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Chapter 6: Problem 69

Suppose \(V=U+W\), and suppose \(U\) and \(V\) are each invariant under a linear operator \(F: V \rightarrow V\). Also, suppose \(\operatorname{dim} U=r\) and \(\operatorname{dim} W=S\). Show that \(F\) has a block diagonal matrix representation \(M=\left[\begin{array}{ll}A & 0 \\ 0 & B\end{array}\right]\), where \(A\) and \(B\) are \(r \times r\) and \(s \times s\) submatrices.

Short Answer

Expert verified
Given the linear operator \(F: V \rightarrow V\) where \(V = U + W\), and \(F\) preserves the invariance of both \(U\) and \(V\), with \(\operatorname{dim} U = r\) and \(\operatorname{dim} W = s\), we construct bases \(B_1\) for \(U\) and \(B_2\) for \(W\), and form a basis \(B\) for \(V\). The matrix representation of \(F\) corresponding to the basis \(B\) is given by \(M\), with (i, j)-th entries \(M_{ij} = F(b_j)_i = b_{ij}\). Using the information that \(F(U) \subseteq U\) and \(F(V) \subseteq V\), we determine the action of \(F\) on both the bases of \(U\) and \(V\). By combining these results, we obtain the block diagonal matrix \(M = \left[\begin{array}{c|c} A & 0 \\ \hline 0 & B \end{array}\right]\) where \(A\) and \(B\) are \(r \times r\) and \(s \times s\) submatrices, respectively.

Step by step solution

01

1. Construct bases for U, W and V

Let's take basis \(B_1 = \{u_1, u_2, \cdots, u_r\}\) for U, and basis \(B_2 = \{w_1, w_2, \cdots, w_s\}\) for W. We can construct a basis for V, say \(B = B_1 \cup B_2 = \{u_1, u_2, \cdots, u_r, w_1, w_2, \cdots, w_s\}\).
02

2. Define the matrix representation of F

We want to find the matrix representation of \(F\) corresponding to the basis \(B\). Let's denote that matrix as \(M\). The (i, j)th entry of the matrix M is given by: \(M_{ij} = F(b_j)_i = b_{ij}\), where \(b_j\) denotes the jth element in the basis B and \(\{b_{ij}\}\) are the coordinates of \(F(b_j)\) in the basis B.
03

3. Determine the action of F on the basis of U and V

Given that \(F(U) \subseteq U\) and \(F(V) \subseteq V\), it's clear that the action of F on the basis vectors of U and W will result in vectors that still belong to U and W, respectively. Mathematically, we can write this as: \[F(u_i) = \sum_{k=1}^{r} a_{ki} u_k\] \[F(w_j) = \sum_{k=1}^{s} b_{kj} w_k\]
04

4. Compute the matrix M using (i, j)th entries from Step 2

Now, let's find the (i, j)th entry of M using the results from the previous step: - For columns 1 to r, i.e., the vectors in basis \(B_1\): \[F(u_i) = \sum_{k=1}^{r} a_{ki} u_k \longrightarrow M_{ij} = \left\{ \begin{array}{ll} a_{ki} & 1 \leq i,j \leq r \\ 0 & r+1 \leq i \leq r+s \\ 0 & 1 \leq j < i \leq r \end{array} \right.\] - For columns r+1 to r+s, i.e., the vectors in basis \(B_2\): \[F(w_j) = \sum_{k=1}^{s} b_{kj} w_k \longrightarrow M_{ij} = \left\{ \begin{array}{ll} 0 & 1 \leq i \leq r \\ b_{kj} & r+1 \leq i, j \leq r+s \\ 0 & 1 \leq i < j \leq r+s \end{array} \right.\]
05

