Logo

AskSia

Plus

Let HH and KK be subspaces of a vector space VV. The intersection of HH and...
Sep 24, 2024
Solution by Steps
step 1
To show that H∩K H \cap K is a subspace of V V , we need to verify three properties: the zero vector is in H∩K H \cap K , it is closed under addition, and it is closed under scalar multiplication
step 2
First, we check if the zero vector 0 \mathbf{0} is in H∩K H \cap K . Since both H H and K K are subspaces, they both contain the zero vector. Thus, 0∈H \mathbf{0} \in H and 0∈K \mathbf{0} \in K implies 0∈H∩K \mathbf{0} \in H \cap K
step 3
Next, we check closure under addition. Let u,v∈H∩K \mathbf{u}, \mathbf{v} \in H \cap K . This means u∈H \mathbf{u} \in H and u∈K \mathbf{u} \in K , and v∈H \mathbf{v} \in H and v∈K \mathbf{v} \in K . Since H H and K K are subspaces, u+v∈H \mathbf{u} + \mathbf{v} \in H and u+v∈K \mathbf{u} + \mathbf{v} \in K . Therefore, u+v∈H∩K \mathbf{u} + \mathbf{v} \in H \cap K
step 4
Finally, we check closure under scalar multiplication. Let u∈H∩K \mathbf{u} \in H \cap K and c c be a scalar. Since u∈H \mathbf{u} \in H and u∈K \mathbf{u} \in K , and both H H and K K are subspaces, we have cu∈H c\mathbf{u} \in H and cu∈K c\mathbf{u} \in K . Thus, cu∈H∩K c\mathbf{u} \in H \cap K
Answer
H∩K H \cap K is a subspace of V V
Key Concept
The intersection of two subspaces is also a subspace.
Explanation
The intersection H∩K H \cap K satisfies all the properties required for a subspace, confirming it is indeed a subspace of V V .
(a) Let P be the plane in R3 with equation x + y − 2z = 4. Find two vectors in P and check that their sum is not in P. (b) Let P0 be the plane through (0, 0, 0) and parallel to P. Write the equation for P0. Find two vectors in P0 and check that their sum is in P0.
Solution by Steps
step 1
For part (a), the equation of the plane P P is given by x+y−2z=4 x + y - 2z = 4 . To find two vectors in P P , we can choose values for x x and y y and solve for z z . Let’s choose x=0 x = 0 and y=4 y = 4 : substituting these values gives 0+4−2z=4 0 + 4 - 2z = 4 which simplifies to z=0 z = 0 . Thus, one vector is v1=(0,4,0) \mathbf{v_1} = (0, 4, 0) . Now, let’s choose x=2 x = 2 and y=0 y = 0 : substituting gives 2+0−2z=4 2 + 0 - 2z = 4 which simplifies to z=−1 z = -1 . Thus, another vector is v2=(2,0,−1) \mathbf{v_2} = (2, 0, -1)
step 2
Now, we check that the sum of these two vectors is not in P P . The sum is v1+v2=(0,4,0)+(2,0,−1)=(2,4,−1) \mathbf{v_1} + \mathbf{v_2} = (0, 4, 0) + (2, 0, -1) = (2, 4, -1) . Substituting into the plane equation gives 2+4−2(−1)=2+4+2=8 2 + 4 - 2(-1) = 2 + 4 + 2 = 8 , which does not equal 4. Thus, the sum is not in P P
step 3
For part (b), the plane P0 P_0 is parallel to P P and passes through the origin, so its equation is the same as P P but with the constant term set to 0: x+y−2z=0 x + y - 2z = 0
step 4
To find two vectors in P0 P_0 , we can again choose values for x x and y y . Let’s choose x=0 x = 0 and y=0 y = 0 : substituting gives 0+0−2z=0 0 + 0 - 2z = 0 which simplifies to z=0 z = 0 . Thus, one vector is u1=(0,0,0) \mathbf{u_1} = (0, 0, 0) . Now, let’s choose x=2 x = 2 and y=4 y = 4 : substituting gives 2+4−2z=0 2 + 4 - 2z = 0 which simplifies to z=3 z = 3 . Thus, another vector is u2=(2,4,3) \mathbf{u_2} = (2, 4, 3)
step 5
Now, we check that the sum of these two vectors is in P0 P_0 . The sum is u1+u2=(0,0,0)+(2,4,3)=(2,4,3) \mathbf{u_1} + \mathbf{u_2} = (0, 0, 0) + (2, 4, 3) = (2, 4, 3) . Substituting into the plane equation gives 2+4−2(3)=2+4−6=0 2 + 4 - 2(3) = 2 + 4 - 6 = 0 , which satisfies the equation of P0 P_0 . Thus, the sum is in P0 P_0
Answer
Two vectors in P P are (0,4,0) (0, 4, 0) and (2,0,−1) (2, 0, -1) , and their sum (2,4,−1) (2, 4, -1) is not in P P . Two vectors in P0 P_0 are (0,0,0) (0, 0, 0) and (2,4,3) (2, 4, 3) , and their sum (2,4,3) (2, 4, 3) is in P0 P_0 .
