# Tutor profile: Gaddam S.

## Questions

### Subject: Linear Algebra

Q) Let the linear transformation $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be defined as: $T(x, y, z) = (x, x+y, x+y+z)$. Show that $T$ is non-singular.

A) The linear operator $T$ is said to be non-singular if and only if Kernel of $T$ contains only the zero vector. Kernel of $T$, say K(T), is defined as: $K(T) := \{(x,y,z) \in \mathbb{R}^3: T(x,y,z) = (0,0,0) \in\mathbb{R}^3\}$. Now by definition of $T$, it is clear that $T(x,y,z) = (0,0,0)$ implies that $(x,x+y,x+y+z) = (0,0,0)$. This further implies that $x=0, y=0, z=0$. So $K(T)=\{(0,0,0)\}$. Hence the result.

### Subject: Numerical Analysis

Q) Suppose that $g:\mathbb{R}\rightarrow \mathbb{R}$ is a continuous function and let for $a < b \in\mathbb{R}$, $g(a)\cdot g(b) < 0$. Show that there is a $c$ with $a < c < b$ such that $f(c) = 0$.

A) This is an application of Intermediate value theorem (IVT) of a continuous function. This the idea is the key to the Bisection Method. Since $g$ is a continuous function, then the restriction of the function $g$ to the interval $[a,b]$ is also continuous. Also since $g(a)\cdot g(b) < 0$, this implies that $g(a)$ and $g(b)$ have opposite signs and being a continuous function, $g$ should intersect with the x-axis at least one point between $(a,b)$. This implies, by using IVT, that there exists an intermediate function value $0$ at some point, say $c$, in the interval $(a,b)$.

### Subject: Calculus

Let $g(x)$ be a given function such that $\lim_{x\rightarrow 0^+} g(x)=R$ and $\lim_{x\rightarrow 0^-} g(x)=L$, then compute the following limits: a) $\lim_{x\rightarrow 0^+} g(x^5-x)$, b) $\lim_{x\rightarrow 0^-} g(f(x))$, where $f(x)$ is a strictly increasing and continuious function and $f(0)=0$, and c) $\lim_{x\rightarrow 0^+} g(\exp(x)-\exp(-x))$

a) For $0\leq x< 1$, we have $x^5<x$. So as $x$ approaches $0$ from the right, $x^5-x$ approaches 0 from the left. This implies $\lim_{x\rightarrow 0^+} g(x^5-x)=L$. b) Since $f(x)$ is a strictly increasing function and $f(0)=0$, this implies that $f(x)<0$ whenever $x<0$. As $x$ approaches $0$ from the left, and also by continuity $f(x)$ approaches $0$ from the left. This implies $\lim_{x\rightarrow 0^+} g(f(x))=L$. c) For $0\leq x$, we have $\exp(x)\geq\exp(-x))$ and equality holds only at $x=0$, that is, $\exp(0)=\exp(-0)=1$. So as $x$ approaches $0$ from the right, and because of continuity$\exp(x)-\exp(-x))$ approaches 0 from the right. This implies $\lim_{x\rightarrow 0^+} g(\exp(x)-\exp(-x))=R$

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