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Tutor profile: Youtong W.

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Youtong W.
1+ year of tutoring experience, undergraduate student in mathematics
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Questions

Subject: Linear Algebra

TutorMe
Question:

Given a nonzero vector $$\vec{w} ∈ R^2$$ Define the linear transformation $$F\vec{w} : R^2 → R^2$$ as the function that reflects the plane across the line $$Span( \vec{w})$$. Compute the standard matrix $$[F\vec{w}]$$.

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Youtong W.
Answer:

Our goal is to compute $$[F\vec{w}]$$, given that the function $$F\vec{w}$$ is a linear transformation. Geometrically, it is easy to see how this function preserves addiction and scaler multiplication. But you might want an algebraic illustration of why this function is a linear transformation. This question can ultimately be answered as we derive the standard matrix of $$F\vec{w}$$. Let us first think about the function $$F\vec{w}$$. We can fix $$a, b\in R$$ such that $$\vec{w}= \begin{pmatrix} a\\b \end{pmatrix}$$. By definition of $$[F\vec{w}]$$, we know that for any vector $$\vec v$$, $$[F\vec{w}]\vec v$$ is the result of reflecting $$\vec v$$ across the line W. Let $$\vec {v'} = F\vec{w}(\vec v)$$. It is at first glance pretty hard to derive the formula for $$\vec {v'}$$. Instead of digging directly into the 'reflecting action', we will think about projection first. Just like reflection, projection of points or vectors to a line is a vital action in linear algebra and serves as a stepstone for us to go to projection as we will see in this example. Define a function $$P\vec{w}$$ as following. For all $$\vec{w} ∈ R^2$$, let $$P\vec{w}(\vec v)$$ be the resulting vector of projecting $$\vec{v}$$ onto $$Span(\vec w)$$. Again, the reason why this function is a linear transformation will be clear in a moment. Let $$\begin{pmatrix} x\\y \end{pmatrix} = \vec v$$. We know that the projection of $$\vec v$$ onto $$\vec w$$ is the vector that has $$\frac{\vec v \cdot \vec w}{\| \vec w\|}$$ as its length in the direction of $$\vec w$$. Thus, $$\begin{equation} \begin{split} P\vec{w}(\vec v) &= \frac{\vec v \cdot \vec w}{{\| \vec w\|}^2} \cdot \vec w\\ &= \frac{ax+by}{a^2+b^2} \cdot \begin{pmatrix} a\\b \end{pmatrix}\\ &= \begin{pmatrix} \frac{a^2x+aby}{a^2+b^2} \\ \frac{abx+b^2y}{a^2+b^2} \end{pmatrix}\\ &= \begin{pmatrix} \frac{a^2}{a^2+b^2} x+ \frac{ab}{a^2+b^2}y \\ \frac{ab}{a^2+b^2}x+ \frac{b^2}{a^2+b^2}y\end{pmatrix} \end{split} \end{equation}$$ Now we know that $$P\vec w$$ is indeed a linear transformation, since a function f with the form $$f\begin{pmatrix} x\\y \end{pmatrix}=\begin{pmatrix} ax+by\\cx+dy \end{pmatrix}$$ is always a linear transformation. If you are not satisfied with this property, notice that we can easily check that $$P\vec{w}$$ preserves addition and multiplication with this expression at hand. Now we can use $$P\vec w$$ to get to the reflecting action $$F\vec w$$. The key insight is that, we can get to $$\vec {v'}$$ by adding $$[P\vec{w}]\vec v - \vec v$$ twice to $$\vec v$$, because the vector $$[P\vec{w}]\vec v - \vec v$$ is the "distance vector" from $$\vec v$$ to $$Span(\vec w)$$. Thus, $$\begin{equation} \begin{split} F\vec{w}(\vec v) &= \begin{pmatrix} x\\y \end{pmatrix} + 2 \times (\begin{pmatrix} \frac{a^2x+aby}{a^2+b^2} \frac{abx+b^2y}{a^2+b^2} \end{pmatrix}- \begin{pmatrix} x\\y \end{pmatrix})\\ &= 2 \times (\begin{pmatrix} \frac{a^2x+aby}{a^2+b^2} \\ \frac{abx+b^2y}{a^2+b^2} \end{pmatrix})- \begin{pmatrix} x\\y \end{pmatrix}\\ &= \begin{pmatrix} \frac{2a^2x+2aby}{a^2+b^2} -x \\ \frac{2abx+2b^2y}{a^2+b^2}-y \end{pmatrix}\\ &= \begin{pmatrix} \frac{a^2x+2aby-b^2x}{a^2+b^2} \\ \frac{2abx+b^2y-a^2y}{a^2+b^2}\end{pmatrix}\\ &= \begin{pmatrix} \frac{a^2-b^2}{a^2+b^2}x+\frac{2ab}{a^2+b^2}y \\ \frac{2ab}{a^2+b^2}x +\frac{b^2-a^2}{a^2+b^2}y\end{pmatrix} \end{split} \end{equation}$$ From the formula, we know that $$F\vec w$$ is a linear transformation just as $$P\vec w$$, and that $$\begin{equation} [F \vec w] = \begin{pmatrix} \frac{a^2-b^2}{a^2+b^2} & \frac{2ab}{a^2+b^2} \\ \frac{2ab}{a^2+b^2} & \frac{b^2-a^2}{a^2+b^2} \end{pmatrix} \end{equation}$$

