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Chapter 10: Problem 42

Suppose dim \(V=n .\) Show that \(T: V \rightarrow V\) has a triangular matrix representation if and only if there exist \(T\) -invariant subspaces \(W_{1} \subset W_{2} \subset \cdots \subset W_{n}=V\) for which \(\operatorname{dim} W_{k}=k, k=1, \ldots, n\).

Short Answer

Expert verified
T: V 鈫� V has a triangular matrix representation if and only if there exist T-invariant subspaces W1 鈯� W2 鈯� ... 鈯� Wn = V for which dimWk = k, k = 1,..., n. This is shown by proving two parts: 1) Triangular matrix representation implies the existence of T-invariant subspaces of the given dimensions, and 2) the existence of T-invariant subspaces of the given dimensions implies a triangular matrix representation. We define suitable subspaces and bases and show the T-invariance and triangular matrix representation accordingly.

Step by step solution

01

Show existence of T-invariant subspaces

Consider the triangular matrix representation of T with respect to some basis B. Define Wk as the subspace of V spanned by the first k basis vectors of B. Note that Wk has dimension k and W1 鈯� W2 鈯� ... 鈯� Wn = V.
02

Show T-invariance of Wk

To show T-invariance of Wk, we need to prove that for every vector w in Wk, Tw is also in Wk. Since T has a triangular matrix representation, by applying T to the basis vectors of Wk, we can confirm that Tw will remain combinations of the first k basis vectors of B. Therefore, Tw lies in Wk, showing the T-invariance of Wk. ##Part 2: T-invariant subspaces imply triangular matrix representation##
03

Define basis B'

For each T-invariant subspace Wk, choose a basis Bk. Then, we can form a basis B' for V by taking the union of all bases B1, B2, ..., Bn.
04

Show triangular matrix representation with respect to B'

Now, consider the matrix representation of T with respect to the basis B'. Due to the T-invariance of Wk, applying T to the kth basis vector will result in a linear combination of the first k basis vectors of B'. Therefore, in the matrix representation of T with respect to B', all entries below the main diagonal are zero, meaning T has a triangular matrix representation with respect to B'. Thus, we have shown that T: V 鈫� V has a triangular matrix representation if and only if there exist T-invariant subspaces W1 鈯� W2 鈯� ... 鈯� Wn = V for which dimWk = k, k = 1,..., n.

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Key Concepts

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

T-invariant Subspaces
In the context of linear algebra, a T-invariant subspace is a vector space that is preserved under the application of a linear transformation. Simply put, if we have a transformation T that acts on a vector space V, and we find a subspace W within V such that applying T to any vector in W results in another vector that is also in W, then W is considered T-invariant.

Understanding T-invariant subspaces is crucial because they offer a way to simplify complex linear transformations. They allow us to break down the transformation into more manageable parts related to each subspace. As seen in the exercise, a series of nested T-invariant subspaces, with the property that each subspace has one dimension greater than the previous, can lead to a triangular matrix representation for T.

A triangular matrix is particularly useful because it can simplify calculations such as determining eigenvalues, solving linear equations, and computing powers of matrices.
Matrix Representation of Linear Transformations
Matrix representation is an effective tool for handling linear transformations in vector spaces. It is essentially a way of encoding the information about a linear transformation as a matrix, allowing for easier computational manipulation. In our previous exercise, the transformation T is represented by a triangular matrix when certain conditions are met.

A triangular matrix, as the name suggests, is a type of square matrix where all entries above or below the main diagonal are zero. There are two primary types of triangular matrices: upper and lower. In this case, when we have a set of T-invariant subspaces that form a chain, W1 ⊂ W2 ⊂ ... ⊂ Wn = V, with corresponding dimensions increasing from 1 to n, the matrix representation of T becomes upper triangular.

This structure greatly simplifies many theoretical and practical applications since the eigenvalues of T can be read off the diagonal, and many matrix-related computations become less complex.
Dimension of Vector Spaces
The dimension of a vector space is a fundamental concept in linear algebra, representing the number of vectors in a basis of the space. Essentially, it measures the size or complexity of a vector space. In the exercise, we talk about an n-dimensional space V, meaning that there is a set of n linearly independent vectors that span the entire space. Each subspace Wk mentioned in the solution has a dimension k, which determines the number of vectors that span that subspace.

The importance of considering dimensions in this context lies in constructing the increasing chain of T-invariant subspaces. It allows us to form a step by step approach to dealing with the entire space by examining one dimension at a time. This paves the way for a systematic method of achieving a triangular matrix representation for the transformation T, highlighting the integral role that the dimension plays in understanding the structure of vector spaces and their transformations.

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Most popular questions from this chapter

The subspaces \(W_{1}, \ldots, W_{r}\) are said to be independent if \(w_{1}+\cdots+w_{r}=0, w_{i} \in W_{i},\) implies that each \(w_{i}=0 .\) Show that \(\operatorname{span}\left(W_{i}\right)=W_{1} \oplus \cdots \oplus W_{r}\) if and only if the \(W_{i}\) are independent. [Here \(\operatorname{span}\left(W_{i}\right)\) denotes the linear span of the \(\left.W_{i} .\right]\)

Let \(W\) be invariant under \(T_{1}: V \rightarrow V\) and \(T_{2}: V \rightarrow V .\) Prove \(W\) is also invariant under \(T_{1}+T_{2}\) and \(T_{1} T_{2}\)

