I'll use Maple syntax for mathematical notation on this page.
All section and page numbers refer to sections from Lay, Updated 2nd Edition.
Due Monday 12/3 at 8am
Section 6.5: Least Squares Problems
Reading questions:
RZ say:
The point of this section is to find a way to best approximate Ax=b when it has no solution. We are able to do this using the general least-spuares problem. This is a way of making the space between b and Ax the smallest.
BH and AJ say:
Every system Ax=b has at least one least-squares solution.
The solution is not necessarily unique, because Theorem 14 says the solution is unique only if the columns of A are linearly independent.
Due Friday 11/30 at 8am
Section 6.3: Orthogonal Projections
Reading questions:
I say:
We know how to project onto coordinate axes already, we don't need anything fancy for this.Because y=1e_{1}+2e_{2}+3e_{3}, we know that the orthogonal projection of y onto the xz-plane is the vector y-hat=(1,0,3).
Note that the projection is a vector and not a scalar.
MT says:
There is no point in W that is closer to y than the orthogonal projection of it. This is justified by the Best Approximation Theorem which says that ||y-^y|| < ||y-v|| for all v in W distinct from ^y.
Due Wednesday 11/14 at 8am
Section 5.6: Discrete Dynamical Systems
Reading questions:
MB says:
(The characteristic equation for A is)
h^2 - 8h + 15 = 0.
So h = 5,3.The eigenvalues are greater than one so the origin is a repellor.
Due Monday 11/12 at 8am
Section 5.3: Diagonalization
Reading questions:
MT says:
Diagonalization allows us to easily find A^k, no matter what the value of k is.
JM says:
Yes, A is diagonalizable because an n x n matrix with n distinct eigenvalues is diagonalizable.æ
Due Wednesday 11/7 at 8am
Section 5.2: The Characteristic Equation
Reading questions:
I say:
The characteristic equation is det(A-(lambda)I)=0. A-(lambda)I has 7-lambda and 1-lambda on the diagonal, and is otherwise unchanged.Thus det(A-(lambda)I)=(7-lambda)(1-lambda)+8, and therefore, the characteristic equation is
(7-lambda)(1-lambda)+8=0.
JL, RZ, and JA combine to say:
They are related because lambda is an eigenvalue of an n X n matrix A if and only if lambda satisfies the characteristic equation det(A- lambda *I) =0.Characteristic equations make it so that there is only one variable in the (A-kI)x=0 equation.
The multiplicity of an eigenvalue lambda is its multiplicity as a root of the characteristic equation.
Due Monday 11/5 at 8am
Section 5.1: Eigenvectors and Eigenvalues
Reading questions:
RZ says:
The dimension of Nul(A) is 0 becasue of the Invertible Matrix Theorem.
I add:
A is a square matrix and 0 is not an eigenvalue. That means that A is invertible by the final statement of the invertible matrix theorem. Hence A has a pivot in every column. Since dim(Nul(A))=the number of free variables, dim(Nul(A))=0.
MB says:
The dimension of the eigenspace of A is 3.
I add:
For each eigenvalue, there exists an eigenvector. Since the eigenvalues are distinct, Theorem 2 tells us that these three eigenvectors are linearly independent. That meanst that dim(eigenspace(A)) is at least 3. But since the eigenvectors are in R^3, the dimension of the eigenspace must be no more than 3. Hence the dimension of the eigenspace is exactly 3. -->
Due Friday 11/2 at 8am
Section 4.6: Rank
Introduction to Chapter 5
Section 5.1: Eigenvectors and Eigenvalues
Reading Questions:
RZ says:
The dimension of NulA when A is a 4x7 matrix with 3 pivots is 4 because the dimension of NulA is the number of free variables in a matrix. So this means if A has 3 pivots then there must be 4 columns without pivots making them free variables.
GC says:
The dimension of Nul A is four. By Theorem 13, the number of columns in a matrix equals the number of pivots plus the dimension of Nul A. 3+4=7
MT says:
A= [(7, 2)], [(-4,1)] and u= (1, -2).We multiply Au and get the vector (3, -6), which can be reduced to 3(1, -2).
(Therefore, it is true that Au=3u,), and so (1, -2) is an eigenvector and 3 is an eigenvalue.