5. Combine the results to form the block diagonal matrix M

When we combine these results, we get the block diagonal matrix M with r x r submatrix A and s x s submatrix B as follows: \[ M = \left[\begin{array}{c|c} A & 0 \\ \hline 0 & B \end{array}\right] \] The block diagonal matrix representation for the linear operator F with submatrices A and B has been found as required.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Operator
A linear operator is a function that operates on vectors within a vector space, maintaining specific properties. For a linear operator, like the one in our exercise, two essential rules apply:
  • Additivity: It respects vector addition, meaning if you apply the operator to the sum of two vectors, it's the same as applying the operator to each vector individually and then adding the results.
  • Scalar Multiplication: It respects scalar multiplication, so when a vector is multiplied by a scalar and then the operator is applied, it's the same as applying the operator first and then scaling the resulting vector.
In the context of our problem, the linear operator \(F: V \rightarrow V\) acts on vector spaces \(U\) and \(W\), meaning it takes any vector from the combined space \(V\) and returns a vector still in \(V\). This structural integrity is crucial when explaining how the operator interacts with bases of different subspaces.
Invariant Subspaces
Invariant subspaces are a fundamental concept when discussing linear operators. A subspace is considered invariant under a linear operator if applying the operator to any vector within the subspace keeps the result within that same subspace.
For example, if we have a vector space \(V\) built from subspaces \(U\) and \(W\), and a linear operator \(F\) where \(F(U) \subseteq U\) and \(F(W) \subseteq W\), it means:
  • Any operation that \(F\) performs on \(U\) will result in vectors that remain in \(U\).
  • Similarly, operations on \(W\) result in vectors that stay within \(W\).
Invariant subspaces make analyzing the behavior of operators easier, particularly in creating matrix representations that highlight their structure, such as the block diagonal matrix seen in this exercise. Understanding invariance ensures that the structured behavior of vectors remains consistent under transformations.
Matrix Representation
Matrix representation is a technique to express a linear operator in the form of a matrix. This is incredibly helpful for calculations involving linear transformations, as it allows us to perform operations like additions and multiplications systematically.
In our exercise, we transform the operator \(F\) into a matrix \(M\) by using a basis of \(V\). The matrix is constructed so that its action on vectors can be calculated through matrix multiplication. This matrix is said to be block diagonal if it consists of smaller square or rectangular blocks of matrices (like \(A\) and \(B\) in the given problem) along the diagonal and zeros elsewhere.
  • Block diagonal matrices save computation effort, as their complexity is reduced due to zeros in the off-diagonal blocks.
  • This structure also illustrates the separability of actions of \(F\) on the invariant subspaces \(U\) and \(W\), encoded into submatrices \(A\) and \(B\).
By using a suitable basis, expressing \(F\) as a block diagonal matrix depicts the operations on each subspace independently.
Dimensions of Subspaces
Dimensions of subspaces quantify the number of independent vectors that span a subspace. For any vector space, its dimension is crucial because it dictates the maximum number of linearly independent vectors.
In this problem, we are given that the subspace \(U\) has dimension \(r\), and subspace \(W\) has dimension \(s\). Together, they span the space \(V\) with dimension \(r + s\).
  • The dimension directly informs the size of the basis: for \(U\), the basis consists of \(r\) vectors, and for \(W\), \(s\) vectors.
  • These dimensions are critical during matrix representation because the size of blocks \(A\) and \(B\) in the block diagonal matrix \(M\) are determined by \(r\) and \(s\) respectively.
Understanding dimensions is essential not only for forming effective bases but also to ensure that mathematical operations and transformations retain their proper structure within vector spaces.

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Most popular questions from this chapter

Consider a finite sequence of vectors \(S=\left\\{u_{1}, u_{2}, \ldots, u_{n}\right\\} .\) Let \(S^{\prime}\) be the sequence of vectors obtained from \(S\) by one of the following "elementary operations": (1) Interchange two vectors. (2) Multiply a vector by a nonzero scalar. (3) Add a multiple of one vector to another vector. Show that \(S\) and \(S^{\prime}\) span the same subspace \(W\). Also, show that \(S^{\prime}\) is linearly independent if and only if \(S\) is linearly independent. Observe that, for each operation, the vectors \(S^{\prime}\) are linear combinations of vectors in \(S\). Also, because each operation has an inverse of the same type, each vector in \(S\) is a linear combination of vectors in \(S^{\prime}\). Thus, \(S\) and \(S^{\prime}\) span the same subspace \(W\). Moreover, \(S^{\prime}\) is linearly independent if and only if \(\operatorname{dim} W=n\), and this is true if and only if \(S\) is linearly independent.

Let \(\mathbf{D}\) denote the differential operator on the vector space \(V\) of functions with basis \(S=\\{\sin \theta, \cos \theta\\}\) (a) Find the matrix \(A=[\mathbf{D}]_{s}\) (b) Use \(A\) to show that \(\mathbf{D}\) is a zero of \(f(t)=t^{2}+1\)

Prove Theorem 6.7: Let \(P\) be the change-of-basis matrix from a basis \(S\) to a basis \(S^{\prime}\) in a vector space \(V .\) Then, for any linear operator \(T\) on \(V,[T]_{S}=P^{-1}[T]_{S} P\).

Let \(G: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) be defined by \(G(x, y, z)=(2 x+3 y-z, 4 x-y+2 z)\). (a) Find the matrix \(A\) representing \(G\) relative to the bases \(S=\\{(1,1,0),(1,2,3),(1,3,5)\\} \quad\) and \(\quad S^{\prime}=\\{(1,2),(2,3)\\}\) (b) For any \(v=(a, b, c)\) in \(\mathbf{R}^{3}\), find \([v]_{S}\) and \([G(v)]_{S^{\prime}}\). (c) Verify that \(A[v]_{S}=[G(v)]_{S^{\prime}}\).

For each of the following linear transformations (operators) \(L\) on \(\mathbf{R}^{2},\) find the matrix \(A\) that represents \(L\) (relative to the usual basis of \(\mathbf{R}^{2}\) ): (a) \(L\) is defined by \(L(1,0)=(2,4)\) and \(L(0,1)=(5,8)\) (b) \(L\) is the rotation in \(\mathbf{R}^{2}\) counterclockwise by \(90^{\circ}\) (c) \(L\) is the reflection in \(\mathbf{R}^{2}\) about the line \(y=-x\)
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