math <3
Suppose a 4×7 matrix A has three pivot columns. What is dim Nul A? Is Col A=ℝ3? Why or why not? Is Col A=ℝ3? Why or why not?
7 - 3 = 4 dim Nul A=4 No, because Col A is a subspace of ℝ4.
If the null space of a 5×8 matrix A is 4-dimensional, what is the dimension of the column space of A?
8 - 4 = 4 Col A = 4
If a 3×9 matrix A has rank 3, find dim Nul A, dim Row A, and rank AT.
9 - 3 = 6 dim Nul A=6 dim Row A=3 rank AT=3
Let A=a1, a2, a3 and D=d1, d2, d3 be bases for V, and let P=d1Ad2Ad3A. Which of the following equations is satisfied by P for all x in V? (i) [x]A = P[x]D (ii) [x]D = P[x]A
Equation (i) is satisfied by P for all x in V.
Let U=u1, u2 and W=w1, w2 be bases for V, and let P be a matrix whose columns are u1W and u2W. Which of the following equations is satisfied by P for all x in V? (i) [x]u = P[x]w (ii) [x]w = P[x]u
Equation (ii) is satisfied by P for all x in V.
A matrix with orthonormal columns is an orthogonal matrix.
False. A matrix with orthonormal columns is an orthogonal matrix if the matrix is also square.
. If v1, v2, v3is an orthogonal basis for W, then multiplying v3 by a scalar c gives a new orthogonal basis v1, v2, cv3.
False. If the scale factor is zero it will not give a new orthogonal basis.
If the vectors in an orthogonal set of nonzero 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.
If L is a line through 0 and if y is the orthogonal projection of y onto L, then y gives the distance from y to L.
False. The distance from y to L is given by ||y−y^||.
How can vectors be shown to be linearly dependent?
Form a matrix using the vectors as columns and determine the number of pivots in the matrix.
Let B=b1, ..., bn and C=c1, ..., cnbe bases of a vector space V. Then there is a unique n×n matrix PC ← B such that [x]C=PC ← B[x]B. The columns of PC ← B are the C-coordinate vectors of the vectors in the basis B. That is, PC ← B=b1Cb2C...bnC.
Given v in V, there exists scalars x1, ..., xn, such that v=x1b1+x2b2+...+xnbn because B is a basis for V. Apply the coordinate mapping determined by the basis C, and obtain [v]C=x1b1C+x2b2C+...+xnbnC because the coordinate mapping is a linear transformation. This equation may be written in the form [v]C=b1Cb2C...bnC x1...xn by the definition of the product of a matrix and a vector. This shows that the matrix PC ← B=b1Cb2C...bnC satisfies [v]C=PC ← B[x]B for each v in V, because the vector x1...xn is the coordinate vector of v relative to Upper B .
Which theorem could help prove one of these criteria from another?
If S=u1, ..., up is an orthogonal set of nonzero vectors in ℝn, then S is linearly independent and hence is a basis for the subspace spanned by S.
What is satire's purpose?
Its a technique to influence individuals of the corruption and foolishness in society through exaggerated humor .
A is a 3×3 matrix with two eigenvalues. Each eigenspace is one-dimensional. Is A diagonalizable? Why?
No. The sum of the dimensions of the eigenspaces equals 2 and the matrix has 3 columns. The sum of the dimensions of the eigenspace and the number of columns must be equal.
Without calculation, find one eigenvalue and two linearly independent eigenvectors of A=222222222. Justify your answer.
One eigenvalue of A is λ=0 because the columns of A are linearly dependent. Two linearly independent eigenvectors of A are ..... because the entries of each vector sum to 0.
How do these calculations show that u1, u2, u3 is an orthogonal basis for ℝ3?
Since each inner product is 00, the vectors form an orthogonal set. From the theorem above, this proves that the vectors are also a basis. Express x as a linear combination of the u's.
What is the dimension of the vector space ℙ3?
Since the dimension of ℙ3 is equal to the number of elements in the linearly independent set formed by the given polynomials, the given set of polynomials forms a basis for ℙ3.