Prove Theorem 10.16: Suppose \(W\) is a subspace invariant under a linear operator \(T: V \rightarrow V .\) Then \(T\) induces a linear operator \(\bar{T}\) on \(V / W\) defined by \(\bar{T}(v+W)=T(v)+W\). Moreover, if \(T\) is a zero of any polynomial, then so is \(\bar{T}\). Thus, the minimal polynomial of \(\bar{T}\) divides the minimal polynomial of \(T\) We first show that \(\bar{T}\) is well defined; that is, if \(u+W=v+W\), then \(T(u+W)=\bar{T}(v+W)\). If \(u+W=v+W,\) then \(u-v \in W,\) and, as \(W\) is \(T\) -invariant, \(T(u-v)=T(u)-T(v) \in W .\) Accordingly, \\[ \bar{T}(u+W)=T(u)+W=T(v)+W=\bar{T}(v+W) \\] as required. We next show that \(\bar{T}\) is linear. We have \\[ \begin{aligned} \bar{T}((u+W)+(v+W)) &=\bar{T}(u+v+W)=T(u+v)+W=T(u)+T(v)+W \\ &=T(u)+W+T(v)+W=\bar{T}(u+W)+\bar{T}(v+W) \end{aligned} \\] Furthermore, \\[ \bar{T}(k(u+W))=\bar{T}(k u+W)=T(k u)+W=k T(u)+W=k(T(u)+W)=k \hat{T}(u+W) \\] Thus, \(\bar{T}\) is linear. Now, for any coset \(u+W\) in \(V / W\) \\[ \overline{T^{2}}(u+W)=T^{2}(u)+W=T(T(u))+W=\bar{T}(T(u)+W)=\bar{T}(\bar{T}(u+W))=\bar{T}^{2}(u+W) \\] Hence, \(\overline{T^{2}}=\bar{T}^{2}\). Similarly, \(\overline{T^{n}}=\bar{T}^{n}\) for any \(n\). Thus, for any polynomial \\[ \begin{aligned} f(t) &=a_{n} t^{n}+\cdots+a_{0}=\sum a_{i} t^{i} \\ \overline{f(T)}(u+W) &=f(T)(u)+W=\sum a_{i} T^{i}(u)+W=\sum a_{i}\left(T^{i}(u)+W\right) \\ &=\sum a_{i} \overline{T^{i}}(u+W)=\sum a_{i} \bar{T}^{i}(u+W)=\left(\sum a_{i} \bar{T}^{i}\right)(u+W)=f(\bar{T})(u+W) \end{aligned} \\] and so \(\overline{f(T)}=f(T) .\) Accordingly, if \(T\) is a root of \(f(t)\) then \(\overline{f(T)}=\overline{0}=W=f(T) ;\) that is, \(\bar{T}\) is also a root of \(f(t) .\) The theorem is proved.

Prove Theorem 10.10: Let \(T: V \rightarrow V\) be a nilpotent operator of index \(k .\) Then \(T\) has a unique block diagonal matrix representation consisting of Jordan nilpotent blocks \(N .\) There is at least one \(N\) of order \(k,\) and all other \(N\) are of orders \(\leq k .\) The total number of \(N\) of all orders is equal to the nullity of \(T\) Suppose \(\operatorname{dim} V=n .\) Let \(W_{1}=\operatorname{Ker} T, W_{2}=\operatorname{Ker} T^{2}, \ldots, W_{k}=\operatorname{Ker} T^{k} .\) Let us set \(m_{i}=\operatorname{dim} W_{i},\) for \(i=1, \ldots, k .\) Because \(T\) is of index \(k, W_{k}=V\) and \(W_{k-1} \neq V\) and so \(m_{k-1}1 \\ 0 & \text { for } j=1 \end{array}\right. \\] Now it is clear [see Problem \(10.13(\mathrm{d})]\) that \(T\) will have the desired form if the \(v(i, j)\) are ordered lexicographically: beginning with \(v(1,1)\) and moving up the first column to \(v(1, k),\) then jumping to \(v(2,1)\) and moving up the second column as far as possible. Moreover, there will be exactly \(m_{k}-m_{k-1}\) diagonal entries of order \(k\). Also, there will be \(\left(m_{k-1}-m_{k-2}\right)-\left(m_{k}-m_{k-1}\right)=2 m_{k-1}-m_{k}-m_{k-2} \quad\) diagonal entries of order \(k-1\) \(2 m_{2}-m_{1}-m_{3} \quad\) diagonal entries of order 2 \\[ 2 m_{1}-m_{2} \\] diagonal entries of order as can be read off directly from the table. In particular, because the numbers \(m_{1}, \ldots, m_{k}\) are uniquely determined by \(T,\) the number of diagonal entries of each order is uniquely determined by \(T .\) Finally, the identity \\[ m_{1}=\left(m_{k}-m_{k-1}\right)+\left(2 m_{k-1}-m_{k}-m_{k-2}\right)+\cdots+\left(2 m_{2}-m_{1}-m_{3}\right)+\left(2 m_{1}-m_{2}\right) \\] shows that the nullity \(m_{1}\) of \(T\) is the total number of diagonal entries of \(T\)

Show that \(V=W_{1} \oplus \cdots \oplus W_{r}\) if and only if (i) \(V=\operatorname{span}\left(W_{i}\right)\) and (ii) for \(k=1,2, \ldots, r\) \(W_{k} \cap \operatorname{span}\left(W_{1}, \ldots, W_{k-1}, W_{k+1}, \ldots, W_{r}\right)=\\{0\\}\).
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