Due Wednesday 10/31 at 8am
Section 4.5
Reading questions:
MT says:
Seeing as how the basis for R^{n} has n vectors, the dimension of R^{n} must be n, and therefore the dimension of R^{3} is 3. This makes sense geometrically because R^{3} is the 3rd dimension, go figure!JL adds:
Of course this makes geometric sense because there are three different direction sets.
RZ says:
This question does not make sence to me but I might have missed something in the reading.If you meant to have the set {v1, v2, ..., v12} then it must span R^{12}, because according to Theorem 12 (The Basis Theorem), the set is automatically a basis for R^{12}, which means that the set spans R^{12}.
Due Friday 10/26 at 8am
Section 4.3: Linearly Independent Sets; Bases
Reading questions:
I say:
Since v_{3}=v_{1}+v_{2}, v_{3} is a linear combination of v_{1} and v_{2}, and so is redundant.Thus we can span the same space with just v_{1} and v_{2}. Now the question is, can we eliminate another one?
When you're only dealing with two vectors, the only way they can be linearly dependent is if one is a multiple of the other. Since that is not the case here, a basis for H is {v_{1}, v_{2}}.
MT says:
According to theorem 6, the pivot columns of a matrix A form a basis for Col A. A is a 4x5 matrix with 3 pivot positions so three vectors form a basis for ColA.
Due Wednesday 10/24 at 8am
Section 4.2: Null Spaces, Column Spaces, and Linear Transformations
Reading questions:
False. In this case, Nul A is a subspace of R5, and Col A is a subspace of R3.
2x_{1}=0, 4x_{2}=0
=> Nul A is {(0,0)}
Due Monday 10/22 at 8am
Introduction to Chapter 4
Section 4.1: Vector Spaces and Subspaces
Reading questions:
SB says:
The subset of R^{3} that consists of the all the scalar multiples of the vector (5, 6, -3) is a subspace of R^{3} because it passes through the zero vector and it is closed under both vector multiplication and addition.
RZ says:
An example of a subset of R2 that is not a a subspace of R2 is a line that does not go through (0,0).
Due Wednesday 10/17 at 8am
Introduction to Chapter 3
Section 3.1: Introduction to Determinants
Section 3.2: Properties of Determinants
Reading questions:
MT says:
A determinant can give us formulas for the volume of different polyhedra. They also give us information about matrices, information that is very useful in linear algebra.MB adds:
If det(A) does not equal zero and is a square matrix, then it is invertible.
SB says:
A cofactor expansion down the first column will make
detA=2*matrix( [ [ 4, -4], [0, 1] ]) = [=2*(4*1-(-4)*0) ] = 2*4=8
Due Monday 10/15 at 8am
Section 2.8: Applications to Computer Graphics
Reading questions:
GC says
Using homogeneous coordinates is advantageous because it allows the mathematician to use a matrix to represent a translation.SB adds:
This is especially useful in biology where biologists will use computer graphics to similate a protein. The visualizing of potential chemical reactions and changes in view can be done by using homogeneous coordinates.
Due Friday 10/12 at 8am
Section 1.9: Linear Models in Business, Science, and Engineering
Section 4.9: Applications to Markov Chains
Reading questions:
JM says
A steady state vector for a stochastic matrix P is a probability vector such that Pq=q.
SB says:
Each regular Stochastic Matrix has a steady state vector.
Due Wednesday 10/10 at 8am
Section 1.9: Linear Models in Business, Science, and Engineering
Section 4.9: Applications to Markov Chains
Reading questions:
MB says:
The point of studying Markov chains is to take current trends of a specific situation and use this to predict what will happen in the future.MT says:
Markov chains are a way for the Linear Algebra we have been doing to be applied to everyday situations. The way that they are set up enables us to predict certain outcomes with specific results. If we know what x0 is, then the nature of the chain allows us to find xk.JL says:
The point in studying Markov chains is that they are used in many fields. The examples from the book are biology, business, chemistry, engineering, and physics. They are good at representing the data of an experiment or measurement.
