The Proof is Trivial!

Problem 369***(easy)

Prove that any diagonalisable matrix satisfies it's own characteristic polynomial

Problem 369***(hard)

This time, without the assumption of diagonalisability.

The hard version is the Caley-Hamilton theorem; the easy one just a special case which is much more straightforward. Getting boring just seeing number theory and integrals.

Reply 2341

Mladenov

1

Solution 369

What's a matrix? It's a morphism between vector spaces. So we identify our matrix with the endomorphism of vector spaces

A : V \to V

where

V

is a finite dimensional vector space over some algebraically closed field, call it

K

. If we do not assume the field algebraically closed, we simply choose a larger algebraically closed field which contains our field.
Now view

V

as a module over

K[t]

under the map

f(t)v = f(A)v

, where

v \in V

. Trivial observation shows that

V_{A}

is a torsion module. Hence, we can apply the structure theorem for modules over principal rings. Thus, we obtain the Jordan canonical form

V = V_{1} \oplus V_{2} \oplus \cdots \oplus V_{q}

, where

V_{i}

are

A

-invariant which can be written as

Unparseable latex formula:

W_{V_{i}}_{1} \oplus \cdots \oplus W_{V_{i}}_{r(V_{i})}

where

Unparseable latex formula:

W_{V_{i}}_{j}

are subspaces isomorphic to

Unparseable latex formula:

K[t]v_{i}_{j}

for some

Unparseable latex formula:

v_{i}_{j} \in V_{i}

.
Let

P_{A} = (t-\alpha_{1})^{m_{1}}\cdots (t-\alpha_{q})^{m_{q}}

be the characteristic polynomial. Then the kernel of the map

Unparseable latex formula:

f(t) \mapsto f(A)W_{V_{i}}_{j}

is

Unparseable latex formula:

(t-\alpha_{i})^{s_{i}_{j}}

, where

Unparseable latex formula:

s_{i}_{j}

depends on

i

and

j

. Also,

Unparseable latex formula:

s_{i}_{j} \le m_{i}

and

Unparseable latex formula:

\displaystyle \sum_{j} s_{i}_{j} = m_{i}

. By

Unparseable latex formula:

(A-\alpha_{i}I)^{m_{i}}v_{i}_{j}=0

and the fact that

V

is generated by the elements of the form

Unparseable latex formula:

f(A)v_{i}_{j}

, for which

Unparseable latex formula:

P_{A}(A)f(A)v_{i}_{j}=f(A)P_{A}(A)v_{i}_{j}=0

, we get for all

v \in V

,

P_{A}(A)v=0

and consequently

P_{A}(A)=0

.

We can think of this proof as follows: firstly, we decompose

V

as a direct sum of torsion modules over the principal ring

K[t]

such that these torsion modules have exponents which are prime powers. Then, we use the structure theorem to decompose each of the summands in a direct sum of cyclic modules. Thus, under the corresponding basis our endomorphism is composed of cells in which all the elements are zero, except these on the diagonal (they are equal) and these on the super diagonal (they are not relevant). So we are done if we show that for an endomorphism composed of a single such cell, the proposition is true. But this is obvious since the vector space viewed as a module over the polynomial ring is cyclic and the ring of polynomials is commutative.

Spoiler

Edit: I forgot to note the obvious connection between linear algebra and algebra established by the structure theorems for modules over principal rings. I quite like this link. I can post much more interesting propositions; however, I restrain myself from doing it since I do not know whether people would like them (these would be mainly from my current reading list which includes some category theory and algebraic topology).

(edited 10 years ago)

Reply 2342

Arieisit

3

Why is everyone boycotting my question 364? :frown:

Posted from TSR Mobile

Reply 2343

Mladenov

1

Original post by Arieisit

Why is everyone boycotting my question 364? :frown:

Posted from TSR Mobile

I have just seen the problem. I personally am not going to do it for it requires too much calculations (for me), and I dislike them passionately.

Reply 2344

Arieisit

3

Problem 370**/***

Prove that in the real number system that the inverse of addition is represented by

x + (-x) = 0

Posted from TSR Mobile

(edited 10 years ago)

Reply 2345

james22

16

Original post by Arieisit

Problem 370*/**

Prove that in the real number system that the inverse of addition is represented by

x + (-x) = 0

Posted from TSR Mobile

From what starting point? One perfectly valid definition of the reals is as an ordered field (with a few other properties). Under this definition each numeber, x, has an additive inverse which is normally denoted by -x.

Reply 2346

Arieisit

3

Original post by james22

From what starting point? One perfectly valid definition of the reals is as an ordered field (with a few other properties). Under this definition each numeber, x, has an additive inverse which is normally denoted by -x.

