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Consider the following subspaces of R4\mathbb{R}^{4} : \[ U_{1}:=\left\{(a, b, ...
Aug 8, 2024
Solution by Steps
step 1
To find the matrix A1 A_1 for the subspace U1={(a,b,b,c)T:a,b,c∈R} U_1 = \{(a, b, b, c)^T : a, b, c \in \mathbb{R}\} , we express the vectors in U1 U_1 in terms of the parameters a,b,c a, b, c . This gives us the following representation: (abbc)=a(1000)+b(0110)+c(0001). \begin{pmatrix} a \\ b \\ b \\ c \end{pmatrix} = a \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + b \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} + c \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}. Thus, the matrix A1 A_1 can be formed by taking the coefficients of these vectors. The corresponding matrix is: A1=(1amp;0amp;00amp;1amp;10amp;0amp;00amp;0amp;1). A_1 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}.
step 2 ⋮ For the subspace U2={(0,a,−a,0)T:a∈R} U_2 = \{(0, a, -a, 0)^T : a \in \mathbb{R}\} , we can express the vectors in U2 U_2 as: (0a−a0)=a(01−10). \begin{pmatrix} 0 \\ a \\ -a \\ 0 \end{pmatrix} = a \begin{pmatrix} 0 \\ 1 \\ -1 \\ 0 \end{pmatrix}. Thus, the matrix A2 A_2 is: A2=(0amp;00amp;10amp;−10amp;0). A_2 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \\ 0 & -1 \\ 0 & 0 \end{pmatrix}.
step 3
Next, we stack the matrices A1 A_1 and A2 A_2 on top of each other to form the augmented matrix: A=(1amp;0amp;00amp;1amp;10amp;0amp;00amp;0amp;10amp;00amp;10amp;−10amp;0). A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 \\ 0 & 1 \\ 0 & -1 \\ 0 & 0 \end{pmatrix}.
step 4 ⋮ We now apply row reduction to the augmented matrix A A to find the intersection U1∩U2 U_1 \cap U_2 . After performing row operations, we find: (1amp;0amp;00amp;1amp;10amp;0amp;00amp;0amp;00amp;0amp;00amp;1amp;−10amp;0amp;00amp;0amp;0)→(1amp;0amp;00amp;1amp;10amp;0amp;00amp;0amp;0). \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. This indicates that the intersection U1∩U2 U_1 \cap U_2 is spanned by the vector (0,0,0,0)T (0, 0, 0, 0)^T , which means the intersection is trivial.
Answer
The intersection U1∩U2 U_1 \cap U_2 is the zero vector space, {(0,0,0,0)T} \{(0, 0, 0, 0)^T\}
Key Concept
The intersection of subspaces involves finding common solutions to their defining equations.
Explanation
The intersection U1∩U2 U_1 \cap U_2 is found by solving the combined system of equations derived from the matrices A1 A_1 and A2 A_2 , leading to the conclusion that the only common vector is the zero vector.
row reduction
To find the intersection U1∩U2 U_1 \cap U_2 through row reduction, we first need to set up the matrices A1 A_1 and A2 A_2 based on the definitions of the subspaces U1 U_1 and U2 U_2 . Step 1: Define the matrices For the subspace U1={(a,b,b,c)T:a,b,c∈R} U_1 = \{(a, b, b, c)^T : a, b, c \in \mathbb{R}\} , we can express it in matrix form as: A1=(1amp;0amp;0amp;00amp;1amp;1amp;00amp;0amp;0amp;1) A_1 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} This matrix represents the equations that define U1 U_1 . For the subspace U2={(0,a,−a,0)T:a∈R} U_2 = \{(0, a, -a, 0)^T : a \in \mathbb{R}\} , we can express it as: A2=(0amp;1amp;−1amp;0) A_2 = \begin{pmatrix} 0 & 1 & -1 & 0 \end{pmatrix} Step 2: Stack the matrices Now, we stack A1 A_1 and A2 A_2 on top of each other: A=(1amp;0amp;0amp;00amp;1amp;1amp;00amp;0amp;0amp;10amp;1amp;−1amp;0) A = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & -1 & 0 \end{pmatrix} Step 3: Row reduction Next, we perform row reduction on the matrix A A : 1. The first row is already in the correct form. 2. We can use the first row to eliminate the first entry of the fourth row (which is already zero). 3. Now, we focus on the second row. We can use it to eliminate the second entry of the fourth row. After performing the row operations, we get: (1amp;0amp;0amp;00amp;1amp;1amp;00amp;0amp;0amp;10amp;0amp;0amp;0) \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} Step 4: Interpret the result The row-reduced form indicates that the system of equations has a free variable. The last row being all zeros suggests that there are solutions other than the trivial solution. To find the intersection U1∩U2 U_1 \cap U_2 , we can set up the equations from the row-reduced matrix: 1. x1=0 x_1 = 0 2. x2+x3=0 x_2 + x_3 = 0 (which implies x3=−x2 x_3 = -x_2 ) 3. x4=0 x_4 = 0 Thus, the intersection can be expressed as: U1∩U2={(0,b,−b,0)T:b∈R} U_1 \cap U_2 = \{(0, b, -b, 0)^T : b \in \mathbb{R}\} This means that the intersection is a line in the direction of the vector (0,1,−1,0)T (0, 1, -1, 0)^T . If you have any further questions or need clarification, feel free to ask!