Key Concept
Understanding the properties of planes in three-dimensional space and how to determine if vectors lie within them.
Explanation
The solution demonstrates how to find vectors in a plane and check their sums against the plane's equation, illustrating the concept of parallel planes and vector addition.
Solution by Steps
For Question 6:
step 1
To find a nonzero vector in N(D) \mathbf{N}(D) , we need to solve the equation Dx=0 D \mathbf{x} = \mathbf{0} . The matrix D D is given by:
\( D = \begin{bmatrix} 2 & -6 \\ -1 & 3 \\ -4 & 12 \\ 3 & -9 \end{bmatrix} \)
step 2
We can row-reduce the matrix D D to find its null space. Performing row operations, we get:
\( \begin{bmatrix} 1 & -3 \\ 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{bmatrix} \)
step 3
The null space is spanned by the vector [31] \begin{bmatrix} 3 \\ 1 \end{bmatrix} . Thus, a nonzero vector in N(D) \mathbf{N}(D) is:
\( \begin{bmatrix} 3 \\ 1 \end{bmatrix} \)
step 4
To find a nonzero vector in C(D) \mathbf{C}(D) , we can take the columns of D D . The first column [2−1−43] \begin{bmatrix} 2 \\ -1 \\ -4 \\ 3 \end{bmatrix} is a nonzero vector in C(D) \mathbf{C}(D)
Answer
Nonzero vector in N(D) \mathbf{N}(D) : [31] \begin{bmatrix} 3 \\ 1 \end{bmatrix} ; Nonzero vector in C(D) \mathbf{C}(D) : [2−1−43] \begin{bmatrix} 2 \\ -1 \\ -4 \\ 3 \end{bmatrix}
Key Concept
The null space N(D) \mathbf{N}(D) consists of all vectors that are mapped to the zero vector by the matrix D D , while the column space C(D) \mathbf{C}(D) consists of all linear combinations of the columns of D D .
Explanation
The solution involves finding vectors that satisfy the conditions for the null space and column space of the matrix D D . The row reduction helps identify the null space, while the columns of the original matrix provide the column space.
--- For Question 7:
step 1
The equation of the plane is given by x+2y+z=0 x + 2y + z = 0 . To find a basis for this plane in R3 \mathbb{R}^3 , we can express z z in terms of x x and y y :
\( z = -x - 2y \)
step 2
We can choose two parameters, say x=1,y=0 x = 1, y = 0 and x=0,y=1 x = 0, y = 1 , to find two vectors in the plane:
For \( x = 1, y = 0 \): \( \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \) and for \( x = 0, y = 1 \): \( \begin{bmatrix} 0 \\ 1 \\ -2 \end{bmatrix} \)
step 3
Thus, the basis for the set of vectors in the plane x+2y+z=0 x + 2y + z = 0 is:
\( \left\{ \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ -2 \end{bmatrix} \right\} \)
Answer
Basis for the plane x+2y+z=0 x + 2y + z = 0 : {[10−1],[01−2]} \left\{ \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ -2 \end{bmatrix} \right\}
Key Concept
A basis for a plane in R3 \mathbb{R}^3 consists of two linearly independent vectors that lie in that plane.
Explanation
The vectors found are linearly independent and satisfy the plane equation, thus forming a basis for the plane defined by x+2y+z=0 x + 2y + z = 0 .