There are a few ways I can think of in which this can be proven but its up to you to decide what to do (if you can).

Posted from TSR Mobile

Reply 2347

james22

16

Original post by Arieisit

There are a few ways I can think of in which this can be proven but its up to you to decide what to do (if you can).

Posted from TSR Mobile

My problem is that I could say that -x is the additive inverse of x (by definition of -x) and so x+(-x)=0 as 0 is defined as the additive identity. This is the definition of the reals that I was taught.

Reply 2348

henpen

Not really a problem, but

\large \sum_{n\ge1}\frac{1}{\binom{n+m+1}{m+1}}

can be evaluated using Beta functions. I'm not sure how well known this is, so apologies if this is a little mundane.

Reply 2349

Felix Felicis

13

Original post by henpen

Not really a problem, but

\large \sum_{n\ge1}\frac{1}{\binom{n+m+1}{m+1}}

can be evaluated using Beta functions. I'm not sure how well known this is, so apologies if this is a little mundane.

\displaystyle \binom{n+m+1}{m+1}^{-1} = \frac{\Gamma (m+2) \Gamma (n+1)}{\Gamma (m+n+2)}

.

Now,

\Gamma (z+1) = z \Gamma (z)

and

\displaystyle \frac{\Gamma (x) \Gamma (y) }{\Gamma (x+y)} = \text{B} (x,y)

.

Solution

(edited 10 years ago)

Reply 2350

MW24595

Original post by Mladenov

I can post much more interesting propositions; however, I restrain myself from doing it since I do not know whether people would like them (these would be mainly from my current reading list which includes some category theory and algebraic topology).

Go for it.
How's the self-study coming along now? Also, did you finish with Lang's "Algebra" yet?

Reply 2351

henpen

Original post by Felix Felicis

\displaystyle \binom{n+m+1}{m+1}^{-1} = \frac{\Gamma (m+2) \Gamma (n+1)}{\Gamma (m+n+2)}

.

Now,

\Gamma (z+1) = z \Gamma (z)

and

\displaystyle \frac{\Gamma (x) \Gamma (y) }{\Gamma (x+y)} = \text{B} (x,y)

.

Solution

Indeed. Any other nice problems solvable with Beta?

Reply 2352

Hodor

3

Original post by FireGarden

Problem 369***(easy)

Prove that any diagonalisable matrix satisfies it's own characteristic polynomial

Problem 369***(hard)

This time, without the assumption of diagonalisability.

The hard version is the Caley-Hamilton theorem; the easy one just a special case which is much more straightforward. Getting boring just seeing number theory and integrals.

"Solution" 369:

The characteristic polynomial of a square matrix A is p(x) = det(A-xI). So p(A) = det(A-AI) = det(A-A) = det(0) = 0.

:smile:

Reply 2353

Indeterminate

3

Time for some number theory

Problem 371 ***

Prove that for any integer

M \geq 1

\displaystyle \sum_{m=1}^{M} \frac{1}{m} \leq \prod_{p\leq M} \left(1-\frac{1}{p}\right)^{-1}

where

p

is a prime number

(edited 10 years ago)

Reply 2354

arkanm

5

Let

p\in \mathbb{P}

. Observe that

RHS=\prod_{p\le M}\left(\frac{1}{1-1/p}\right)=\prod_{p\le M}\left(1+\frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}+\cdots \right)

(infinite geometric series: 1+x^2+x^3+...=1/(1-x), given that |x|<1, which is the case here since all prime are greater than 1)

Thus, by the fundamental theorem of arithmetic, when

p\le M

the RHS of the inequality is

1+\frac{1}{2}+\frac{1}{3}+\cdots

,

where the

\cdots

indicates that the sum goes on forever, since there are infinitely many numbers that have factors in

p\le M

. A simple example would be

\frac{1}{2^a\times 3^a\times 5^a\times \cdots \times p^a}

(p being the largest prime

\le M

), where

a

take on infinitely many values.

Then the problem boils down to showing that

1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{M-1}+\frac{1}{M}\le 1+\frac{1}{2}+\frac{1}{3}+\cdots

.

No idea now. This seems more like an analytic number theory problem to me, which I've never studied in my life. (in A2 at the moment; don't even know single variable calculus properly yet lol..)

I can note that the LHS is a finite sum while the RHS is an infinite one. I don't know how to show if it diverges or not though, and how then to deploy that fact on the problem. :s-smilie:

EDIT: I also don't know exactly how to deal with the equality case here.

