[UnSA] An Integers Matrix Lemma that sometimes being forgotten

[UnSA] An Integers Matrix Lemma that sometimes being forgotten

21:53:00   UnSA,

I am going to prove the following useful lemma and wrapping the proof of inverse-adjugate formula inside my proof. I hope I put enough details on it.

Lemma
Let $A\in \mathbb{Z}^{n\times n}$ with $\text{det } A \neq 0$. Then there exists a unique matriks $B\in \mathbb{Q}^{n\times n}$ such that $AB=BA=(\text{det } A) I$, and moreover $B$ has integer entries.

Proof:
Suppose that $A=(a_{ij})$, where $a_{ij}\in \mathbb{Z}$. Let $C_{ij}$ denotes the $(i,j)$ cofactor of $A$, that is $C_{ij}=(-1)^{i+j}M_{ij}$ where $M_{ij}$ is the determinant of the $(n-1) \times (n-1)$ matrix obtained by deleting the $i$-th row and $j$-th column of $A$. Since $A$ has integers entries, $M_{ij}$ is the determinant of a matrix with integers entries, thus $M_{ij}$ is also an integer for all $i,j \in \{1,2,\cdots,n\}$.  So we have $C_{ij} \in \mathbb{Z}$ for all $i,j\in\{1,2,\cdots,n\}$.

Define the matrix $C=(C_{ij})$, that is the $ij$-th entry of $C$ is $C_{ij}$, and let $B=C^{T}$. Since all $C_{ij}$ are integers,  we conclude that $B$ has integer entries.

Next we try to compute $AB$. Denote the $ij$-th entry of $AB$ by $z_{ij}$, therefore by the rule of multiplication of matrix we have
\[z_{ij}=\sum_{k=1}^n a_{ik} C_{jk}\]
If $i=j$ then by cofactor expansion for determinant of $A$ we have
\[z_{ii}=\sum_{k=1}^n a_{ik} C_{ik}= \text{det } A\]
If $i\neq j$, consider the matrix
\[A_{ij}=\left(\begin{array}{ccccc} a_{11} &a_{12} &\cdots &\cdots &a_{1n}\\ \vdots &\vdots &\vdots &\vdots &\vdots \\ a_{i1} &a_{i2} &\cdots &\cdots &a_{in} \\ \vdots &\vdots &\vdots &\vdots &\vdots \\  a_{i1} &a_{i2} &\cdots &\cdots &a_{in} \\ \vdots &\vdots &\vdots &\vdots &\vdots \\ a_{n1} &a_{n2} &\cdots &\cdots &a_{nn}\end{array}\right)\]
here is the explanation of the above matrix, the two identical rows are the $i$-th row and the $j$-th row, and since we have these two identical rows, the determinant of this new matrix is zero. Also we have $C_{jk}(A_{ij})=C_{jk}$, that is the cofactor of $A_{ij}$ amongst $j$-th row is the same as the cofactor of $A$ amongst $j$-th row. Using these facts and cofactor expansion, for $i\neq j$ we have
\[0=\text{det }(A_{ij})=\sum_{k=1}^n a_{ik} C_{jk}(A_{ij})=\sum_{k=1}^n a_{ik} C_{jk}=z_{ij}\]
We conclude that
\[z_{ij} = \left\{ \begin{array}\text{det } A & i= j \\ 0 & i\neq j \end{array}\right.\]

and we have managed to prove that $AB=(\text{det } A) I$, where $B$ has integer entries. Now, since $\text{det } (A) \neq 0$ the inverse matrix $A^{-1}$ exists, thus
\[\begin{align*}AB=(\text{det } A) I &\Rightarrow ABA= (\text{det } A)IA=A (\text{det } A) \\ &\Rightarrow A^{-1}A(BA)=A^{-1}A(\text{det } A) \end{align*}\]
so that we have $BA=(\text{det } A) I$ as well. It remains to prove that $B$ is unique.
Suppose there exists $B_1\neq B$ such that $AB_1 = B_1A= (\text{det } A) I$. Therefore \[AB_1 = (\text{det } A) I = AB\]
multiplying both sides by $A^{-1}$ we have $B_1=B$, a contradiction, thus $B$ is unique and we are done.

One Corollary that follows from this lemma is:
The inverse of the integer matrix with determinant equals $+1$ or $-1$ is again the integer matrix.

[sometimes we just forgot we have this fact]

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[UnSA] The Condition for existence of minimum in Survival-Related Function.

[UnSA] The Condition for existence of minimum in Survival-Related Function.