--- For Question 8:
step 1
We are given three vectors v1=[4−37],v2=[19−2],v3=[7116] \mathbf{v}_{1} = \begin{bmatrix} 4 \\ -3 \\ 7 \end{bmatrix}, \mathbf{v}_{2} = \begin{bmatrix} 1 \\ 9 \\ -2 \end{bmatrix}, \mathbf{v}_{3} = \begin{bmatrix} 7 \\ 11 \\ 6 \end{bmatrix} and the relation 4v1+5v2−3v3=0 4 \mathbf{v}_{1} + 5 \mathbf{v}_{2} - 3 \mathbf{v}_{3} = \mathbf{0}
step 2
To find a basis for H=Span⁡{v1,v2,v3} H = \operatorname{Span}\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\} , we need to determine if these vectors are linearly independent. We can form a matrix with these vectors as columns and row-reduce it:
\( \begin{bmatrix} 4 & 1 & 7 \\ -3 & 9 & 11 \\ 7 & -2 & 6 \end{bmatrix} \)
step 3
After row-reducing, we find that the rank of the matrix is 2, indicating that two of the vectors are linearly independent. We can take v1 \mathbf{v}_{1} and v2 \mathbf{v}_{2} as a basis for H H
step 4
Thus, a basis for H H is:
\( \left\{ \begin{bmatrix} 4 \\ -3 \\ 7 \end{bmatrix}, \begin{bmatrix} 1 \\ 9 \\ -2 \end{bmatrix} \right\} \)
Answer
Basis for H H : {[4−37],[19−2]} \left\{ \begin{bmatrix} 4 \\ -3 \\ 7 \end{bmatrix}, \begin{bmatrix} 1 \\ 9 \\ -2 \end{bmatrix} \right\}
Key Concept
A basis for a span of vectors is a set of linearly independent vectors that can be combined to express any vector in that span.
Explanation
The vectors chosen are linearly independent and span the same space as the original set of vectors, thus forming a valid basis for H H .
--- For Question 9:
step 1
We have the polynomials p1(t)=1+t,p2(t)=1−t,p3(t)=2 \mathbf{p}_{1}(t) = 1 + t, \mathbf{p}_{2}(t) = 1 - t, \mathbf{p}_{3}(t) = 2 . To find a linear dependence relation, we can express one polynomial as a linear combination of the others
step 2
By inspection, we can see that:
\( \mathbf{p}_{1}(t) + \mathbf{p}_{2}(t) - \mathbf{p}_{3}(t) = (1 + t) + (1 - t) - 2 = 0 \)
step 3
This shows that p1,p2,p3 \mathbf{p}_{1}, \mathbf{p}_{2}, \mathbf{p}_{3} are linearly dependent. To find a basis for Span⁡{p1,p2,p3} \operatorname{Span}\{\mathbf{p}_{1}, \mathbf{p}_{2}, \mathbf{p}_{3}\} , we can exclude one of the polynomials
step 4
A basis can be formed by taking p1 \mathbf{p}_{1} and p2 \mathbf{p}_{2} :
\( \left\{ \mathbf{p}_{1}(t), \mathbf{p}_{2}(t) \right\} \)
Answer
Basis for Span⁡{p1,p2,p3} \operatorname{Span}\{\mathbf{p}_{1}, \mathbf{p}_{2}, \mathbf{p}_{3}\} : {1+t,1−t} \left\{ 1 + t, 1 - t \right\}
Key Concept
A set of polynomials is linearly dependent if at least one polynomial can be expressed as a linear combination of the others.
Explanation
The identified linear dependence relation allows us to exclude one polynomial, resulting in a basis that spans the same space as the original set.
--- For Question 10:
step 1
We need to find the B \mathcal{B} -coordinate of the vector x=[2−6] \mathbf{x} = \begin{bmatrix} 2 \\ -6 \end{bmatrix} with respect to the basis B={[3−5],[−46]} \mathcal{B} = \left\{ \begin{bmatrix} 3 \\ -5 \end{bmatrix}, \begin{bmatrix} -4 \\ 6 \end{bmatrix} \right\}
step 2
We can set up the equation:
\( c_1 \begin{bmatrix} 3 \\ -5 \end{bmatrix} + c_2 \begin{bmatrix} -4 \\ 6 \end{bmatrix} = \begin{bmatrix} 2 \\ -6 \end{bmatrix} \)
step 3
This gives us the system of equations:
\( 3c_1 - 4c_2 = 2 \) and \( -5c_1 + 6c_2 = -6 \)
step 4
Solving this system, we can use substitution or elimination to find c1 c_1 and c2 c_2 . After solving, we find:
\( c_1 = 2, c_2 = 1 \)
Answer
[x]B=[21] [\mathbf{x}]_{\mathcal{B}} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}
Key Concept
The B \mathcal{B} -coordinate of a vector expresses the vector as a linear combination of the basis vectors in B \mathcal{B} .