When are polynomials linearly dependent?
Since the matrix has a pivot in each column, its columns (and thus the given polynomials) are linearly independent.
If A is a 12×9 matrix, what is the largest possible dimension of the row space of A? If A is a 9×12 matrix, what is the largest possible dimension of the row space of A? Explain.
The dimension of the row space of A is equal to the number of pivot positions in A. Since there are only 9 columns in a 12×9 matrix, and there are only 9 rows in a 9×12 matrix, there can be at most 9 pivot positions for either matrix. Therefore, the largest possible dimension of the row space of either matrix is 9.
If the distance from u to v equals the distance from u to −v, then u and v are orthogonal.
The given statement is false. By the theorem of orthogonal complements, it is known that vectors in Col A are orthogonal to vectors in Nul AT. Using the definition of orthogonal complements, vectors in Col A are orthogonal to vectors in Nul A if and only if the rows and columns of A are the same, which is not necessarily true.
For any scalar c, u•(cv)=c(u•v).
The given statement is true because this is a valid property of the inner product.
v•v = ||v||^2
The given statement is true. By the definition of the length of a vector v, ||v||=sqrt(v•v).
If vectors v1,...,vp span a subspace W and if x is orthogonal to each vj for j=1,...,p, then x is in W⊥.
The given statement is true. If x is orthogonal to each vj, then x is also orthogonal to any linear combination of those vj. Since any vector in W can be described as a linear combination of vj, x is orthogonal to all vectors in W.
Row operations preserve the linear dependence relations among the rows of A. Is this statement true or false?
The statement is false. Row operations may change the linear dependence relations among the rows of A.
The columns of the change-of-coordinates matrix PC ← B are B-coordinate vectors of the vectors in C.
The statement is false because the columns of the matrix PC ← B are the C-coordinate vectors of the vectors in B.
If B is any echelon form of A, then the pivot columns of B form a basis for the column space of A. Is this statement true or false?
The statement is false. The columns of an echelon form B of A are often not in the column space of A.
If V=ℝn and C is the standard basis for V, then PC ← B is the same as the change-of-coordinates matrix PB that satisfies x=PB[x]B for all x in V.
The statement is true because if C is the standard basis for ℝn, then biC=bi for 1≤i≤n, and PB=b1...bn.
The row space of AT is the same as the column space of A. Is this statement true or false?
The statement is true because the rows of AT are the columns of ATT=A.
If A and B are row equivalent, then their row spaces are the same. Is this statement true or false?
The statement is true. If B is obtained from A by row operations, the rows of B are linear combinations of the rows of A and vice-versa.
The dimension of the null space of A is the number of columns of A that are not pivot columns. Is this statement true or false?
The statement is true. The dimension of Nul A equals the number of free variables in the equation Ax=0.
Which of the following criteria are necessary for a set of vectors to be an orthogonal basis for a subspace W of ℝn? Select all that apply.
The vectors must span W. The vectors must form an orthogonal set.
The Gram-Schmidt process produces from a linearly independent set x1, ... , xp an orthogonal set v1, ... , vp with the property that for each k, the vectors v1, ... , vk span the same subspace as that spanned by x1, ... , xk.
True. For Wk=Span{x1, ... , xk}, v1=x1, and some v1, ... , vk where {v1, ... , vk} is an orthogonal basis for Wk, vk+1=xk+1−projWkxk+1 is orthogonal to Wkand is in Wk+1. Also, vk+1≠0. Hence, {v1, ... , vk+1} is an orthogonal basis for Wk+1.
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•ujuj•uj, where j=1, . . . , p.
Not every linearly independent set in ℝn is an orthogonal set.
True. For example, the vectors 01 and 11 are linearly independent but not orthogonal.
If A=QR, where Q has orthonormal columns, then R=QTA.
True. Since Q has orthonormal columns then QTQ=I. So QTA=QT(QR)=IR=R.
Is λ=2 an eigenvalue of A=−36−58? Why or why not?
Yes, λ is an eigenvalue of A because Ax=λx has a nontrivial solution.