22:18:00   UnSA,

Today, I was asked about the proof  stated on this paper  (theorem 2.1 page 98), (the proof is on page 110), we are dazed with the argument given there, and started to think that it's probably an errata (mistyped but not serious) or we just missed the point.

Let $S_X(t)$ be a survival function, define for each $d>0$ and $0<\alpha < S_X(0)$ the function

\[\text{VaR}_T(d,\alpha) := \left\{\begin{array}{lr} d+\delta(d)\qquad &  \text{for }d \in (0,S_X^{-1}(\alpha)\\ S_X^{-1}(\alpha)+\delta(d) \qquad & \text{for } d > S_X^{-1}(\alpha)
\end{array}\right.\]

Where $\delta(d)=(p+1)\int_d^{\infty} S_X(t) \, dt$ and $p^*=\frac{1}{1+p}$.

The value $d^*>0$ such that $\text{VaR}_T(d^*,\alpha)=\min_{d>0}(\text{VaR}_T(d,\alpha))$ exists if and only if 

\[S_X^{-1}(\alpha)\geq S_X^{-1}(p^*)+\delta(S_X^{-1}(p))\]

and 

\[\alpha < p^* < S_X(0)\]

Proof:

Recall that, the survival function is monotone decreasing function positive. Also it can be proved that $\text{VaR}_T(d,\alpha)$ is continuous. For $d>S_X^{-1} (\alpha)$, the derivative of $\text{VaR}_T(d,\alpha)$ is 

\[\delta'(d)=-(1+p) S_X(d) < 0\] 

(note that the differentiation under integral sign is verified since $S_X(d)$ is bounded, hence clearly dominated).
Therefore $\text{VaR}_T(d,\alpha)$ is decreasing on $(S_X^{-1}(\alpha),\infty)$, with limit $S_X^{-1}(\alpha)$ as $d\rightarrow \infty$. Also,  for $p^*>S_X(0)$ we have the derivative of $d+\delta(d)$ is

\[1-(p+1)S_X(d)<0\]

thus $d+\delta(d)$ is decreasing when $p^* > S_X(0)$ until $d=S_X^{-1}(p^*)$. Therefore the function $d+\delta(d)$ attains minimum value on $d_0=S_X^{-1}(p^*)$. If the both conditions hold then $d_0 \in (0,S_X^{-1}(\alpha))$  and the minimum point is now also the minimum point of $\text{VaR}_T(d,\alpha)$, therefore $d^*=d_0>0$.

Conversely, If $\alpha\geq p^*$ then since $S_X(t)$ is decreasing its inverse function is also  decreasing, it follows that $S^{-1}_X(\alpha) \leq S_X^{-1}(p^*)=d_0$, and for all $d\leq S^{-1}_X(\alpha)$ we have $S_X(d) \geq p^*$, therefore the derivative of the function $\text{VaR}_T(d,\alpha)$ is equal to $1-(p+1)S_X(d)$ on $d\in (0,S_X^{-1}(\alpha))$ and equal to $-(p+1)S_X(d)$ when $d>S_X^{-1}(\alpha)$ which is consistently negative. Thus $\text{VaR}_T(d,\alpha)$ is decreasing on $(0,\infty)$ and as $d\rightarrow \infty$ it has limiting value $S_X^{-1}(\alpha)$, while no such $d^*$ with $\text{VaR}_T(d^*,\alpha)=S_X^{-1}(\alpha)$, so no such $d^*$ exists.


Now suppose that $p^* \geq S_X(0)$, then we have $d=S_X^{-1}(p^*) \leq 0$ thus $d_0 =0$. Since $d_0$ is the only stationery point of the function $d+\delta(d)$, it follows that this function does not have a stationery point when $d>0$,  therefore on $0< d\leq S_X^{-1}(\alpha)$ the derivative of the function is either strictly positive or strictly negative (if it ever been positive in one point and then negative in other point, then since the derivative is continuous by Intermediate Value Theorem it has zero, thus it has stationery point.)
From the above argument it follows that $d+\delta(d)$ is either monotone decreasing or monotone increasing on $0< d \leq S_X^{-1}(\alpha)$.
[In the paper it is stated only the decreasing case] 
It it is increasing, then $VaR_T(d,\alpha)$ is increasing on $(0,S_X^{-1}(\alpha))$ and eventually starts decreasing on $(S_X^{-1}(\alpha),\infty)$ with limited value $S_X^{-1}(\alpha)$, while it has no minimum value on $(0,S_X^{-1}(\alpha))$ the limited value $S_X^{-1}(\alpha)$ is the smallest the function $\text{VaR}_T(d,\alpha)$ can approach, but as before no such $d^*$ with  $\text{VaR}_T(d^*,\alpha)=S_X^{-1}(\alpha)$. If $\delta(d)+d$ is decreasing, then $\text{VaR}_T(d,\alpha)$ is decreasing on $(0,\infty)$ with limiting value $S_X^{-1}(\alpha)$ as $d\rightarrow \infty$, thus no such $d^*$ exists.