M*(0, x, -x, 0)=0. What is matrix M?
Pattern Sequences
Pattern 1
To find the matrix M M such that M(0x−x0)=0 M \begin{pmatrix} 0 \\ x \\ -x \\ 0 \end{pmatrix} = 0 , we need to consider the properties of matrix multiplication. The result must be the zero vector
Pattern 2
The matrix M M must have a structure that ensures the second and third components of the vector (0x−x0) \begin{pmatrix} 0 \\ x \\ -x \\ 0 \end{pmatrix} are mapped to zero. This implies that the rows of M M corresponding to these components must be linear combinations that yield zero
Pattern 3
A suitable choice for M M can be M=(0amp;0amp;0amp;00amp;1amp;−1amp;00amp;0amp;0amp;00amp;0amp;0amp;0) M = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} . This matrix will ensure that M(0x−x0)=(0000) M \begin{pmatrix} 0 \\ x \\ -x \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}
Pattern Rule
The matrix M M must be structured to annihilate the vector (0x−x0) \begin{pmatrix} 0 \\ x \\ -x \\ 0 \end{pmatrix} by ensuring that the output is the zero vector.
Answer
The matrix M M is (0amp;0amp;0amp;00amp;1amp;−1amp;00amp;0amp;0amp;00amp;0amp;0amp;0) \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} .
Key Concept
Matrix multiplication and the properties of linear transformations.
Explanation
The matrix M M is designed to ensure that the specific vector (0x−x0) \begin{pmatrix} 0 \\ x \\ -x \\ 0 \end{pmatrix} results in the zero vector when multiplied, demonstrating the concept of annihilation in linear algebra.
G*(p, q, q, r)=0. What is the matrix G?
Solution by Steps
step 1
To find the matrix M M such that M(0x−x0)=0 M \begin{pmatrix} 0 \\ x \\ -x \\ 0 \end{pmatrix} = 0 , we need to ensure that the multiplication results in the zero vector
step 2
The matrix M M must be structured to eliminate the contributions from the second and third components of the vector (0x−x0) \begin{pmatrix} 0 \\ x \\ -x \\ 0 \end{pmatrix}
step 3
A suitable form for M M could be M=(aamp;bamp;camp;d0amp;0amp;0amp;00amp;0amp;0amp;00amp;0amp;0amp;0) M = \begin{pmatrix} a & b & c & d \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} , where a,b,c,d a, b, c, d are any real numbers
step 4
This structure ensures that the second and third rows of M M will yield zero when multiplied by the vector, resulting in M(0x−x0)=(ax+b(−x)+c(0)+d(0)000)=0 M \begin{pmatrix} 0 \\ x \\ -x \\ 0 \end{pmatrix} = \begin{pmatrix} ax + b(-x) + c(0) + d(0) \\ 0 \\ 0 \\ 0 \end{pmatrix} = 0
Answer
The matrix M M can be any matrix of the form (aamp;bamp;camp;d0amp;0amp;0amp;00amp;0amp;0amp;00amp;0amp;0amp;0) \begin{pmatrix} a & b & c & d \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} where a,b,c,d a, b, c, d are real numbers.
Key Concept
Matrix multiplication and the zero vector condition
Explanation
The matrix M M must be structured to ensure that the output is the zero vector when multiplied by the given vector, which can be achieved by having rows that eliminate the contributions from the non-zero components.
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