senior Teaching Assistant on Linear Algebra, Probability, Calculus
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Linear Algebra
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Question:

\begin{thm}(Steinitz Replacement Theorem) $\cV$ is a vector space over $\bF$, $x_1,x_2,\ldots ,x_n \in \cV$, and $y_1,y_2,\ldots ,y_m \in span(\{x_1,x_2,\ldots ,x_n\}):= \cW$. If $\{y_1,y_2,\ldots ,y_m\}$ is linearly independent, $m\leq n$. \end{thm}

Alan L.

Suppose by contradiction that $m>n$. Since $y_1 \in span(\{x_1,x_2,\ldots ,x_n\})$, we have $y_1=a_1x_1+\dots +a_nx_n$, $a_i \in \bF$. $y_1$ can not be zero because $\{y_1,y_2,\ldots ,y_m\}$ is L.I. So $a_i \neq 0$, let $a_1 \neq 0$. We have $span(\{x_1,x_2,\ldots ,x_n\})=span(\{y_1,x_2,\ldots ,x_n\})$. Similarly, $y_2\in span(\{y_1,x_2,\ldots ,x_n\})$, we have $y_2=b_1y_1+\dots +b_nx_n$, $b_i \in \bF$. If $b_i=0,i=2\dots n$, $y_2=b_1y_1$, $\{y_1,y_2,\ldots ,y_m\}$ is L.D. which is a contradiction. Hence we assume $b_2 \neq 0$. We have $span(\{y_1,x_2,\ldots ,x_n\})=span(\{y_1,y_2,\ldots ,x_n\})$. If we repeat this way, we can get $span(\{x_1,x_2,\ldots ,x_n\})=span(\{y_1,y_2,\ldots ,y_n\})$. Since $m>n$, we have $y_m \in span(\{x_1,x_2,\ldots ,x_n\})=span(\{y_1,y_2,\ldots ,y_n\})$. Therefore, $y_m=c_1y_1+\dots +c_ny_n,c_i \in \bF$. Which is mean $\{y_1,y_2,\ldots ,y_m\}$ is L.D., that is a contradiction. Hence we have $m \leq n$.

Calculus
TutorMe
Question:

In general, the derivative of a real-valued function, $f : \mathbb{R} \rightarrow \mathbb{R}$ at some point $t_0 \in \mathbb{R}$ can be approximated as $$$\frac{df}{dt}|_{t_0} \approx \frac{f(t_0+h)-f(t_0)}{h}$$$ \label{eq1} if the real quantity $h$ is small enough. If we can regard the independent variable $t$ as time(maybe in seconds, in hours, etc), please find an expression based on (\ref{eq1}) that can be used as a prediction approximation for $f(x_0+h).$

Alan L.

We may achieve our goal by simply changing the arrangement on the both sides of (\ref{eq1}). That is, $$$f(t_0+h) \approx f(t_0) + }\frac{df}{dt}|_{t_0} h.$$$

Differential Equations
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Question:

Solve $\dot{x}=\left[\begin{array}{ccc}2 & -2 & 3\\ 1 & 1 & 1\\ 1 & 3 & -1\end{array}\right]x,x(0)=\left[\begin{array}{ccc}2\\1\\3\end{array}\right].$

Alan L.

Sol: Let $A=\left[\begin{array}{ccc}2 & -2 & 3\\ 1 & 1 & 1\\ 1 & 3 & -1\end{array}\right].$ $|A - \lambda I|= \left|\begin{array}{ccc}2-\lambda & -2 & 3\\ 1 & 1-\lambda & 1\\ 1 & 3 & -1-\lambda \end{array}\right|=(3-\lambda)(1-\lambda)(-2-\lambda).$ $\lambda=3, (A - \lambda I)v_1=(A - 3I)v_1=0, v_1=\left[\begin{array}{ccc}1\\1\\1\end{array}\right];$ $\lambda=1, (A - \lambda I)v_2=(A - I)v_2=0, v_2=\left[\begin{array}{ccc}1\\-1\\-1\end{array}\right];$ $\lambda=-2, (A - \lambda I)v_3=(A - (-2)I)v_3=0, v_3=\left[\begin{array}{ccc}-11\\-1\\14\end{array}\right]$ $Q=\left[\begin{array}{ccc}1 & 1 &-11\\1 &-1 &-1\\1 &-1 &14\end{array}\right]$ Let $x=Qy, y= Q^{-1}x. \therefore \dot{x}=Q\dot{y}=AQy \rightarrow \dot{y}=Q^{-1}AQy= \left[\begin{array}{ccc}3 & 0 &0\\0 & 1 &0\\0 & 0 & -2\end{array}\right]\left[\begin{array}{ccc}y_1\\y_2\\y_3\end{array}\right].$ Then $y=\left[\begin{array}{ccc}e^{3t} & 0 &0\\0 & e^t &0\\0 & 0 & e^{-2t}\end{array}\right]\left[\begin{array}{ccc}c_1\\c_2\\c_3\end{array}\right]$, since $x=Qy$, we have $x=c_1\left[\begin{array}{ccc}1\\1\\1\end{array}\right]e^{3t}+c_2\left[\begin{array}{ccc}1\\-1\\-1\end{array}\right]e^{t}+ c_3\left[\begin{array}{ccc}-11\\-1\\14\end{array}\right]e^{-2t}$. $x(0)=\left[\begin{array}{ccc}2\\1\\3\end{array}\right]=\left[\begin{array}{ccc}1 & 1 &-11\\1 &-1 &-1\\1 &-1 &14\end{array}\right]\left[\begin{array}{ccc}c_1\\c_2\\c_3\end{array}\right]$ $\therefore \left[\begin{array}{ccc}c_1\\c_2\\c_3\end{array}\right]=\left[\begin{array}{ccc}\frac{23}{10}\\ \frac{7}{6}\\ \frac{2}{15}\end{array}\right],$ we obtain $x=\frac{23}{10}\left[\begin{array}{ccc}1\\1\\1\end{array}\right]e^{3t}+\frac{7}{6}\left[\begin{array}{ccc}1\\-1\\-1\end{array}\right]e^{t}+ \frac{2}{15}\left[\begin{array}{ccc}-11\\-1\\14\end{array}\right]e^{-2t}$

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