Explanation
The coefficients c1 c_1 and c2 c_2 represent how the vector x \mathbf{x} can be constructed from the basis vectors, providing its coordinates in the new basis.
--- For Question 11:
step 1
We are given the vectors [2−5],[−410],[−36] \begin{bmatrix} 2 \\ -5 \end{bmatrix}, \begin{bmatrix} -4 \\ 10 \end{bmatrix}, \begin{bmatrix} -3 \\ 6 \end{bmatrix} . To find the dimension of the subspace H H spanned by these vectors, we can form a matrix with these vectors as columns and row-reduce it:
\( \begin{bmatrix} 2 & -4 & -3 \\ -5 & 10 & 6 \end{bmatrix} \)
step 2
After row-reducing, we find the rank of the matrix, which gives us the dimension of the subspace. The row-reduced form is:
\( \begin{bmatrix} 1 & -2 & -\frac{3}{2} \\ 0 & 0 & 0 \end{bmatrix} \)
step 3
The rank is 1, indicating that the dimension of the subspace H H is:
\( \operatorname{dim}(H) = 1 \)
Answer
Dimension of the subspace H H : 1
Key Concept
The dimension of a subspace is the number of vectors in a basis for that subspace, which corresponds to the rank of the matrix formed by the spanning vectors.
Explanation
The row-reduction process reveals the number of linearly independent vectors, which determines the dimension of the subspace spanned by the given vectors.
--- For Question 12:
step 1
We are given the matrix A=[1amp;0amp;9amp;50amp;0amp;1amp;−4] A = \begin{bmatrix} 1 & 0 & 9 & 5 \\ 0 & 0 & 1 & -4 \end{bmatrix} . To find the dimensions of N(A) \mathbf{N}(A) and C(A) \mathbf{C}(A) , we first determine the rank of A A
step 2
The rank of A A is the number of leading 1's in its row echelon form, which is 2. Thus, we have:
\( \operatorname{rank}(A) = 2 \)
step 3
The dimension of the null space N(A) \mathbf{N}(A) can be found using the formula:
\( \operatorname{dim} \mathbf{N}(A) = n - \operatorname{rank}(A) \) where \( n \) is the number of columns. Here, \( n = 4 \):
\( \operatorname{dim} \mathbf{N}(A) = 4 - 2 = 2 \)
step 4
The dimension of the column space C(A) \mathbf{C}(A) is equal to the rank of A A :
\( \operatorname{dim} \mathbf{C}(A) = \operatorname{rank}(A) = 2 \)
Answer
dim⁡N(A)=2 \operatorname{dim} \mathbf{N}(A) = 2 ; dim⁡C(A)=2 \operatorname{dim} \mathbf{C}(A) = 2
Key Concept
The dimensions of the null space and column space are related to the rank of the matrix and the number of columns.
Explanation
The rank-nullity theorem helps us find the dimensions of the null space and column space based on the rank of the matrix and the total number of columns.
--- For Question 13:
step 1
We have a 3×8 3 \times 8 matrix A A with rank⁡(A)=3 \operatorname{rank}(A) = 3 . To find dim⁡N(A) \operatorname{dim} \mathbf{N}(A) , we use the formula:
\( \operatorname{dim} \mathbf{N}(A) = n - \operatorname{rank}(A) \) where \( n = 8 \):
\( \operatorname{dim} \mathbf{N}(A) = 8 - 3 = 5 \)
step 2
The dimension of the column space C(A) \mathbf{C}(A) is equal to the rank of A A :
\( \operatorname{dim} \mathbf{C}(A) = \operatorname{rank}(A) = 3 \)
step 3
The rank of the transpose AT A^{T} is equal to the dimension of the column space of A A :
\( \operatorname{rank}(A^{T}) = \operatorname{dim} \mathbf{C}(A) = 3 \)
Answer
dim⁡N(A)=5 \operatorname{dim} \mathbf{N}(A) = 5 ; dim⁡C(AT)=3 \operatorname{dim} \mathbf{C}(A^{T}) = 3 ; rank⁡(AT)=3 \operatorname{rank}(A^{T}) = 3
Key Concept
The rank-nullity theorem relates the rank of a matrix to the dimensions of its null space and column space.