Due Wednesday 10/3 at 8am
Section 2.2 : The Inverse of a Matrix
Section 2.3: Characterization of Invertible Matrices
Reading Questions:
SB says:
The inverse of matrix A is
Matrix([[1/6, -2, -9], [0, 1/2, 2], [0, 0, 1]]); or æææThis was attained by (augmenting) matrix A with the Identity matrix as shown in example 7 of Section 2.2, and row reducing the far left side, until the identity matrix could be seen in the place of where matrix A was. æThat is, reducing it to reduced echelon form on the far left.
GC says:
The span of {u, v, w} is R3, according to Theorem 8.I add:
This question is so simply asked, and yet it is surprisingly layered. Since A is invertible, we know it is square. Since there are 3 columns, there must be 3 rows, and hence u, v, and w are all in R^{3}. Also because A is invertible,we know that there's a pivot in every row. Thus we know that Span(u,v,w)=R^{3}.
MT says:
The Invertible Matrix Theorem points out that either all of its statements are true or all of it's statements are false. So if A is invertible, then by the IMT, the linear transformation T maps Rn onto RnI add:
To understand why this is true, we can go through some of the logic: T is one-to-one, so the columns of the associated matrix A are linearly independent, and so A has a pivot in every column. Because T goes from R^n, A has n columns and so A has n pivots. But since T goes to R^n, A also has n rows, and so there's a pivot in every row and thus T is onto.
Due Monday 10/1 at 8am
Introduction to Chapter 2
Section 2.1: Matrix Operations
Section 2.2 : The Inverse of a Matrix
Reading questions:
GC says:
AB does not exist, because the 7 columns of A do not match the 2 rows of B, and it is impossible to multiply the two matrices together.However, BA does exist, and by definition, it has 7 columns.
JM says:
Matrix multiplication differs from multiplication of real numbers in that the cancellation laws do not hold true, for example, if AB=AC, B=C may not be true.RZ adds:
(Another) way that matrix multipllication differs from multiplication of real numbers is that it is not commutative: AB does not always equal BA.JL finishes with:
(Finally), multiplying matrices is different then multiplying regular numbers because with matrices, it may be possible to multiply x by y but not y by x, where as with real numbers, it doesn't matter which order you put them in.
BH says:
No, it does not have an infinite number of solutions. It has the unique solution x=A^{-1 }b, by theorem 5.
Due Friday 9/28 at 8am
Section 1.8: The Matrix of a Linear Transformation
Reading questions:
T is not one-to-one, because of the existence of free variables. By definition, T is one to one only if each b in Rm has either a unique solution (to T(x)=b) or none at all.
BH says:
Since there is a pivot in evey row, A spans R^m which makes T onto, because of theorem 12.
Due Wednesday 9/26 at 8am
Section 1.8: The Matrix of a Linear Transformation
Reading questions:
SB says:
matrix( [ [1], [ 0] ]) rotates onto matrix([ [cos(pi/4)] , [sin(pi/4) ] ]).
matrix( [ [0], [ 1] ]) rotates onto matrix([ [-sin(pi/4)], [cos(pi/4) ] ]).Therefore A = matrix( [ [cos(pi/4), -sin(pi/4)], [ sin(pi/4) , cos(pi/4)] ¾
RZ says:
Matrix([[2, 0], [0, 1]]) is the matrix A for the linear transformation T: R2--> R2 that expands horizontally by a factor of 2.
Due Monday 9/24 at 8am
Section 1.7: Introduction to Linear Transformations
Reading Questions:
<0L>
MB says:
T does not appear to be a linear transformation because it doesn't preserve the operations of vector addition and scalar multiplication.I add:
In order to be linear, we need that T(u+v)=T(u)+T(v) and that T(ru)=rT(u). T(u+v)=T(u_{1}+v_{1}, u_{2}+v_{2})
=( u_{2}+v_{2}-3, 4(u_{1}+v_{1})+10)
= (u_{2}-3, 4u_{1}+10)+(v_{2},4v_{1})
=T(u)+some other stuff that's not T(v)Hence T fails our first criterion, and so it's not a linear transformation.
JL says:
A must have five columns for Ax to be defined, and A must have three rows for the codomain of T to be R^3.
Due Friday 9/21 at 8am
Section 1.7: Introduction to Linear Transformations
Reading questions:
JL says (more or less)
(-4 x 1) + (10 x 5)=46,
(-4 x 7) + (10 x -3) = -58Therefore T(-4,10)=(46,-58).
