# Tutor profile: Sarah B.

## Questions

### Subject: Linear Algebra

If $$A$$ and $$B$$ are linear transformations (on the same vector space), then a necessary and sufficient condition that both $$A$$ and $$B$$ be invertible is that both $$AB$$ and $$BA$$ be invertible.

Let us show that if $$V$$ is a vector space and $$A,B\in L(V,V)$$, then $$A,B$$ are invertible iff $$AB, BA$$ are invertible. To do this, lets show that if $$f:X\rightarrow Y, g:Y\rightarrow Z$$ and if $$g\circ f$$ is injective, then $$f$$ is injective and if $$g\circ f$$ is surjective then $$g$$ is surjective. First, let us suppose that $$g\circ f$$ is injective, but that $$f$$ is not. Then there exists $$x\neq y$$ in $$X$$ such that $$f(x)=f(y)$$. We then have that $$g(f(x))=g(f(y))$$, contradicting our assumptions that $$g\circ f$$ is injective. Thus $$f$$ must be injective. Now, let us suppose that $$g\circ f$$ is surjective but that $$g$$ is not. Then we have that $$g(f(X))\subseteq g(Y)\subset Z$$, which contradicts that $$g\circ f$$ is surjective. Thus, $$g$$ must be surjective. From what we just proved, we can show that if $$f,g: X\rightarrow X$$ and if $$g\circ f$$ and $$f\circ g$$ are bijections, then $$f$$ and $$g$$ are bijections. As $$g\circ f$$ is injective, we know that $$f$$ is injective and as $$f\circ g$$ is surjective, we know that $$f$$ is surjective. Thus $$f$$ is a bijection. Similarly, we have that as $$f\circ g$$ is injective, we have that $$g$$ is injective and as $$g\circ f$$ is surjective we know $$g$$ is surjective. And so we have that $$g$$ is a bijection. And so from all of this, we can see that if $$AB$$ and $$BA$$ are invertible (bijective), then so are $$A$$ and $$B$$. And, for the other direction, if $$A$$ and $$B$$ are invertible, then there exists matrices $$A^{-1}$$ and $$B^{-1}$$ such that $$AA^{-1}=I$$ and $$BB^{-1}=I$$. And so consider the matrices $$A^{-1}B^{-1}$ and $B^{-1}A^{-1}$$. we then have that $$(AB)(B^{-1}A^{-1})=I$$ and $$(BA)(A^{-1}B^{-1})=I$$.

### Subject: Calculus

Prove that the series $$\sum_{n=1}^{\infty} (-1)^n \frac{x^2+n}{n^2}$$ converges uniformly in every bounded interval, but does not converge absolutely for any value on $$x$$.

We have that $$\sum |(-1)^n\frac{x^2+n}{n^2}|=\sum |\frac{x^2}{n^2}+\frac{1}{n}|\geq \sum |\frac{1}{n}|.$$ We cna see that the rightmost sum is the harmonic series, which diverges. Thus, by the comparison test we have that the leftmost series must also diverge.\ Now, to show that the series converges uniformly in every bounded interval $$[a,b]$$, let $$\epsilon >0$$ be given. Define $$X$$ to be $$\sup\{|a|,|b|\}$$. Define the partial sum $$f_m$$ to be $$f_m=\sum_{n=1}^m (-1)^n \frac{x^2+n}{n^2}.$$ We can rearrange this to get $$f_m=\sum_{n=1}^m (-1)^n \frac{1}{n}+x^2\sum_{n=1}^m (-1)^n \frac{1}{n^2}.$$ We know that the alternating harmonic series and $$\sum\frac{1}{n^2}$$ converges; thus, by the Cauchy criterion for convergence we can find $$N$$ such that $$p>q>N\Rightarrow |\sum_{n=q}^p (-1)^n\frac{1}{n}|<\frac{\epsilon}{2}$$ and we can find an $$M$$ such that $$p>q>M\Rightarrow |\sum_{n=q}^p (-1)^n\frac{1}{n^2}|<\frac{\epsilon}{2|X|^2}.$$ Thus, we can choose $$p>q>\max\{N,M\}$$ so that we have $$|f_p(x)-f_q(x)|=|\sum_{n=q}^p (-1)^n \frac{1}{n}+x^2\sum_{n=q}^p (-1)^n \frac{1}{n^2}|$$ $$\leq |\sum_{n=q}^p (-1)^n\frac{1}{n}|+|x^2\sum_{n=q}^p (-1)^n \frac{1}{n^2}|$$ $$\frac{\epsilon}{2}+\frac{\epsilon x^2}{2X^2}\leq \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$

### Subject: Algebra

Let $$p$$ be a prime. Show that in $$\mathbb{Z}_p$$ we have $$(a+b)^p=a^p+b^p$$.

We know that the order of $$\mathbb{Z}_p$$ is $$p$$. And as $$p$$ is a prime, we know that $$\mathbb{Z}_p$$ is a commutative ring. As the binomial expansion is valid in a commutative ring, we know that $$(a+b)^p=\sum_{k=0}^{p} \frac{p!}{k!(p-k)!} a^{p-k}b^k$$. For $$k=0$$ or $$k=p$$, we have that $$\frac{p!}{k!(p-k)!}=1$$. This means that $$(a+b)^p=a^p$$ for $$k=0$$, and $$(a+b)^p=b^p$$ for $$k=p$$. For any other value for $$k$$, we know that $$\frac{p!}{k!(p-k)!}$$ is an integer. This means that its' denominator, $$k!(p-k)!$$, divides $$p!$$. But since $$p$$ is a prime and $$p$$ is not a factor of $$k!(p-k)!$$, we can deduce that $$k!(p-k)!$$ divides $$(p-1)!$$. In other words, $$\frac{p!}{k!(p-k)!}=1$$ is a multiple of $$p$$ when $$k$$ is not $$0$$ or $$P$$. Thus, $$\frac{p!}{k!(p-k)!}=1\cong 0 \mod p$$. And so the only nonzero terms of $$(a+b)^p=\sum_{k=0}^{p} \frac{p!}{k!(p-k)!} a^{p-k}b^k$$ occur when $$k=0$$ and $$k=p$$, giving us that $$(a+b)^p=a^p+b^p$$.

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