Explanation
The calculations show how the rank of the original matrix informs us about the dimensions of both the null space and the column space of the original and transposed matrices.
--- For Question 14:
step 1
We have a homogeneous system of 5 linear equations in 6 unknowns, and the solutions are all multiples of one nonzero solution. This indicates that the solution space is one-dimensional
step 2
Since there is a nonzero solution, the system is consistent. For every choice of constants on the right sides of the equations, the system will have a solution if the constants do not contradict the equations
step 3
Given that the system has a unique solution up to scalar multiples, it implies that the system will not necessarily have a solution for every possible choice of constants
Answer
No, the system will not necessarily have a solution for every possible choice of constants.
Key Concept
A homogeneous system of equations has solutions that depend on the consistency of the equations with the constants on the right-hand side.
Explanation
The existence of a unique solution does not guarantee that all possible constants will yield a consistent system; thus, some choices may lead to no solution.
--- For Question 15:(a)
step 1
We need to find the change-of-coordinates matrix from basis A \mathcal{A} to basis B \mathcal{B} . The vectors in A \mathcal{A} are expressed in terms of the vectors in B \mathcal{B} :
\( \mathbf{a}_{1} = 4 \mathbf{b}_{1} - \mathbf{b}_{2} \)
\( \mathbf{a}_{2} = -\mathbf{b}_{1} + \mathbf{b}_{2} + \mathbf{b}_{3} \)
\( \mathbf{a}_{3} = \mathbf{b}_{2} - 2 \mathbf{b}_{3} \)
step 2
We can express these equations in matrix form to find the change-of-coordinates matrix P P :
\( P = \begin{bmatrix} 4 & -1 & 0 \\ -1 & 1 & 1 \\ 0 & -2 & 1 \end{bmatrix} \)
(b)
step 3
To find [x]B [\mathbf{x}]_{\mathcal{B}} for x=3a1+4a2+a3 \mathbf{x} = 3 \mathbf{a}_{1} + 4 \mathbf{a}_{2} + \mathbf{a}_{3} , we first express x \mathbf{x} in terms of B \mathcal{B} :
\( \mathbf{x} = 3(4 \mathbf{b}_{1} - \mathbf{b}_{2}) + 4(-\mathbf{b}_{1} + \mathbf{b}_{2} + \mathbf{b}_{3}) + (\mathbf{b}_{2} - 2 \mathbf{b}_{3}) \)
step 4
Simplifying this expression gives us:
\( \mathbf{x} = (12 - 4) \mathbf{b}_{1} + (3 + 4 - 1) \mathbf{b}_{2} + (4 - 2) \mathbf{b}_{3} = 8 \mathbf{b}_{1} + 6 \mathbf{b}_{2} + 2 \mathbf{b}_{3} \)
step 5
Thus, the coordinates of x \mathbf{x} in basis B \mathcal{B} are:
\( [\mathbf{x}]_{\mathcal{B}} = \begin{bmatrix} 8 \\ 6 \\ 2 \end{bmatrix} \)
Answer
Change-of-coordinates matrix P P : [4amp;−1amp;0−1amp;1amp;10amp;−2amp;1] \begin{bmatrix} 4 & -1 & 0 \\ -1 & 1 & 1 \\ 0 & -2 & 1 \end{bmatrix} ; [x]B=[862] [\mathbf{x}]_{\mathcal{B}} = \begin{bmatrix} 8 \\ 6 \\ 2 \end{bmatrix}
Key Concept
The change-of-coordinates matrix allows us to convert coordinates of vectors from one basis to another.
Explanation
The calculations show how to express vectors in terms of different bases, providing the necessary transformation for coordinate representation.
© 2023 AskSia.AI all rights reserved