Linear Final

The alternative notation for a (column) vector is:

(-4,3), using parentheses and commas/

Let A be a 3x3 matrix with two pivot positions. Does Ax=0 have a nontrivial solution? Does the equation Ax=b have at least one solution for every possible b?

-Yes. Since A has 2 pivots, there is one free variable. So, Ax=o has a nontrivial solution. -No. A has one free variable. so the free variable can equal any value such that there is at least one solution for every possible b.

If the columns of a 7x7 matrix D are linearly independent, what can you say about the solutions of Dx=b? why?

Equation Dx=b has a solution for each b in R7. According to the invertible matrix theorem, a matrix is invertible if the columns of the matrix form a linearly independent set; this would mean that the equation Dx=b has at least one solution for each b in Rn.

A subset of H of Rn is a subspace if the zero vector is in H.

False, For each u and v in H and each scalar c, the sum u+v and the vector cu must also be in H

Det(A+B)=detA+detB

False, If A = 1 0,0 1 and B= -1 0, 0 -1, then det(A+B)= 0 and det A+detB=2

If one row in an echelon form of an augmented matrix is [ 0 0 0 5 0], then the associated linear system is inconsistent.

False, The indicated row corresponds to the equation 5x4=​0, which does not by itself make the system inconsistent.

Finding a parametric description of the solution set of a linear system is the same as solving the system. T/F

False, The solution set of a linear system can only be expressed using a parametric description if the system has at least one solution.

Two matrices are row equivalent if they have the same number of rows T/F?

False, because if two matrices are row equivalent it means that there exists a sequence of row operations that transforms one matrix to the other

If A is an n×n ​matrix, then the equation Ax=b has at least one solution for each b in ℝn.

False, by the invertible matrix theorem Ax=b has at least one solution for each b in Rn only if a matrix is invertible.

in some cases, a matrix may be row reduced to more than one matrix in reduced echelon form, using different sequences of row operations. T/F?

False, each matrix is row equivalent to one and only one reduced echelon matrix.

The columns of a matrix A are linearly independent if the equation Ax=0 has the trivial solution. T/F

False, for every matrix A, Ax=0 has the trivial solution. The columns of A are independent only if the equation has no solution other than the trivial solution.

If S is a linearly dependent set, then each vector is a linear combination of the other vectors in S. T/F

False, if S is linearly dependent then there is at least one vector that is not a linear combination of the other vectors, but the others may be linear combinations of each other.

The equation Ax=b is consistent if the augmented matrix [A b] has a pivot position in every row. T/F

False, if the augmented matrix [A b] has a pivot position in every row, the equation Ax=b may or may not be consistent. One pivot position may be in the column representing b.

If A is invertible, then A is diagonalizable.

False, invertible matrices always have a maximum of n linearly independent eigenvectors, making it not diagonalizable.

If the vectors in an orthogonal set of non zero vectors are normalized then some of the new vectors may not be orthogonal

False, normalization changes all nonzero vectors to have unit length, but does not change their relative angles. therefore orthogonal vectors will always remain orthogonal after they are normalized.

The determinant of A is the product of the pivots in any echelon form U of A, multiplied by (-1)r, where r is the number of row interchanges made during row reduction from A to U.

False, reduction to an echelon form may also include scaling a row by a nonzero constant which can change the value of the determinant.

The row reduction algorithm applies only to augmented matrices for a linear system. T/F

False, the algorithm applies to any matrix, whether or not the matrix is viewed as an augmented matrix for a linear system.

The column space of a matrix A is the set of solutions of Ax=b

False, the column space of A is the set of all b for which Ax=b has a solution

If B is an echelon form of a matrix A, then the pivot columns of B form a basis for Col A

False, the columns of an echelon form a matrix are often not in the column space of the original matrix.

The equation x=p+tv decribes a line through v parallel to p. T/F

False, the effect of adding p to v is to move v in a direction parallel to the line through p and 0. So the equation x = p + tv describes a line through p parallel to v.

The equation Ax=0 gives an explicit description of its solution set T/F

False, the equation Ax=0 gives an implicit description of its solution set. Solving the equation amounts to finding an explicit description of its solution set.

