math

standard matrix for the linear transformation T

A=[T(e1)...T(en)=

If the vector equation has the same solution as the augmented matrix what do we know?

That b is in Span{v1...vp}

One-to-One

one to one: Let R^n to R^m be a linear transformation. Then T is one-to-one if and only if the equation T(x)=0 has only the trivial solution

linear dependence relation

when the weights are not all zero, you have a linear dependence relationship among the set of vectors {v1...vp} To find a linear dependence relation among a set of v1,v2,v3 completely row reduce the matrix and choose any non zero value for the free variable, then substitute the values in for the equation of v1+v2+v3=0

Suppose Ax=b has a solution. Explain why the solution is unique precisely when Ax=0 has only the trivial solution.

(Proof using free variables) If Ax b has a solution, then the solution is unique if and only if there are no free variables in the corresponding system of equations, that is, if and only if every column of A is a pivot column. This happens if and only if the equation Ax 0 has only the trivial solution.

Criteria of Thm 4 relating if A is m x n matrix span{v1....vp}=Rm

1. For each b in Rm, Ax=b has a solution 2. Each b in Rm is a linear combo of the columns of A 3. Columns of A span Rm 4. A has a pivot in each row

Could a set of three vectors in R4 span all of R4? Explain. What about n vectors in Rm when n is less than m?

A set of 3 vectors cannot span R^4 because the matrix A whose columns are these three vectors has four rows. To have a pivot in each row, A would have to have at least four columns (one for each pivot), which is not the case. Since A does not have a pivot in every row, its columns do not span R^4, by theorem 4. A set of n vectors in R^m cannot span Rm when n is less than m

What does it mean for a set of vectors {v1,...vp} in Rm to span Rm

According to Thm. 4, it means that every vector in Rm spans Rm if every vector in Rm is a linear combination of v1....vp. That is, if span{v1,...vp}=Rm

How many pivot columns must a 7 x 5 matrix have if its columns are linearly independent? Why?

All five columns of the 7x5 matrix A must be pivot columns. Otherwise the equation Ax=0 would have a free variable, in which case the columns of A would be linearly dependent.

What would you have to know about the pivot columns in an augmented matrix in order to know that the linear system is consistent and has a unique solution?

Every column in the augmented matrix except the rightmost column is a pivot column, and the rightmost column is not a pivot column.

What does it mean for a set of vectors to be Linearly Independent?

An indexed set of vectors {v1,...vp} in Rn is said to be linearly independent if the vector equation x1v1....+xpvp=0 has ONLY the trivial solution

Why is f called a linear function?

F is called a linear function because the graph of f is a line

(T/F) The weights c1,...,cp in a linear combination c1v1+...+cpvp cannot all be zero.

False

Suppose vectors v1,...,vp span Rn, and let T:Rn to Rn be a linear transformation. Suppose T(vi)=0 for i=1,....p. Show that T is the zero transformation. Show that if x is any given vector in Rn, then T(x)=0.

Given any x in R^n, there are constants c1,...,cp such that x=c1v1 + cpvp because v1,... vp span Rn. Then, from property (5) of a linear transformation, T(x)=c1T(v1)+...+cpT(vp)=c10+...

Let A be an m x n matrix, and let u and v be vectors in Rn with the property that Au=0 and Av=0. Explain why A(u+v) must be the zero vector. Then explain why A(cu+dv)=0 for each pair of scalars c and d.

Suppose Au=0 and Av=0. Then, since A(u+v)=Au+Av byTheorem5(a) in Section1.4, A(u + v) = Au + Av = 0 + 0 = 0. Now, let c and d be scalars. Using both parts of Theorem 5, A(cu + dv) = A(cu) + A(dv) = cAu + dAv = c0 + d0 = 0.

Let A be a 5 x 3 matrix, let y be a vector in R3, and let z be a vector in R5. Suppose Ay=z. What fact allows you to conclude that the system Ax=4z is consistent?

Suppose that y and z satisfy Ay=z. Then 4z=4Ay. By Theorem 5(b), 4Ay=A(4y). So 4z=A(4y), which shows that 4y is a solution of Ax=4z. Thus, the equation Ax=4z is consistent

How would you determine whether or not a given set of vectors {v1,...vp} in Rm spans Rm?

If A is an m x n matrix, Span{v1....vp}=Rm

If a linear transformation T:Rn to Rm maps onto Rm, can you give a relation between m and n? If T is one-to-one, what can you say about m and n?

If T: Rn to Rm maps Rn onto Rm then its standard matrix A has a pivot in each row, by Theorem 12 and by Theorem 4 in Section 1.4. So A must have at least as many columns as rows. So that the columns of A span R^m and thus map onto R^m. When T is one-to-one, the standard matrix A for T must have only the trivial solution for the equation T(x)=0. A must have a pivot in each column, by Theorem 12, so m > n.

How would you determine whether or not a given vector b is in Span{v1...,vp}?

If a solution exists to the linear system corresponding to the matrix [a1 a2 ... an b] (Thus indicating that the system is consistent)

Uniqueness

If an equation has a unique solution, then the system of equations has no free variables. (A pivot position is needed in each row)

How many pivot columns must a 5 x 7 matrix have if its columns span R5? Why?

If the columns of a 5x7 matrix span R5, then A has a pivot in each row, by Theorem 4. Since each pivot position is in a different column, A has five pivot columns.

