linear algebra - If $AB = I$ then $BA = I$

If $A$ and $B$ are square matrices such that $AB = I$, where $I$ is the identity matrix, show that $BA = I$.

I do not understand anything more than the following.

1. Elementary row operations.


2. Linear dependence.



3. Row reduced forms and their relations with the original matrix.

If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem?

P.S.: Please avoid using the transpose and/or inverse of a matrix.

Answer

Dilawar says in 2. that he knows linear dependence! So I will give a proof, similar to that of TheMachineCharmer, which uses linear independence.

Suppose each matrix is $n$ by $n$. We consider our matrices to all be acting on some $n$-dimensional vector space with a chosen basis (hence isomorphism between linear transformations and $n$ by $n$ matrices).

Then $AB$ has range equal to the full space, since $AB=I$. Thus the range of $B$ must also have dimension $n$. For if it did not, then a set of $n-1$ vectors would span the range of $B$, so the range of $AB$, which is the image under $A$ of the range of $B$, would also be spanned by a set of $n-1$ vectors, hence would have dimension less than $n$.

Now note that $B=BI=B(AB)=(BA)B$. By the distributive law, $(I-BA)B=0$. Thus, since $B$ has full range, the matrix $I-BA$ gives $0$ on all vectors. But this means that it must be the $0$ matrix, so $I=BA$.

complex variable integral using residue theorem

I am asked to calculate a complex integral. how to compute $\displaystyle \int \limits_{-\infty}^{\infty}\frac{x^4}{1+x^8}dx$ with residue theorem?

cardinals - Why are the cardinality of $mathbb{R^n}$ and $mathbb{R}$ the same?

As I study the first part of abstract algebra, I have a question: why $\left\vert{\mathbb R}\right\vert = \left\vert{\mathbb R^2}\right\vert$?

And moreover, I knew the fact that $\left\vert{\mathbb R}\right\vert = \left\vert{\mathbb R^n}\right\vert$.

Does the equality apply to any infinite set?

How can I prove them?

real analysis - A discontinuous function such that $f(x + y) = f(x) + f(y)$

Is it possible to construct a function $f \colon \mathbb{R} \to \mathbb{R}$ such that $$f(x + y) = f(x) + f(y)$$ and $f$ is not continuous?

Answer

"Fix a basis for $\mathbb R$ as a $\mathbb Q$-vector space" (which exists under the axiom of choice, but under weaker axioms I have no idea what happens). The condition $f(x+y) = f(x) + f(y)$ is equivalent to $f$ being $\mathbb Q$-linear, so you're asking if there exists a non-trivial discontinuous map of $\mathbb Q$-vector spaces between $\mathbb R$ and itself. If you map the basis elements to other basis elements in a discontinuous way, you will obtain such a map.

Added : A quick way to see that "a discontinuous way of doing it" exists, is that the set of $\mathbb Q$-linear maps that are also $\mathbb R$-linear has cardinality of the reals, where as the set of all $\mathbb Q$-linear maps (or in other words, the number of maps between the basis elements) has cardinality $|\mathbb R|^{|\mathbb R|}$. To understand why, well of course the basis for $\mathbb R$ as a $\mathbb Q$-vector space has cardinality $\le |\mathbb R|$, but if it was countable, then $\mathbb R$ would be countable because all of its elements would be a linear combination of elements in a countable set with countably many possible coefficients. Since the basis for $\mathbb R$ as a $\mathbb Q$-vector space is uncountable, the set of all maps from a basis to another also is. Therefore there must be at least one map between the two bases which generates a discontinuous linear map.

Hope that helps,

sequences and series - Is $sum_{n=3}^inftyfrac{1}{nlog n}$ absolutely convergent, conditionally convergent or divergent?

Classify
$$\sum_{n=3}^\infty \frac{1}{n\log(n)}$$
as absolutely convergent, conditionally convergent or divergent.

Is it,
$$\sum_{n=3}^\infty \frac{1}n$$ is a divergent $p$-series as $p=1$, and

$$\lim_{n\to\infty} \frac{1}{n\log(n)}{n} = 0$$ by comparison test. And this converges to $0$.

So,
$$\sum_{n=3}^\infty \frac{1}{n\log (n)}$$
is conditionally convergence?

I'm not sure if I'm doing right or not. Could you guide me?

Thanks in advance! :)

Answer

Note that if your given series is convergent then it's also absolutely convergent since $\frac{1}{n\log n}\geq 0\quad \forall n\geq 3$.

Now, since the sequence $(\frac{1}{n\log n})$ is decreasing to $0$ so by the integral test your series has the same nature that the improper integral
$$\int_3^\infty\frac{dx}{x\log x}=\left[\log(\log(x))\right]_3^{\to\infty}=\infty$$
hence the series is divergent.

algebra precalculus - Prove that $(1+2+3+cdots+n)^2=1^3+2^3+3^3+cdots+n^3$ $forall n in mathbb{N}$.

Prove that $(1+2+3+\cdots+n)^2=1^3+2^3+3^3+\cdots+n^3$ for every $n \in \mathbb{N}$.

I'm trying to use induction on this one, but I'm not sure how to. The base case is clearly true. But when I add $n+1$ to the right and left hand sides, I don't know what I'm supposed to get. For example, when I extend the right hand side's sequence by $n+1$, I get $n^3+(n+1)^3$ at the end of the sequence, which is $2n^3 +3n^2+3n+1$.

This doesn't seem that meaningful, so I don't know how to proceed.

polynomials - How do I come up with a function to count a pyramid of apples?

My algebra book has a quick practical example at the beginning of the chapter on polynomials and their functions. Unfortunately it just says "this is why polynomial functions are important" and moves on. I'd like to know how to come up with the function (the teacher considers this out of scope). Even suggesting google search terms to find out more information on this sort of thing would be helpful.

The example

Consider a stack of apples, in a pyramid shape. It starts with one apple at the top. The next layer is 2x2, then 3x3, and so on. How many apples are there, given x number of layers?

The polynomial function

$$f(x) = \frac{2x^3 + 3x^2 + x}{6}$$

What I do and don't understand

Thanks to @DJC, I now know this is a standard function to generate Square Pyramidal Numbers, which is part of Faulhaber's formula. Faulhaber's formula appears to be about quickly adding sequential coefficients which all have the same exponent. Very cool. But how does one get from:

$$\sum_{k=1}^{n} k^p$$

to the spiffy function above? If I'm sounding stupid, how do I make the question better?

Fwiw, I'm in intermediate algebra in the USA. The next course would be Trigonometry or Calculus. (to help people place my current knowledge level)