The homogenous equation Ax=0 has the trivial solution if and only if the equation has at least one free variable. T/F

False, the homogenous equation Ax=0 always has the trivial solution.

The solution set of Ax=b is the set of all vectors of the form w=p+vh, where vn is an solution of the equation Ax=0.

False, the solution set could be empty. The statement is only true when the equation Ax=b is consistent for some given b, and there exists a vector p such that p is a solution.

The set span {u,v} is always visualized as a plane through the origin. T/F?

False, this statement is often true but Span {u,v} is not a plane when v is a multiple of u or when u is the zero vector.

is it possible for a 5x5 matrix to be an invertible when its columns do not span R5? Why or Why not?

It is not possible; according to the invertible matrix theorem an nxn matrix cannot be invertible when its column does not span Rn.

For a 3x5 matrix with three pivot columns, Nul A= R2?

No, because the null space of a 3x 5 matrix is a subspace of R5. Althgouh dim null A=2, it is not strictly equal to R2 because each vector in Nul A has five components. Each vector in R2 has two components. Therefore, Nul A is isomorphic to R2, but not equal.

Could a set of three vectors R4 span all R4? Explain.

No, the matrix A whose columns are the 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)

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

No. The matrix A whose columns are the n vectors has m rows. To have a pivot in each row, A would have to have at least m columns. (one for each pivot)

For any scalar c, abso(vc)=c*abso(v)

The given statement is false, since length is always positive, the value of abso(cv) will always be positive. By the dame logic, when c is negative, the values of c abso(v) is negative

u*v-v*u=0

The given statement is true. Since the inner product is cumulative, u*v=v*u. Subtracting v*u from each side of the equation gives u*v-v*u=0

For each y and each subspace W, the vector y-projwy is orthogonal to W.

The statement is true because y can be written uniquely in the form y=projwy+z where projxy is in W and z is in W and it follows that z=y-projqy

If A is an mxn matrix and if the equation Ax=b is inconsistent for some b in Rm , then A cannot have a pivot position in every row. T/F

True, If a is an mxn matrix and if the equation Ax=b is inconsistent for some b in Rm, then the equation Ax=b has no solution for some b in Rm.

The columns of any 4x5 matrix are linearly dependent. T/F

True, a 4x5 matrix has more columns than rows, and if a set contains more vectors than there are entries in each vector, then the set is linearly dependent

A homogenous equation is always consistent. T/F

True, a homogenous equation can be written in the form Ax=0, where A is an mxn matrix and 0 is the zero vector in Rm. Such a system Ax=0 always has at least one solution, namely, x=0. So, a homogenous equation is always consistent.

In order for a matrix B to be the inverse of A, both equations AB=I and BA=I must be true. T/F

True, by definition of invertible

If At is not invertible, then A is not invertible

True, by the invertible matrix theorem if At is not invertible all statements in the theorem are false, including A is invertible. Therefore, A is not invertible.

If the equation Ax=0 has a nontrivial solution, then A has fewer than n pivot positions.

True, by the invertible matrix theorem if the equation Ax=0 has a nontrivial solution, then matrix A is not invertible. Therefore, A has fewer than n pivot positions.

If the equation Ax=0 has only the trivial solution, then A is row equivalent to the nxn identity matrix. T/F

True, by the invertible matrix theorem if the equation Ax=0 has only the trivial solution, then the matrix is invertible. so, A must also be row equivalent to the nxn identity matrix

The null space of an mxn matrix is a subspace of Rn

True, for an mxn matrix A, the solutions of Ax=0 are in vectors in Rn and satisfy the properties of a vector space.

If y is a linear combination of nonzero vectors from an orthogonal set, then the weights in the linear combination can be computed without row operations on a matrix.

True, for each y in W, the weights in the linear combination y=c1u1+....+cpup can be computed by cj=y*uj?uj*uj where j=1,....,p.

a row replacement operation does not affect the determinant of a matrix

True, if a multiple of one row of a matrix A is added to another to produce a matrix B, then det B equals det A.

If the columns of A are linearly dependent, then det A=0

True, if the columns of A are linearly dependent, then A is not invertible.

