Chapter 8: Problem 9
Suppose \(T \in \mathcal{L}(V)\) and \(m\) is a nonnegative integer such that range \(T^{m}=\) range \(T^{m+1}\). Prove that range \(T^{k}=\) range \(T^{m}\) for all \(k > m\).
Short Answer
Expert verified
We demonstrated that range \(T^k\) is a subset of range \(T^m\) and range \(T^m\) is a subset of range \(T^k\) for all \(k > m\), which implies that range \(T^k = T^m\) for all \(k > m\).
Step by step solution
01
Prove that range \(T^k\) is a subset of range \(T^m\) for all \(k > m\)#
Let's consider an element \(y \in\) range \(T^k\), for some \(k > m\). This means that there exists \(x \in V\) such that \(y = T^k(x)\).
Now, since range \(T^m = T^{m+1}\), by definition, there exists \(z \in V\) such that \(T^m(z) = T^{m+1}(z)\).
Since \(T^m = T^{m+1}\), we can apply the transformation \(T^{-1}\) (inverse) to both sides of the equation:
\[
T^{-1}(T^m(z)) = T^{-1}(T^{m+1}(z))
\]
Simplifying, we get \(T^{m-1}(z) = T^m(z)\).
Now, returning to \(y = T^k(x)\), since \(k > m\), we can rewrite it as:
\[
y = T^m(T^{k-m}(x))
\]
Using the fact that \(T^{m-1}(z) = T^m(z)\), we can rewrite this expression as:
\[
y = T^{m-1}(T^{k-m+1}(x))
\]
Since \(k > m\), \(k-m+1 > 1\), so that \(T^{k-m+1}(x) \in V\). Therefore, \(y \in\) range \(T^m\). This implies that the range \(T^k\) is a subset of range \(T^m\) for all \(k > m\).
02
Prove that range \(T^m\) is a subset of range \(T^k\) for all \(k > m\)#
Now, let's consider an element \(y \in\) range \(T^m\). This means that there exists \(x \in V\) such that \(y = T^m(x)\).
Since range \(T^m = T^{m+1}\), we have \(y = T^{m+1}(x) = T^m(T(x))\).
Now, let \(z = T(x)\). Then, \(y = T^m(z)\), where \(z \in V\), since \(T(x) = z\) is just applying the linear transformation \(T\) to \(x\). It implies that \(y = T^m(z) = T^{k}(T^{m-k}(z))\), where \(k > m\).
Recall that \(k > m\), meaning that \(m-k < 0\). However, since \(T^m = T^{m+1}\), the transformation is invertible (i.e., it has an inverse). Therefore, we can write \(T^{-(k-m)}(z) = T^{-1}(T^{k-m}(z))\). With this, we can rewrite the expression as:
\[
y = T^k(T^{-(k-m)}(z))
\]
This means that \(y \in\) range \(T^k\) since \(T^{-(k-m)}(z) \in V\). Therefore, the range \(T^m\) is a subset of range \(T^k\) for all \(k > m\).
03
Conclusion#
Since range \(T^k\) is a subset of range \(T^m\) and range \(T^m\) is a subset of range \(T^k\) for all \(k > m\), we can conclude that range \(T^k = T^m\) for all \(k > m\). This completes the proof.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Transformation
A linear transformation is a fundamental concept in linear algebra. It is a function between two vector spaces that preserves the operations of vector addition and scalar multiplication. This means if you have two vectors, say \( u \) and \( v \), in your vector space along with a scalar \( c \), then a linear transformation \( T \) from one vector space to another will satisfy two main properties:
- Additivity: \( T(u + v) = T(u) + T(v) \).
- Homogeneity: \( T(cu) = cT(u) \).
Range of an Operator
The range of an operator, particularly a linear operator, is an essential concept that describes the set of all possible outputs – or images – of a linear transformation. If \( T: V \rightarrow W \) is a linear operator where \( V \) and \( W \) are vector spaces, then the range of \( T \), denoted as \( ext{range}(T) \), is defined as:
- \( ext{range}(T) = \{ T(v) \mid v \, \in \, V \} \).
Subset in Vector Spaces
Understanding subsets within vector spaces is crucial when discussing linear transformations and their properties. A subset is simply a collection of vectors that altogether form part of a larger vector space. Let's consider two vector spaces: \( U \) and \( V \); \( U \) is a subset of \( V \) if every vector in \( U \) is also a vector in \( V \). Subsets are not necessarily vector spaces on their own unless they satisfy the conditions for being a vector space (closure under addition and scalar multiplication, etc.). When dealing with operators, like the linear operator \( T \) from the exercise, the range and the domain, which are themselves subsets, gain significance. To prove that one range is a subset of another, we show that applying the transformation \( T \) or its power, maps vectors into a particular subset of another vector space, implying containment. As described in the solution steps of the exercise, the demonstration that one range is a subset of another is a preliminary step in proving equality between ranges when they're impacted by the transformation's repeated application.