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Chapter 10: Problem 38

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 \(T_{1} T_{2}\)

Short Answer

Expert verified
We have proven that if subspace W is invariant under transformations T鈧 and T鈧, it is also invariant under their sum T鈧 + T鈧 and their product T鈧乀鈧. To show this, we first proved that W is invariant under the sum T鈧 + T鈧 by demonstrating that for any \(w \in W\), \((T_1 + T_2)(w) = T_1(w) + T_2(w) \in W\). Then, we proved that W is invariant under the product T鈧乀鈧 by showing that for any \(w \in W\), \(T_1(T_2(w)) \in W\).

Step by step solution

01

Proving invariance under the sum T鈧 + T鈧

Given that W is invariant under T鈧 and T鈧, we have: \(T_1(w) 鈭 W\) and \(T_2(w) 鈭 W\), \(\forall w 鈭 W\) To prove W is invariant under T鈧 + T鈧, we need to show that: \((T_1 + T_2)(w) 鈭 W\), for all \(w 鈭 W\). Consider \(w 鈭 W\), we can now calculate (T鈧 + T鈧)(w) as follows: \((T_1 + T_2)(w) = T_1(w) + T_2(w)\) Since W is invariant under T鈧 and T鈧, both T鈧(w) and T鈧(w) are in W. Since W is a subspace, it must be closed under addition. Therefore, their sum must also be in W: \(T_1(w) + T_2(w) 鈭 W\) Thus, we have proven that W is invariant under the sum T鈧 + T鈧.
02

Proving invariance under the product T鈧乀鈧

Given that W is invariant under T鈧 and T鈧, we have: \(T_1(w) 鈭 W\) and \(T_2(w) 鈭 W\), \(\forall w 鈭 W\) To prove W is invariant under T鈧乀鈧, we need to show that: \(T_1(T_2(w)) 鈭 W\), for all \(w 鈭 W\). Consider \(w 鈭 W\), we first apply the transformation T鈧 on w: \(T_2(w) 鈭 W\) (since W is invariant under T鈧) Now, we apply T鈧 on the result of T鈧(w), which is also in W: \(T_1(T_2(w)) 鈭 W\) (since W is invariant under T鈧) Thus, we have proven that W is invariant under the product T鈧乀鈧. In conclusion, W is invariant under both the sum T鈧 + T鈧 and the product T鈧乀鈧.

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Key Concepts

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

Linear Transformation
Linear transformations are, in essence, functions that map vectors from one vector space to another while preserving the operations of vector addition and scalar multiplication. This means, if you have a linear transformation, say, let's call it either of \( T \) or \( T_{1} \) and \( T_{2} \), and any two vectors \( u \) and \( v \) in space \( V \), and a scalar \( c \), the following always holds true:
  • \( T(u + v) = T(u) + T(v) \) - This means the transformation of a sum is the sum of the transformations.
  • \( T(cu) = cT(u) \) - This means the transformation of a scalar multiple is the scalar multiple of the transformation.
When watching these transformations in action, it's like seeing a seamless choreography where vectors glide from one space to another, always adhering to their inherent linear rules. This is crucial in many areas of mathematics and engineering, as it ensures consistency and predictability in the behavior of systems under linear transformations.

In the exercise's context, both \( T_{1} \) and \( T_{2} \) are linear transformations operating on the vector space \( V \). Their properties are being used to explore and ultimately prove the invariance of a subspace \( W \) under their sum and product. Understanding these transformations individually is the first step to unraveling the problem at hand.
Vector Space
A vector space can be visualized as a playground where vectors live and interact according to specific rules. The technical definition speaks of a collection of objects known as vectors, which can be added together and multiplied by scalars, forming a space that satisfies ten axioms laid out in linear algebra. These axioms include the existence of an additive identity (a zero vector), additive inverses (negative vectors), and the distributive property, among others.

Visualizing Vector Spaces

Imagine you could move around in this playground in any linear fashion鈥攕traight lines, scaling up or down. Each movement you make is a representation of vector addition or scalar multiplication. And, no matter what linear move you make, you're never leaving the bounds of the playground鈥攖hat鈥檚 the essence of a vector space.

