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Chapter 7: Problem 5

Describe all linear operators_ \(T\) on \(R^{2}\) such that \(T\) is diagonalizable and \(\mathrm{T}^{3}-2 \mathrm{~T}^{2}+\mathrm{T}=\mathrm{T}_{0}\).

Short Answer

Expert verified
In summary, the diagonalizable linear operators T on \(R^2\) that satisfy the equation \(T^{3}-2T^{2}+T=T_{0}\) are those with the characteristic polynomial \(\lambda(\lambda^2-3\lambda+2)=0\), yielding eigenvalues \(\lambda_1=0\), \(\lambda_2=1\), and \(\lambda_3=2\). These operators have a minimal polynomial that is a product of distinct linear factors, confirming their diagonalizability.

Step by step solution

01

Find the characteristic polynomial of T

Let \(\lambda\) be an eigenvalue of T. Then \((T - \lambda I)v = 0\) for some nonzero vector \(v\), where \(I\) is the identity operator. Multiplying this equation with \((T^2 - 2T + I)\) from the left, we get \((T^2 - 2T + I)(T - \lambda I)v = (T^3 - 2T^2 + T)v = T_0v = 0\) Since \(v\) is nonzero, we can divide by \(v\) and obtain the characteristic polynomial of T: \((T - \lambda I)(T^2 - 2T + I) = T^3 - \lambda T^2 -(2 - \lambda)T + \lambda(\lambda - 1)I = 0\)
02

Find the eigenvalues of T

We set the determinant of the characteristic polynomial to zero, i.e., \(det(T^3 - \lambda T^2 -(2 - \lambda)T + \lambda(\lambda - 1)I) = 0\) Calculating the determinant, we get \(\lambda(\lambda^2 - 3\lambda + 2) = 0 \) This gives us three eigenvalues: \(\lambda_1 = 0\), \(\lambda_2 = 1\), and \(\lambda_3 = 2\).
03

Check diagonalizability

As the minimal polynomial of T is \((\lambda - 0)(\lambda - 1)(\lambda - 2)\), which is a product of distinct linear factors, T is diagonalizable.
04

Verify the equation

Now we can substitute the operator T in the given equation to verify if it holds true for all the diagonalizable matrices. \((T^3 - 2T^2 + T)v = (T^2(T - 2I) + I)v = 0\) Since T is diagonalizable, and the equation holds true for all eigenvectors, it also holds true for all the vectors in R^2. Thus, we've described all such linear operators T on R^2 that are diagonalizable and satisfy the given equation.

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Key Concepts

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

Characteristic Polynomial
The characteristic polynomial of a linear operator is a fundamental concept in linear algebra as it encapsulates important information about the operator, particularly its eigenvalues. To put it simply, the characteristic polynomial of a matrix or linear operator is obtained by taking the determinant of the matrix minus a scalar multiple of the identity matrix and setting it to zero. In mathematical terms, for a linear operator T, the characteristic polynomial 辫(位) is given by the equation:

det(T - 位I) = 辫(位) = 0

This polynomial equation in 位 will give us the eigenvalues as its roots. Each eigenvalue provides a glimpse into the action of the operator on certain non-zero vectors, known as eigenvectors, which are scaled by these eigenvalues. In the exercise at hand, the characteristic polynomial emerges from a specific operation involving T, and it is critical for understanding how T can be dissected into simpler parts for analysis and application.
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are pairs that play a starring role in the study of linear operators. An eigenvalue, represented often as 位, is a special scalar that indicates how much an eigenvector is stretched or shrunk when a linear transformation is applied.

For a linear operator T and a non-zero vector v, if Tv = 位v, then v is an eigenvector and 位 is the corresponding eigenvalue. Finding these pairs is essential when dealing with diagonalizable operators, as they allow the transformation to be simplified into a diagonal form when employing a suitable basis - typically the set of eigenvectors. This makes computations and further analysis far easier because it decouples the transformation into independent scaling along these new axes, provided by the eigenvectors.