Suppose A is a 3 x 3 matrix and b is a vector in R3 with the property that Ax=b has a unique solution. Explain why the columns of A must span R3

If the equation Ax = b has a unique solution, then the associated system of equations does not have any free variables. If every variable is a basic variable, then each column of A is a pivot column. So the reduced echelon form of a must be [100,010,001]. Now it is clear that A has a pivot position in each row

Let T: Rn to Rn be a linear transformation, and let {v1,v2,v3} be a linearly dependent set in Rn. Explain why the set {T(v1), T(v2), T(v3)} is linearly dependent.

If {v1,v2,v3} is linearly dependent, then we know that there exists scalars c1, c2, c3, not all zero such that c1v1+c2v2+c3v3=0 Then T(c1v1+c2v2+c3v3)=T(0)=0 Since T is linear, C1T(v1)+c2T(v2)+c3T(v3)=0 Since not all the weights are zero, {T(v1),T(v2),T(v3)} is a linearly dependent set.

Let S: Rp to Rn and T: Rn to Rm

T(S(cu + dv)) = T(c⋅S(u) + d⋅S(v)) because S is linear = c⋅T(S(u)) + d⋅T(S(v)) because T is linear This calculation shows that the mapping x →T(S(x)) is linear.

(T/F) The vector u results when a vector u-v is added to the vector v

True

(T/F) When u and v are nonzero vectors, Span {u,v} contains the line through u and the origin

True

Can each vector in R4 be written as a linear combination of the columns of the matrix A above? Do the columns of A span R4?

No. statement d in theorem 4 is false(( A has a pivot position in every row)). Therefore, all four statements in theorem 4 are false. Thus, not all vectors in R4 can be written as a linear combination of the columns of A. The columns of A do not span R4

Given A = [ 2 x 3 ], find one nontrivial solution of Ax=0 by inspection. Hint: think of the equation Ax=0 written as a vector equation

Notice that the second column is 3 times the first. So suitable values for x1 and x2 would be 3 and -1 respectively. Thus x=[3,-1] satisfies Ax=0

et A = a 3 x 3 matrix and b = some set of three numbers. W= Span{a1,a2,a3} a. is b in {a1,a2,a3}? How many vectors are in {a1,a2,a3}? b. is b in W? how many vectors are in W? c. show that a1 is in W. Hint: row operations are unnecessary.

a. There are only three vectors in the set {a1,a2,a3} and b is not one of them. b. there are infinitely many vectors in W=Span{a1,a2,a3} To determine if b is in W, augment and reduce the matrix... if it's consistent we know that b is in W. c.A1=1a1+0a2+0a3

Let A be a 6 x 5 matrix. What must a and b be in order to define T:Ra to Rb by T(x)=Ax?

a=5, the domain of T is R5, because a 6x5 matrix has 5 columns and for Ax to be defined, x must be in R5. b=6, the codomain of T is R6, because Ax is a linear combination of the columns of A, and each column of A is in R6.

What does it mean for a set of vectors to be Linearly Dependent?

The set of vectors {v1....,vp} is said to be linearly dependent if there exist weights c1....cp, not all zero such that c1v1+c2v2+....+cpvp=0

(T/F) Any list of five real numbers is a vector in R5.

True

(T/F) Asking whether the linear system corresponding to an augmented matrix [a1,a2,a3,b] has a solution amounts to asking whether b is in Span {a1,a2,a3}.

True

How do you determine if a set go vectors in Rn is linearly independent?

Use row operations on the augmented matrix to determine if there are any free variables. Each nonzero value in the free variable determines a nontrivial solution

Let A be a 3 x 4 matrix and b is a vector in R3, and let w=y1+y2. Suppose y1=Ax1 and y2=Ax2 for some vectors x1 and x2 in R4. What fact allows you to conclude that the system Ax=w is consistent?

Using theorem 5(a) with x1 and x2 in place of u and v, respectively, Ax1+Ax2=A(x1+x2). So the vector x=x1+x2 is a solution of w=Ax.

Define f: R to R for f(x)=mx+b a. Show that f is a linear transformation when b=0

When b = 0, f (x) = mx. In this case, for all x,y in R and all scalars c and d, f (cx + dy) = m(cx + dy) = mcx + mdy = c(mx) + d(my) = c·f (x) + d·f (y) This shows that f is linear.

Find a property of a linear transformation that is violated when b DNE 0

When f(x)=mx+b, with b nonzero , f(0)=m(0)=b=bDNE0.This shows that f is not linear, because every linear transformation maps the zero vector in its domain into the zero vector in the codomain. (In this case, both zero vectors are just the number 0.) Another argument, for instance, would be to calculate f (2x) = m(2x) + b and 2f (x) = 2mx + 2b. If b is nonzero, then f (2x) is not equal to 2f (x) and so f is not a linear transformation.

Justify linear independent definition

When the definition says you will have only one solution if you in turn have free variables you know that there could be many potential nontrivial solutions.

Horizontal shear matrix

[1k over 01]

Set of one vector linear independent

a set of one vector is linearly independent if and only if the vector v is not the zero vector this is because x1(0)=0 has many non trivial solutions. so thus, it has only the trivial solution for x1v when v DNE 0

Linear Independent

linearly independent and thus one to one you must have as many columns as you do rows. For example, 2 x 4 matrix will not work because it has more columns than rows and is thus linearly dependent. However, it does have a pivot in each row so it does map onto R^2(2 rows)

Mapping onto

maps onto - By theorem 12, if A does have a pivot in each row, the columns of A span R^m. Therefore, it maps onto R^m

set of two vectors

set of two vectors is linearly independent when vectors v1 and v2 are not multiples of each other

What does it mean for a vector b to be in Span {v1,...vp}

x1v1 + x2v2 +...+xpvp=b if the vector equation has a solution, then we know that vector b is in span {v1...vp}