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Chapter 12: Problem 4

We let \(V\) be a finite dimensional vector space over the complex numbers, and we let \(A: V \rightarrow V\) be an operator. If \(A\) is nilpotent and \(\operatorname{not} O\), show that \(A\) is not diagonalizable.

Short Answer

Expert verified
Since A is nilpotent, \(\lambda = 0\) is its only eigenvalue. Suppose for contradiction that A has a basis of eigenvectors \(\{v_1, v_2, ..., v_n\}\), then \(Av_i = 0v_i = 0\) for all i, implying that \(A = 0\), which contradicts A being non-zero. Therefore, A cannot have a basis of eigenvectors and is not diagonalizable.

Step by step solution

01

Define diagonalizable and nilpotent operators

For an operator A: 1. Diagonalizable: Operator A is called diagonalizable if there exists an ordered basis of V consisting of eigenvectors of A. 2. Nilpotent: Operator A is called nilpotent if there exists a positive integer k such that \(A^k = 0\).
02

Eigenvalues of nilpotent operator

Let's analyze the eigenvalues of a nilpotent operator. Let \(\lambda\) be an eigenvalue of A and v be the corresponding eigenvector. Then, we know that \(A^k = 0\) for some positive integer k => \((A^k)v = 0\). Since \(Av = \lambda v\), we get \((A^2)v = A(\lambda v) = \lambda Av = \lambda^2v\), and similarly, we can generalize this pattern to get \((A^k)v = \lambda^k v\). Combining our results, we get \(\lambda^k v = (A^k)v = 0\). Since v is a non-zero eigenvector, we conclude that \(\lambda^k = 0\) and therefore, \(\lambda = 0\) is the only eigenvalue of A, given that A is a nilpotent operator.
03

Show that A cannot have a basis of eigenvectors

Suppose for contradiction that A has a basis of eigenvectors. Denote this basis by \(\{v_1, v_2, ..., v_n\}\). Since each of these is an eigenvector, there exist \(\lambda_i\) for \(i = 1, 2, ..., n\) such that \(Av_i = \lambda_i v_i\). But from Step 2, we know that \(\lambda_i = 0\) for all i. Thus, for each vector in the basis, \(Av_i = 0v_i = 0\). So, applying A to any vector in the basis gives us the zero vector. Therefore, \(A = 0\), which contradicts the assumption that A is not the zero operator. Hence, operator A cannot have a basis of eigenvectors.
04

Conclude that A is not diagonalizable

As we have shown that A cannot have a basis of eigenvectors, we conclude that A is not diagonalizable.

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Key Concepts

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

Diagonalizable Operator
A diagonalizable operator is an operator that can be represented as a diagonal matrix in some basis. This means that the operator has enough eigenvectors to form a basis for the vector space it operates on.
  • If an operator is diagonalizable, there exists a basis in which the operator behaves just like scaling each basis vector by its corresponding eigenvalue.
  • In terms of matrices, being diagonalizable means that a matrix is similar to a diagonal matrix. This implies it can be written in the form \(PDP^{-1}\), where \(D\) is a diagonal matrix and \(P\) is an invertible matrix comprising of eigenvectors as its columns.
  • Diagonalizability simplifies many operations, such as raising matrices to powers, since powers of a diagonal matrix are just powers of its diagonal entries.
Diagonalizability is a powerful tool because it allows you to easily compute functions of operators if they have a diagonal form. However, not all operators have enough distinct eigenvectors to form a basis, and thus are not diagonalizable. This is particularly the case for nilpotent operators which, as we determined, have all zero eigenvalues.
Eigenvalues
Eigenvalues are special scalars associated with a linear operator. They are essentially the factors by which the operator stretches its eigenvectors.
  • If \(A\) is a linear operator acting on a vector space \(V\), and \(v\) is a non-zero vector in \(V\) such that \(Av = \lambda v\), then \(\lambda\) is called an eigenvalue of \(A\), and \(v\) is the corresponding eigenvector.
  • Eigenvalues tell us about the scaling and rotation effects the operator has on the space.
  • In the case of nilpotent operators, all eigenvalues are zero. This is because a nilpotent operator satisfies \(A^k = 0\) for some positive integer \(k\), leading to all potential eigenvalues \(\lambda\) being zero (since \(\lambda^k = 0\) implies \(\lambda = 0\)).
Eigenvalues are critical in determining whether an operator can be diagonalized. If the operator doesn't have enough distinct eigenvalues, it won't have enough eigenvectors to form a basis, which makes it non-diagonalizable.
Finite Dimensional Vector Space
A finite dimensional vector space is a vector space that has a finite basis or, equivalently, a finite number of dimensions.
  • This means there is a set of vectors such that every vector in the space can be expressed as a combination of these basis vectors, and importantly, there is a finite number of these basis vectors.
  • Any vector space is finite dimensional if it is isomorphic to \(\mathbb{C}^n\) for some integer \(n\), in the case of complex numbers, reflecting the fact that it can be described completely by \(n\) coordinates.
  • Operations on finite dimensional spaces, like computing eigenvalues, applying linear operators, and discussing diagonalization, tend to be more manageable compared to infinite dimensional spaces.
In a finite dimensional space, concepts like diagonalization are crucial because they can often convert the operator into a form where calculations become straightforward. Dealing with finite dimensions simplifies many theoretical and practical problems, such as solving systems of linear equations.

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Most popular questions from this chapter

Let \(V=\mathbf{R}^{2}\), let \(S\) consist of the matrix \(\left(\begin{array}{ll}1 & a \\ 0 & 1\end{array}\right)\) viewed as linear map of \(V\) into itself. Here, \(a\) is a fixed non-zero real number. Determine all \(S\) -invariant subspaces of \(V\).

Show that \(t^{n}-1\) is divisible by \(t-1\).

Let \(f, g\) be two polynomials, written in the form $$ f=p_{1}^{i_{1}} \cdots p_{r}^{i} $$ and $$ g=p_{1}^{j_{1}} \cdots p_{r}^{j_{r}} $$ where \(i_{n}, j\), are integers \(\geqq 0\), and \(p_{1}, \ldots, p\), are distinet irreducible polynomials. (a) Show that the greatest common divisor of \(f\) and \(g\) can be expressed as a product \(p_{1}^{k_{1}} \cdots p_{r}^{k,}\) where \(k_{1}, \ldots, k_{r}\) are integers \(\geqq 0 .\) Express \(k_{p}\) in terms of \(i\), and \(j_{r}\) (b) Define the least common multiple of polynomials, and express the least common multiple of \(f\) and \(g\) as a product \(p_{1}^{k_{1}} \cdots p_{r}^{k_{r}}\) with integers \(k, \geq 0 .\) Express \(k\), in terms of \(i\), and \(j_{r}\)

Let \(A\) be an \(n \times n\) matrix over a field \(K\), and let \(J\) be the set of all polynomials \(f(t)\) in \(K[t]\) such that \(f(A)=0 .\) Show that \(J\) is an ideal.

Let \(f(t)=t^{n}+\cdots+a_{0}\) be a polynomial with complex coefficients, of degree \(n\), and let \(\alpha\) be a root. Show that \(|\alpha| \leqq n \cdot \max _{i}\left|a_{i}\right| .\) [Hint: Write \(-\alpha^{n}=a_{n-1} \alpha^{n-1}+\cdots+a_{0} .\) If \(|\alpha|>n \cdot \max _{i}\left|a_{i}\right|\), divide by \(\alpha^{n}\) and take the absolute value, together with a simple estimate to get a contradiction.]
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