If the columns of an m×n matrix A span ℝm​, then the equation Ax=b is consistent for each b in ℝm. T/F

True, if the columns of A span Rm, then the equation Ax=b has a solution for each b in Rm.

If x is orthogonal to every vector in a subspace W, then x is W

True, if x is orthogonal to every vector in W, then x is said to be orthogonal to W. The set of all vectors x that are orthogonal to W is denoted W

If x and y are linearly independent, and if {x,y,z} is linearly dependent, then z is in Span{x,y}. T/F

True, if {x,y,z} is linearly dependent, then z must be a linear combination of x and y because x and y are linearly independent. So z is in Span{x,y}.

A basic variable in a linear system is a variable that corresponds to a pivot column in the coefficient matrix. T/F

True, it is the definition of a basic variable.

If AP=PD, with D diagonal, then the nonzero columns of P must be eigenvectors of A

True, let v be a nonzero column in P and lambda be the corresponding diagonal element in D. Then AP=PD implies that Av=lambda(v), which means that v is an eigenvector of A.

If A is an invertible nxn matrix, then the equation Ax=b is consistent for each b in Rn? T/F

True, since A is invertible, A-1b exists for all b in Rn. Define x= A-1b. Then Ax=b

Each elementary matrix is invertible/F

True, since each elementary matrix corresponds to a row operation, and every row operation is reversible, every elementary matrix has an inverse matrix.

The solution set of the linear system whose augmented matrix is [a1 a 2 a3 b] is the same as the solution set of the equation x1a1+x2a2+x3a3=b. T/F?

True, the augmented matrix for x1a1+x2a2+x3a3=b is [a1 a2 a3 b]

A vector b is a linear combination of the columns of a matrix A if and only if the equation Ax=b has at least one solution. T/F

True, the equation Ax=b has the same solution set as the equation x1a1+x2a2+ .....xnan=b

The first entry in the product Ax is the sum of products. T/F

True, the first entry in Ax is the sum of the products of corresponding entries in x and the first entry in each column of A.

If the columns of A span Rn, then the columns are linearly independent.

True, the invertible matrix theorem states that if the columns of A span Rn, then matrix A is invertible. Therefore, the columns are linearly independent.

Not every linearly independent set in Rn is an orthogonal set.

True, the vectors [0 1] and [1 1] are linearly independent but not orthogonal.

Given vectors v1, .... vp in Rn, the set of all linear combinations of these vectors is a subspace of Rn

True, this set satisfies all properties of a subspace

For a 3 x 5 matrix with three pivot columns, Is Col A=R3?

Yes, because the column space of a 3x5 matrix is a subspace of R3. There is a pivot in each row, so the column of space is 3-dimensional. since any 3-dimensional subspace of R3 is R3.

If a( ) - b( ) does not equal 0 then A is inverible

ad-bc

In order for a system of linear equations to be consistent, it must have ____

at least one solution

If A and B are n x n and invertible, then A-1B-1 is the inverse of AB.TT/F

false, B-1A-1 is the inverse of AB

If A is daigonalizable, then A has n distinct eigenvalues

false, a diagonalizable matrix can have fewer than n eigenvalues and still have n linearly independent eigenvectors.

A matrix A is diagonalizable if A has n eigenvectors.

false, a diagonalizable matrix must have n linearly independent eigenvectorl

The orthogonal projection y hat of y onto subspace W can sometimes depend on the orthogonal basis for W used to compute y hat.

false, because the uniqueness property of the orthogonal decomposition y=yhat +z indicates that, no matter the basis used to find it, it will always be the same.

The equation Ax=b is referred to as a vector equation because:

it contains a matrix A. A vector equation does not contain any matrices.

If A column and B column were on the line through the origin, then Column B should be a ___ of A

multiple

If the augmented matrices of two linear systems are row equivalent, then the two systems have the same ____ soultion set

same

The set of all possible solutions is called the _____. Two linear systems are called ____ if they have the same solution set.

solution set of the linear system equivalent

If z is orthogonal to u1 and u2 and if W=span [u1, u2], then z must be in W.

true, since z is orthogonal to u1 and u2, it is orthogonal to every vector in span {u1, u2}