In the given exercise, knowing the eigenvalues let us derive the minimal polynomial, which helps to check if the operator T is diagonalizable and confirms whether the given equation holds true after transformation.
Minimal Polynomial
The minimal polynomial of a linear operator offers a more refined look at the operator's structure beyond the characteristic polynomial. It is the monic polynomial (leading coefficient of 1) of least degree that T satisfies. Simply put, if you think of the characteristic polynomial as a characteristic DNA of the linear operator revealing potentials (eigenvalues) for transformation, then the minimal polynomial reflects the most 'economical' version of this DNA that still fully defines the operator鈥檚 behavior.

In the context of diagonalizability, if all roots of the minimal polynomial are distinct, we assure that the linear operator T can be diagonalized. This fact is a cornerstone in linear algebra, as it provides a practical method to determine if we can simplify the operator into a very manageable form. For the exercise provided, the minimal polynomial reveals that the operator T is indeed diagonalizable and will conform to the specified equation. This underscores the minimal polynomial's role in identifying key properties for simplifying and solving linear algebra problems.

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Most popular questions from this chapter

The following definitions are used in Exercises 11-19. Definitions. A linear operator \(\mathrm{T}\) on a vector space \(\mathrm{V}\) is called nilpotent if \(\mathrm{T}^{p}=\mathrm{T}_{0}\) for some positive integer \(p .\) An \(n \times n\) matrix \(A\) is called nilpotent if \(A^{p}=O\) for some positive integer \(p\). Prove the converse of Exercise \(13(\mathrm{~d})\) : If \(\mathrm{T}\) is a linear operator on an \(n\) dimensional vector space \(\mathrm{V}\) and \((-1)^{n} t^{n}\) is the characteristic polynomial of \(\mathrm{T}\), then \(\mathrm{T}\) is nilpotent.

Let \(V\) be a vector space and \(\beta_{1}, \beta_{2}, \ldots, \beta_{k}\) be disjoint subsets of \(V\) whose union is a basis for \(\mathrm{V}\). Now suppose that \(\gamma_{1}, \gamma_{2}, \ldots, \gamma_{k}\) are linearly independent subsets of \(\mathrm{V}\) such that \(\operatorname{span}\left(\gamma_{i}\right)=\operatorname{span}\left(\beta_{i}\right)\) for all \(i\). Prove that \(\gamma_{1} \cup \gamma_{2} \cup \cdots \cup \gamma_{k}\) is also a basis for \(\mathrm{V}\).

Exercises 13 and 14 are concerned with direct sums of matrices, defined in Section \(5.4\) on page 318 . Let \(\mathrm{T}\) be a linear operator on a finite-dimensional vector space \(\mathrm{V}\) such that the characteristic polynomial of \(\mathrm{T}\) splits, and let \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\) be the distinct eigenvalues of \(\mathrm{T}\). For each \(i\), let \(J_{i}\) be the Jordan canonical form of the restriction of \(\mathrm{T}\) to \(\mathrm{K}_{\lambda_{i}}\). Prove that $$ J=J_{1} \oplus J_{2} \oplus \cdots \oplus J_{k} $$ is the Jordan canonical form of \(J\).

For each linear operator \(T\) on \(V\), find the minimal polynomial of \(T\). (a) \(\mathbf{V}=\mathbf{R}^{2}\) and \(\mathrm{T}(a, b)=(a+b, a-b)\) (b) \(\mathrm{V}=\mathrm{P}_{2}(R)\) and \(\mathrm{T}(g(x))=g^{\prime}(x)+2 g(x)\) (c) \(\mathrm{V}=\mathrm{P}_{2}(R)\) and \(\mathrm{T}(f(x))=-x f^{\prime \prime}(x)+f^{\prime}(x)+2 f(x)\) (d) \(\mathrm{V}=\mathrm{M}_{n \times n}(R)\) and \(\mathrm{T}(A)=A^{t}\). Hint: Note that \(\mathrm{T}^{2}=1\).

Let \(\mathrm{T}\) be a linear operator on a finite-dimensional vector space. Prove that the rational canonical form of \(T\) is a diagonal matrix if and only if \(T\) is diagonalizable. Visit goo.gl/tK8pru for a solution.
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