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Brendan C.

Graduate Teaching Assistant at University of Minnesota

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Linear Algebra

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Question:

Let $V$ and $W$ be two-dimensional real vector spaces. Show that for any given 5 linear transformations $T_1, \ldots ,T_5$ from $V$ to $W$, there exist scalars $\alpha_1 ,\ldots ,\alpha_5 \in \mathbb{R}$ not all 0 such that $\alpha_1 T_1 + \alpha_2 T_2 + \alpha_3 T_3 + \alpha_4 T_4 + \alpha_5 T_5$ is the zero transformation.

Brendan C.

Answer:

Let $L(V,W)$ denote the space of linear transformations $T: V \to W$. It is straightforward to check that $L(V,W)$ is a real vector space whose $0$ element is the zero linear transformation, since taking linear combinations of linear transformations yields another linear transformation. Furthermore, we show that $L(V,W)$ has dimension $4$. We fix bases $v_1 , v_2$ for $V$ and $w_1, w_2$ for $W$. With respect to these bases, every linear transformation $T: V \to W$ is represented by a unique $2\times 2$ matrix $A_T \in M_{2\times 2}(\mathbb{R})$. Furthermore, if $T_1 , T_2 \in L(V,W)$, the correspondence is bijective and satisfies $A_{T_1 + T_2} = A_{T_1} + A_{T_2}$ and $A_{\alpha T_1} = \alpha A_{T_1}$ for $\alpha \in \mathbb{R}$. It follows that $L(V,W)$ is isomorphic to the vector space $M_{2\times 2}(\mathbb{R})$ of all $2\times 2$ matrices with real entries, which has dimension 4. Finally, any given $T_1, \ldots ,T_5 \in L(V,W)$ must be linearly dependent, so there are scalars $\alpha_1 ,\ldots ,\alpha_5 \in \mathbb{R}$ not all 0 such that $\alpha_1 T_1 + \ldots + \alpha_5 T_5 = 0$.

Pre-Calculus

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Question:

Find the domain of the function $f(x) = \sqrt{\frac{x^2 - 5x}{x^2 - 9}}$

Brendan C.

Answer:

The square root function $\sqrt{x}$ has domain $[0,\infty)$, i.e. the set of all real numbers $x$ with $x\geq 0$. Thus we must determine the set of real numbers $x$ for which the rational function $g(x) = \frac{x^2 - 5x}{x^2 - 9}$ is non-negative. Factoring numerator and denominator, we see that $$ g(x) = \frac{x(x-5)}{(x+3)(x-3)}$$ The denominator is zero when $x = \pm 3$, so $g(x)$ has vertical asymptotes at $x=3$ and $x=-3$, and $g(x) = 0$ when $x = 0$ or $x = 5$. These zeros and vertical asymptotes divide the real line into intervals $(-\infty, -3), (-3, 0), (0,3), (3,5)$, and $(5,\infty)$. On each interval, $g(x)$ assumes either strictly positive or strictly negative values. When $x < -3$, $x(x-5) > 0$ and $(x+3)(x-3) > 0$, so $g(x) > 0$. Through similar reasoning, we see that $g(x)>0$ is positive on the intervals $(-\infty, -3)$, $(0,3)$, and $(5,\infty)$, and $g(x)<0$ on the intervals $(-3,0)$ and $(3,5)$. The points $\pm 3$ are excluded from the domain since we cannot divide by zero, but the points $x = 0$ and $x = 5$ are included since $g(0) = g(5) = 0$. Thus, the domain of the function $f(x) = \sqrt{g(x)}$ is $(-\infty, -3) \cup [0,3) \cup [5,\infty)$. In other words, the domain is the set of all real numbers $x$ except those satisfying $-3 \leq x < 0$ or $3 \leq x < 5$.

Calculus

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Question:

Compute the tangent line to the curve $f(x) = \int_{3x^2 - 3}^{e^x - e} e^{t^2} dt$ at the point $(1,0)$

Brendan C.

Answer:

Using the facts that $\int_a^b g(t) dt + \int_b^c g(t) dt = \int_a^c g(t)dt$ and $\int_a^b g(t)dt = -\int_b^a g(t)dt$ for an integrable function $g$, we can write $$ \int_{3x^2 - 3}^{e^x - e} e^{t^2} dt = \int_{0}^{e^x - e} e^{t^2} dt + \int_{3x^2 - 3}^{0} e^{t^2} dt$$ $$ = \int_{0}^{e^x - e} e^{t^2} dt - \int_{0}^{3x^2 - 3} e^{t^2} dt = h(e^x -e) - h(3x^2 - 3)$$ where $h(x) = \int_0^x e^{t^2} dt$. The first Fundamental Theorem of Calculus assures us that $h^{\prime}(x) = e^{x^2}$, so we can compute $f^{\prime}(x)$ using the chain rule. Using the facts that $\frac{d}{dx}(e^x - e) = e^x$ and $\frac{d}{dx}(3x^2 - 3) = 6x$, we have: $$f^{\prime}(x) = h^{\prime}(e^x -e)(e^x) - h^{\prime}(3x^2 - 3)(6x) = e^{(e^x - e)^2}\cdot e^x - e^{(3x^2 - 3)^2} \cdot 6x $$ so $f^{\prime}(1) = e - 6$. The tangent line to $f(x)$ at $(1,0)$ is then the line $y = (e-6)(x-1)$

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