(edited 10 years ago)

Reply 2355

james22

16

Original post by arkanm

Let

p\in \mathbb{P}

. Observe that

RHS=\prod_{p\le M}\left(\frac{1}{1-1/p}\right)=\prod_{p\le M}\left(1+\frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}+\cdots \right)

(infinite geometric series: 1+x^2+x^3+...=1/(1-x), given that |x|<1, which is the case here since all prime are greater than 1)

Thus, by the fundamental theorem of arithmetic, when

p\le M

the RHS of the inequality is

1+\frac{1}{2}+\frac{1}{3}+\cdots

,

where the

\cdots

indicates that the sum goes on forever, since there are infinitely many numbers that have factors in

p\le M

. A simple example would be

\frac{1}{2^a\times 3^a\times 5^a\times \cdots \times p^a}

(p being the largest prime

\le M

), where

a

take on infinitely many values.

Then the problem boils down to showing that

1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{M-1}+\frac{1}{M}\le 1+\frac{1}{2}+\frac{1}{3}+\cdots

.

No idea now. This seems more like an analytic number theory problem to me, which I've never studied in my life. (in A2 at the moment; don't even know single variable calculus properly yet lol..)

I can note that the LHS is a finite sum while the RHS is an infinite one. I don't know how to show if it diverges or not though, and how then to deploy that fact on the problem. :s-smilie:

EDIT: I also don't know exactly how to deal with the equality case here.

In your last inequality, the RHS diverges and the inequality is obvious enough to assume. You would not be able to prove it formally without knowing some analysis though (not a very intersting problem either and just an application of definitions).

Reply 2356

arkanm

5

Original post by james22

In your last inequality, the RHS diverges and the inequality is obvious enough to assume. You would not be able to prove it formally without knowing some analysis though (not a very intersting problem either and just an application of definitions).

Yeah I don't think it's an interesting problem. I would rather have people posting interesting problems that are elementary (e.g. olympiad problems) than ones that are easy but require some advanced theory.

I didn't know that the RHS would diverge because it isn't the harmonic series as the numbers that are in the denominators depend on M. But if it does diverge then I guess it's easy to prove as the LHS is a finite sum.

What about the equality case?

Reply 2357

Mladenov

1

Solution 371

Note that

\displaystyle \prod_{p \le M} \left(\frac{1}{1-p^{-1}} \right) = \sum_{s \in S} \frac{1}{s}

, where

S

consists of all numbers whose prime factors are less than or equal to

M

. Hence

\begin{aligned} \displaystyle \prod_{p \le M} \left(\frac{1}{1-p^{-1}} \right) = \sum_{s: p|s, p \le M} \frac{1}{s} \ge \sum_{s \le M} \frac{1}{s} \end{aligned}

.

Original post by MW24595

Go for it.
How's the self-study coming along now? Also, did you finish with Lang's "Algebra" yet?

It goes well except for the last week during which I completed my application to read mathematics at Cambridge :biggrin:

. Nowadays, I read from one not-so-well-known book on algebraic topology by Prodanov, Spanier's classic, Hu's Homotopy Theory (I heard that Bott and Tu's Differential Forms in Algebraic Topology is one of the best books on the subject and introduces spectral sequences from the very beginning). I finished Lang's Algebra two months ago, and I dare say that now I find it perfectly clear and not so dry (it has to be supplemented since it is not so heavy on category theory); this is just because I tried Switzer's Homotopy and Homology which is a total dead end (there exists even worse choice - Elements of Homotopy Theory by Whitehead).

Good luck and enjoy your time at Cambridge.

Problem 372***

Let

A

and

B

be subsets of the topological space

X

so that

X = A^{\circ} \cup B^{\circ}

, where

A^{\circ}

is the interior of

A

. Show that there exist natural homomorphisms

\Delta : H_{n}(X) \to H_{n-1}(A \cap B)

such that

\cdots \mathop \to \limits^{\Delta} H_{n}(A \cap B) \mathop \to \limits^{\varphi} H_{n}(A) \oplus H_{n}(B) \mathop \to \limits^{\psi} H_{n}(X) \mathop \to \limits^{\Delta} H_{n-1}(A \cap B) \mathop \to \limits^{\varphi} \cdots

is an exact sequence, where

\varphi

and

\psi

are defined by the induced homomorphisms, namely

\varphi = (i_{\ast},j_{\ast})

and

\psi = k_{\ast} - l_{\ast}

,

i_{\ast} : H_{n}(A \cap B) \to H_{n}(A)

,

j_{\ast} : H_{n}(A \cap B) \to H_{n}(B)

,

k_{\ast} : H_{n}(A) \to H_{n}(X)

and

l_{\ast} : H_{n}(B) \to H_{n}(X)

.
This is a well-known sequence in algebraic topology which plays crucial role in homology theory. Using the same assumptions, show that the homomorphism