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Chapter 9: Problem 10

If \(V\) is finite dimensional, show that a linear operator \(T\) on \(V\) has an inverse if and only if det \(T \neq 0\)

Short Answer

Expert verified
A linear operator \( T \) is invertible if and only if \( \text{det}(T) \neq 0 \).

Step by step solution

01

Define the Problem

We need to show that a linear operator \( T \) on a finite-dimensional vector space \( V \) is invertible if and only if its determinant is not zero (i.e., \( \text{det}(T) eq 0 \)).
02

Determine When T is Invertible

A linear operator \( T \) on \( V \) is invertible if there exists another linear operator \( S \) such that \( ST = TS = I \), where \( I \) is the identity operator on \( V \). This means that for every vector in \( V \), there is a unique solution to the equation \( T(v) = w \) for any \( w \) in \( V \).
03

Apply Determinant Properties on Invertibility

The determinant of a linear operator \( T \), represented by a matrix with respect to a basis, is a scalar real number. If \( \text{det}(T) = 0 \), it means that \( T \) maps at least one non-zero vector to the zero vector, making \( T \) non-invertible. Conversely, if \( \text{det}(T) eq 0 \), \( T \) has full rank and is invertible.
04

Prove \( \text{det}(T) = 0 \) Implies Non-Invertibility

Assume \( \text{det}(T) = 0 \). We know from linear algebra that the kernel of \( T \) (\( \text{ker}(T) \)) is non-trivial, meaning there exist non-zero vectors \( v \) such that \( T(v) = 0 \). Thus, \( T \) is not injective, and therefore cannot be invertible.
05

Prove \( \text{det}(T) \neq 0 \) Implies Invertibility

If \( \text{det}(T) eq 0 \), \( T \) has full rank, meaning it has no non-zero vectors that map to zero (\( \text{ker}(T) = \{0\} \)). This implies that \( T \) is injective and also surjective because \( V \) is finite-dimensional. Hence, \( T \) is invertible.

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Key Concepts

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

Determinant: What Does It Tell Us?
The determinant is a key concept in linear algebra, acting as a bridge between matrices and the properties of the transformations they represent. Simply put, the determinant is a scalar value. You can calculate it for a square matrix, and it holds a very significant meaning!
  • If the determinant of a matrix is zero (\( ext{det}(T) = 0 \)), it indicates that the transformation represented by the matrix squashes the space into a lower dimension. Essentially, this means that the matrix maps vectors into a smaller subspace, often resulting in a loss of information.
  • If the determinant is non-zero (\( ext{det}(T) eq 0 \)), it indicates that the transformation preserves the dimensionality. This means that it does not compress the space into a lower dimension.
Thus, the determinant helps us understand the nature of the transformation: whether it's compressing space or keeping it intact. This understanding is crucial when we discuss matrix invertibility.
Invertibility: More Than Just Reversibility
In linear algebra, saying that a matrix or linear operator is invertible means you can reverse the transformation it performs. This is a pretty big deal!
  • An invertible operator, such as a matrix with \( ext{det}(T) eq 0 \), means that every output vector is mapped from exactly one distinct input vector. In other words, you can "undo" the transformation completely!
  • On the flip side, if an operator is not invertible (\( ext{det}(T) = 0 \)), some vectors might get "mapped together," resulting in overlap and making it impossible to reverse the process fully.
This concept is not just theoretical; it has practical applications in many fields, such as solving systems of linear equations. For instance, if you want to solve a system using matrix operations, you need the coefficient matrix to be invertible. Thus, having a non-zero determinant becomes a gateway to solving such systems efficiently.
Finite-Dimensional Vector Space: Foundations of Linear Operations
A finite-dimensional vector space, such as \( V \) in linear algebra, is simply a vector space that has a limited number of basis vectors. Here's what makes it interesting:
  • Finite-dimensionality ensures that every vector in the space can be expressed as a linear combination of a fixed number of vectors, known as the basis.
  • When dealing with operators like \( T \), within a finite-dimensional vector space, calculations like those involving determinants become well-defined and manageable. This is because matrices representing these operators will have fixed dimensions matched to the space.
  • Moreover, the finite nature of the space means that concepts like linear independence, basis, and dimension play into how operators behave, often simplifying what might otherwise be complex problems in infinite-dimensional contexts.
Understanding finite-dimensional spaces is crucial for studying linear operators, as they provide a structured framework within which we can define, analyze, and apply these operators effectively.

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Most popular questions from this chapter

In each case, show that \(U\) is \(T\) -invariant, use it to find a block upper triangular matrix for \(T,\) and use that to compute \(c_{T}(x)\). a. \(T: \mathbf{P}_{2} \rightarrow \mathbf{P}_{2}\) $$\begin{aligned} T(&\left.a+b x+c x^{2}\right) \\ &=(-a+2 b+c)+(a+3 b+c) x+(a+4 b) x^{2} \\ U &=\operatorname{span}\left\\{1, x+x^{2}\right\\} \end{aligned}$$ b. \(T: \mathbf{P}_{2} \rightarrow \mathbf{P}_{2}\) $$\begin{aligned} T &\left(a+b x+c x^{2}\right) \\ &=(5 a-2 b+c)+(5 a-b+c) x+(a+2 c) x^{2} \\ U&=\operatorname{span}\left\\{1-2 x^{2}, x+x^{2}\right\\} \end{aligned}$$

If \(A=\left[\begin{array}{rrrr}2 & -5 & 0 & 0 \\ 1 & -2 & 0 & 0 \\ 0 & 0 & -1 & -2 \\ 0 & 0 & 1 & 1\end{array}\right],\) show that \(T_{A}: \mathbb{R}^{4} \rightarrow \mathbb{R}^{4}\) has two-dimensional \(T\) -invariant subspaces \(U\) and \(W\) such that \(\mathbb{R}^{4}=U \oplus W,\) but \(A\) has no real eigenvalue.

Let \(E\) be an \(n \times n\) matrix with \(E^{2}=E .\) Show that \(\mathbf{M}_{n n}=U \oplus W,\) where \(U=\\{A \mid A E=A\\}\) and \(W=\\{B \mid B E=0\\} .[\) Hint \(: X E\) lies in \(U\) for every matrix \(X .]\)

Find the standard matrix of the rotation \(R\) about the line through the origin with direction vector \(\mathbf{d}=\left[\begin{array}{lll}2 & 3 & 6\end{array}\right]^{T} \cdot\left[\right.\) Hint: Consider \(\mathbf{f}=\left[\begin{array}{ccc}6 & 2 & -3\end{array}\right]^{T}\) and \(\left.\mathbf{g}=\left[\begin{array}{lll}3 & -6 & 2\end{array}\right]^{T} .\right]\)

Let \(B=\left\\{\mathbf{b}_{1}, \quad \mathbf{b}_{2}, \ldots, \mathbf{b}_{n}\right\\}\) be any ordered basis of \(\mathbb{R}^{n}\), written as columns. If \(Q=\left[\begin{array}{llll}\mathbf{b}_{1} & \mathbf{b}_{2} & \cdots & \mathbf{b}_{n}\end{array}\right]\) is the matrix with the \(\mathbf{b}_{i}\) as columns, show that \(Q C_{B}(\mathbf{v})=\mathbf{v}\) for all \(\mathbf{v}\) in \(\mathbb{R}^{n}\).
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