SB says (more or less):
.
Replace Row 2 by -7(Row1)+ (Row2), to get
.
Since we do not have a situation that equates some non-zero number to zero (i.e. T(4, -2) is consistent), (4, -2) is within the range of T.
Due Wednesday 9/19 at 8am
The Big Picture
Reading questions:
I am not going to display these because none are more right than others -- the point of these reading questions (they'll pop up again) is first of all to encourage you to look for the big picture every now and then, and second of all to give you practice writing synopses, because the ability to pick out a few important points and connect them is both important and difficult.
Due Monday 9/17 at 8am
Linear Independence
Reading questions:
GC says:
No, because for a linear independence to exist, the equation Ax=0 must have only the trivial solution.
I say:
The only way Ax=b can have infinitely many solutions is if A has a free variable, in which case Ax=0 also has infinitely many solutions, and so we're back in the situation of #1. Hence the columns again can not be linearly independent.Another way to say the same thing:
We found in Section 1.5 that the general solution to a consistent system Ax=b has the formx=p+ tx_{h} where tx_{h} is the general solution to the homogeneous equation.So the only way the non-homogeneous equation can have an infinite number of solutions is if the homogeneous equation also has an infinite number of solutions.
JL says:
Since there would be 3 vectors and only two equations to work with, there would be a free variable which would mean that R2 can not be linearly independent.
RZ says:
A set of 3 vectors in R2 cannot be linearly independent because one of the 3 will be in the span of the other 2 making it not linearly independent.
Due Friday 9/14 at 8am
Section 1.5: Solution Sets of Linear Systems
Reading questions:
MB says:
A homogenous system of equations is one that can be written in the form Ax=0 and nonhomogenous systems are of the form Ax = b with b not 0.
JM adds:
A homogeneous system of equations ... always has at least one solution.I add:
Non-homogeneous systems may have no solutions (if the system is inconsistent), a unique solution (corresponding to the homogeneous system having only the trivial solution), or infinitely many solutions (corresponding to the homogeneous system having infinitely many solutions, i.e. when A has at least one free variable.)
JL says
Ax=b would have infinite solutions because since there is at least one ... answer and there is a free variable, there would be an infinite amount of solutions.
Due Wednesday 9/12 at 8am
Section 1.4: The Matrx Equation Ax=b
Reading questions:
MT says:
According to Theorem 4, a matrix with four columns, such as this one must have at least 4 pivot points, not three like the problem suggests.
I say:
To merely simplify it, rather than actually calculating it, we can use Theorem 5, which tells us that
Au+Av=A(u+v)
Thus the above simplifies to matrix[( (7,3),(-1,4) )]*( (-5,4)+(6,3) ), ormatrix[( (7,3),(-1,4) )] *(1,7) .
To then go ahead and calculate it, I'd get:
(7*1+3*7, -1*1+4*7)=(28,27).
Due Monday 9/10 at 8am
Section 1.3 : Vector Equations
Reading questions:
If y is a linear combination of u and v, that means that I can get y by adding scalar multiples of u and v:
y=x_{1}u+x_{2}v This is the same as the system
0x_{1}+12x_{2}=4
1x_{2}+0x_{2}=9
0x_{1}+9x_{2}=3From this we can immediately see that x_1=9 and x_2=1/3. Thus y=9u+(1/3)v.
Span{u,v} is the set of all linear combinations of u and v.
Thus Span{u,v} consists of all vectors of the form
au+bv=(a,0,0)+(0,b,0)=(a,b,0)
where a, b can be any real numbers. Thus Span{u,v} is the xy-plane in R^{3}.
Due Friday 9/7, at 8am
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Introduction to Chapter 1
Section 1.1: Systems of Linear Equations
Section 1.2: Row Reduction and Echelon Form
Reading questions:
MT says:
The matrix A is in echelon form because all nonzero rows are above the zero rows, each leading entry is to the right of the leading entry above it, and all entries stacked below a leading entry are zero.
MB says:
The values in the pivot positions of A are 5, 7, and 4.
SB says:
A is not consistent because in the last row it states that 0=4. This does not make sense, making A inconsistent.
PS adds:
Because the rightmost column of the matrix is a a pivot column the matrix is inconsistent.