If condition (1) does not hold while condition (2) holds then $\inf_{d>0} \text{VaR}_T(d,\alpha)=S_X^{-1}(\alpha)$, in order to have the infimum a minimum, there should be exists $d^*$ with $\text{VaR}_T(d^*,\alpha)=S_X^{-1}(\alpha)$, but this is clearly impossible, and the proof is completed.

Remark: Because I don't know much about the behavior of Survival Function, I tend to avoid the second derivative test for stationery point $d_0$, if we employ this test, we will get that this point is a minima, thus $\text{VaR}_T  (d,\alpha)$ is not always decreasing when $d \in (0,\infty)$, this is  where we think the argument in the paper is fishy.

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[UnSA] Atsuji Space is Complete

[UnSA] Atsuji Space is Complete

09:00:00   Analytic, UnSA,

Today, we were looking at the abstract of a paper which says"It is well-known that Atsuji Space is a complete space" .... I was riddled with this statement, I found the proof, It was based from 1955 Paper by Norman Levine and William G. Saunders.

Recall that a metric space $(X,d)$ is called Atsuji Space iff  every real-valued continuous functions on $X$ are also uniformly continuous on $X$.

We want to prove that:

If $(X,d)$ is an Atsuji space then every Cauchy sequence $(x_n)$  in $X$ converges.

Equivalently, we want to prove that
If there exists a Cauchy sequence in $X$ which does not converge, then there exists a continuous function $f: X \rightarrow \mathbb{R}$  which is not uniformly continuous.

In proving the above assertion, we need a following Lemma
[Uryohn's Lemma]
Let $X$ be a normal space. If $A$ and $B$ are two disjoint closed subsets of X, then there exists a continuous function $f$ from $X$ to the real line $\mathbb{R}$ such that $f(x)=0$ for all $x\in A$ and $f(x)=1$ for all $x\in B$.

The proof of Uryohn's Lemma is rather long, I remembered that the nearly two pages proof is on James Munkers Book, Topology.  Recall also that from elementary topology course: all metric spaces are normal, so Uryohn's Lemma works also for metric space.

It is not hard to prove that: A Cauchy sequence which has at least one convergent subsequence, is a convergent sequence. Therefore a divergent Cauchy sequence would imply that all of its subsequences are also divergent.

Suppose that $\{x_i\}_{i=1}^{\infty}$ is a Cauchy sequence of $X$ which diverges. Consider the set $\{x_1,x_2,\cdots,x_{i},\cdots\}$, since $(x_n)$ diverges this set has no limit point on $X$. Define
\[A:=\{x_{2k} | k\in \mathbb{N}\} \qquad  B:=\{x_{2k-1} | k \in \mathbb{N}\}\]
The element of $A$ will form a subsequence of $(x_n)$ and by our remark on the previous paragraph this subsequence also diverges, thus $A$ also has no limit point. Similarly $B$ also has no limit point. Therefore $A$ and $B$ are disjoint closed subsets of $X$, thus by Uryohn Lemma since $X$ is normal space, there exists a continuous function $f:X\rightarrow \mathbb{R}$ such that  $f(x_{2k-1})=0$ and $f(x_{2k})=1$.  This function is not uniformly continuous, since there exists $\epsilon_0 = \frac{1}{2}$ and two sequences $u_k = x_{2k}$ and $v_k=x_{2k-1}$ with $\lim (u_n - v_n) =0$ (since $x_n$ is Cauchy) but $f(u_n)-f(v_n)=1> \frac{1}{2}=\epsilon_0$.

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New Tag [Unofficial Skripsi Advisor]

New Tag [Unofficial Skripsi Advisor]

06:09:00   UnSA,

Following the new semester, I have some of my junior working on their undergraduate thesis (so called skripsi) with me, since I cannot come to campus quite often, I'm planning to direct them on their skripsi via this blog (LaTex would help).  

The tag is [Unofficial Skripsi Advisor] and will be indicated on its title post by  abbreviation [UnSA]. 

The proofs that will be presented might also being found on the books. But, it will be described in detail. Also, some proofs might be also a new proof based on recent Journal (if it is more elementary than the already known proof).      

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