Subject: Number Theory

TutorMe
Question:

Find, with full explanation, the remainder when dividing $$29^{1761}$$ by 23.

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Youtong W.
Answer:

This is a typical question of Modular arithmetics that directly utilizes Fermat's Little Theorem. We will give a little review of what the theorem says be fore we proceed. Fermat's Little Theorem: If $$p \in N^+ $$ is prime, then: 1. For all $$a\in Z$$ with $$p\nmid a$$, we have $$a^{p-1} \equiv 1$$ (mod p); 2. For all $$a\in Z$$, we have $$a^p \equiv a$$ (mod p). Now 29 is prime, so is 23. So by the first part of Fermat's Little Theorem, $$29^{23-1}=29^{22} \equiv 1 $$(mod 23). Thus, $$(29^{22})^k \equiv 1^k$$ for all $$k \in N$$, which should be a previous conclusion at this point of your number theory study. If you are not comfortable with that, try showing the following theorem by induction. Theorem: If $$a\equiv b (\bmod m)$$, then $$a^k \equiv b^k $$(mod m). Finally, $$29^{1761}=29^{22\times80 +1}=29^{22\times80}\times 29^1\equiv 1^{80} \times29^1 (\bmod 23) \equiv 29^1 \equiv 6(\bmod 23)$$. This just means that the remainder of $$29^{1761}$$ when divided by 23 is 6.

Subject: Calculus

TutorMe
Question:

Show that the equation $$x^3-15x+c=0$$ has at most one root in the interval [-2, 2].

Inactive
Youtong W.
Answer:

This is a typical question that would appear on AP Calc BC test. Let $$f(x)=x^3-15x+c=0$$. We know that $$f'(x)=3x^2-15$$. one the intervel [-2, 2], the largest value of $$x^2$$ is $$2^2=(-2)^2=4$$, which means $$f'(x) \leq 3 \times 4 -15 = -3$$ is always true. This tells us that the value of the derivative function of $$f(x)$$ on [-2, 2] is always negative, which means that $$f(x)$$ is always decreasing. This means that if $$f(a)=0$$ for some$$-2 \leq a \leq 2$$, then we must have $$f(b) > 0$$ for all $$b \in [-2, 2]$$and b > a; $$f(b) < 0$$ for all $$b \in [-2, 2]$$and b < a. In other words, $$f(x)$$ has at most one root.

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