# 数学代写 | math代写 | 线性代数代写 – Linear Algebra (Math 314) Project 1

### Linear Algebra (Math 314) Project 1 数学代写

project – 数学代写 , 这个项目是project代写的代写题目 #### Due Friday, July 1

Problem 1(Interpolating Polynomial). Suppose that experimental data are represented by a set of points in the plane. Aninterpolating polynomialfor the data is a polynomial whose graph passes through every point. In scientific work, such a polynomial can be used, for example, to estimate values between the known data points. Another use is to create curves for graphical images on a computer screen. Find the interpolating polynomialp(t) =a 0 +a 1 t+a 2 t^2 for the data (1,12),(2,15),(3,16) by following the following steps.

(1) Write down the corresponding linear system (with 3 variables and 3 equations) by plugging in each of the data points. (2) Write down the RREF of the corresponding augmented matrix. (3) Write down the solution (a 1 ,a 2 ,a 3 ) to the corresponding linear system. Write down the inter- polating polynomial. (4) Estimate the partial data (2. 5 , ?) and (?,0) (that is, find?).

Problem 2(Interpolating Polynomial Continued). Letnbe any positive number, and let a 1 ,…,anbe any real numbers. Consider thennmatrix

``````An=
``````
``````1 a 1 a^21 a^31 ... an 1 ^1
1 a 2 a^22 a^32 ... an 2 ^1
1 a 3 a^23 a^33 ... an 3 ^1
..
.
``````
##### .
``````1 an a^2 n a^3 n ... ann^1
``````
##### .

(1) Compute det(A 2 ). (2) Show that det(A 3 ) = (a 2 a 1 )(a 3 a 2 )(a 3 a 1 ). Conclude thatA 3 is invertible if and only if … (3) Make an educated guess for det(An).

Note:The matrixAnis very closely related with the matrix you had in Problem 1 (forn= 3). This is the reason why there existsexactly onepolynomial of degree at mostn1 passing throughn differentpoints. This requires just a little bit of thinking, but in a nutshell, beacuse det(An) 6 = 0, we know thatAnwill be invertible, and thus when solving for the coefficients of the polynomial there will be a unique solution.

Problem 3(Inverses for non-square matrices). LetAbe anmnmatrix, wherem < n, and assume its rank is exactlym. Show that there exists a matrixBsuch thatAB= Im, by showing the following steps:

##### 1

(1) Show that, forj= 1,…,mthe matrix equationAx=ejhas a non-trivial solution. Here, as usual,ejdenotes the vector inRmthat has 1 on entryjand zero elsewhere. (2) LetbiinRnbe a non-trivial solution ofAx=ej, and consider the matrixB= [b 1 b 2 bm]. Show thatBdoes the job, that is,AB=Im. (3) Find (show all your steps) a matrixBsuch that

``````[
1 1  1
2 3 4
``````
##### B=I 2.

Note:This was a guided way to prove a not-so-easy statement. Splitting the statement into two mini statements makes the job much easier. Wouldnt you agree? Part (1) could be considered a big hint into proving the entire statement. The matrixB(that always exsits in this case) is called aright inverse. Note the use of a instead of the there could be many right inverses. A similar statement holds forleft inverses.

Problem 4(Markov Chains).In this problem we will cover (and go beyond, using what weve learnt in Chapter 5) concepts from Section 4.9. First some definitions. A vector with non-negative entries that add up to 1 is calledprobability vector. A (square) matrix for which every column is a probability vector is calledstochastic matrix. AMarkov chainis a sequence of probability vectorsx 0 ,x 1 ,x 2 ,…such that

``````xk+1=Pxk, k= 0, 1 , 2 , 3 ,... (0.1)
``````

andPis stochastic matrix^1.

(1) Consider the stochastic matrixP =

##### [
``````1 a b
a 1 b
``````
##### ]

, with 0< a,b <1. Show that= 1 is an eigenvalue ofP and find its corresponding eigenvectoru. (DO NOT PICK YOUR OWN NUMBERS!) (2) Takea= 1/ 2 ,b= 1/4. Find the other eigenvalueofPand corresponding eigenvectorv.

(3) Consider the vectorx 0 =

##### ]

. Finds,tsuch thatx 0 =su+tv.

(4) Show that equation (0.1) now reads as

``````xk=sku+tkv, k= 0, 1 , 2 , 3 ,....
``````

Show that the sequencex 0 ,x 1 ,x 2 ,…,xk,…converges to the vector

##### ]
``````ask.
``````

Note: In this specific instance, we have proved Theorem 18 on page 261. Of course the general case is much more difficult!

(^1) Note here that we are implicitly saying that ifPis a stochastic matrix andxis a probability vector thenPxis also a probability vector. Youre encouraged to convince yourself!