Back to our exercise, we're focusing on a specific vector space \( V \), with \( W \) being a sort of 'special zone' within it that's stable under the influence of the transformations \( T_{1} \) and \( T_{2} \). The goal is to prove that \( W \) remains this 'special zone' even when these transformations combine their effects.
Subspace Closure Properties
Subspaces are the tightly-knit communities within the larger world of a vector space. If a vector space is a city, then a subspace is a neighborhood that's self-sufficient in that it has everything it needs to form a smaller vector space. The technical term for this self-sufficiency is 'closure,' and it refers to the fact that within a subspace, the results of vector addition and scalar multiplication don't lead you out of the neighborhood鈥攖hey keep you within the bounds of that subspace.

What Makes a Subspace?

  • It must contain the zero vector (like having a common ground in the neighborhood).
  • It must be closed under vector addition (neighbors can freely visit each other).
  • It must be closed under scalar multiplication (resizing your house still keeps it in the neighborhood).
For the subspace \( W \) in the exercise, these are the qualities that allow us to think of it as a stable entity under various manipulations by the linear transformations \( T_{1} \) and \( T_{2} \). And they are the precise qualities we utilize to prove that the sum and product of these transformations leave \( W \) as an invariant subspace. By proving closure under these combined operations, we're essentially confirming that \( W \) possesses the quintessential characteristics of a subspace under those circumstances.

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Most popular questions from this chapter

Let \(W\) be the solution space of the linear equation \\[ a_{1} x_{1}+a_{2} x_{2}+\dots+a_{n} x_{n}=0, \quad a_{i} \in K \\] and let \(v=\left(b_{1}, b_{2}, \ldots, b_{n}\right) \in K^{n} .\) Prove that the coset \(v+W\) of \(W\) in \(K^{n}\) is the solution set of the linear equation \\[ a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}=b, \quad \text { where } \quad b=a_{1} b_{1}+\cdots+a_{n} b_{n} \\].

Show that any Jordan nilpotent block matrix \(N\) is similar to its transpose \(N^{T}\) (the matrix with 1 's below the diagonal and \(0^{\prime}\) s elsewhere)

Suppose \(V=W_{1} \oplus \ldots \oplus W_{r} .\) Let \(E_{i}\) denote the projection of \(V\) into \(W_{i} .\) Prove (i) \(E_{i} E_{j}=0, i \neq j\) (ii) \(I=E_{1}+\cdots+E_{r}\)

Prove Theorem 10.11 on the Jordan canonical form for an operator \(T\) By the primary decomposition theorem, \(T\) is decomposable into operators \(T_{1}, \ldots, T_{r} ;\) that is, \(T=T_{1} \oplus \cdots \oplus T_{r},\) where \(\left(t-\lambda_{i}\right)^{m_{i}}\) is the minimal polynomial of \(T_{i} .\) Thus, in particular, \\[ \left(T_{1}-\lambda_{1} I\right)^{m_{1}}=\mathbf{0}, \ldots,\left(T_{r}-\lambda_{r} I\right)^{m_{r}}=\mathbf{0} \\] Set \(N_{i}=T_{i}-\lambda_{i} I .\) Then, for \(i=1, \ldots, r\) \\[ T_{i}=N_{i}+\lambda_{i} I, \quad \text { where } \quad N_{i}^{m^{i}}=\mathbf{0} \\] That is, \(T_{i}\) is the sum of the scalar operator \(\lambda_{i} I\) and a nilpotent operator \(N_{i},\) which is of index \(m_{i}\) because \(\left(t-\lambda_{i}\right)_{i}^{m}\) is the minimal polynomial of \(T_{i}\) Now, by Theorem 10.10 on nilpotent operators, we can choose a basis so that \(N_{i}\) is in canonical form. In this basis, \(T_{i}=N_{i}+\lambda_{i} I\) is represented by a block diagonal matrix \(M_{i}\) whose diagonal entries are the matrices \(J_{i j}\). The direct sum \(J\) of the matrices \(M_{i}\) is in Jordan canonical form and, by Theorem \(10.5,\) is a matrix representation of \(T\) Last, we must show that the blocks \(J_{i j}\) satisfy the required properties. Property (i) follows from the fact that \(N_{i}\) is of index \(m_{i} .\) Property (ii) is true because \(T\) and \(J\) have the same characteristic polynomial. Property (iii) is true because the nullity of \(N_{i}=T_{i}-\lambda_{i} I\) is equal to the geometric multiplicity of the eigenvalue \(\lambda_{i}\) Property (iv) follows from the fact that the \(T_{i}\) and hence the \(N_{i}\) are uniquely determined by \(T\)

Prove that the characteristic polynomial of an operator \(T: V \rightarrow V\) is a product of its elementary divisors.
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