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Magazine Chapter 7: Problem 3
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 \(\mathrm{T}^{2}=1\).
Short Answer
Expert verified
The minimal polynomials of the linear operators are:
(a) \(p(x) = x^2 - 2\)
(b) \(p(x) = x^3 - 2x\)
(c) \(p(x) = (x - 1)^3 - 2(x - 1)\)
(d) \(p(x) = x^2 - 1\)
Step by step solution
01
(a) Matrix representation of T
First, we need to find the matrix representation of T. We can do this by finding the effect of T on the standard basis of \(\mathbf{R}^{2}\).
\[\mathrm{T}(1, 0) = (1, 1)\]
\[\mathrm{T}(0, 1) = (1, -1)\]
So the matrix representation of T is:
\[A_T = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\]
02
(a) Finding the minimal polynomial
To find the minimal polynomial, we need to find the smallest degree polynomial p(x) such that p(A_T) = 0.
Since \(\mathrm{V}\) is of dimension 2, the polynomial can have degree at most 2. We will first try a linear polynomial and then a quadratic polynomial if needed.
Linear polynomial \(p(x) = x - \lambda\), where \(\lambda\) is an eigenvalue of \(A_T\). However, the eigenvalues of \(A_T\) are \(i\) and \(-i\), and since we are working with real coefficients, a linear polynomial won't work.
Thus, we try a quadratic polynomial \(p(x) = x^2 - 2\).
\[p(A_T) = A_T^2 - 2I = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} - 2\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\]
Since p(A_T) = 0, the minimal polynomial of T is \(p(x) = x^2 - 2\).
For problems (b), (c), and (d), we will follow a similar approach by first finding the matrix representation and then finding the minimal polynomial.
03
(b) Matrix representation of T
Given that \(\mathrm{V}=\mathrm{P}_{2}(R)\), the basis can be chosen as \(\{1, x, x^2\}\). Now, we calculate the effect of T on those basis elements.
\[\mathrm{T}(1) = 2\]
\[\mathrm{T}(x) = 1 + 2x\]
\[\mathrm{T}(x^2) = 2x + 2x^2\]
So the matrix representation of T is:
\[A_T = \begin{pmatrix} 0 & 1 & 0 \\ 2 & 0 & 2 \\ 0 & 2 & 0 \end{pmatrix}\]
04
(b) Finding the minimal polynomial
The vector space is of dimension 3. The polynomial p(x) must have degree at most 3. By inspection, we find
\(p(x) = x^3 - 2x\):
\[p(A_T) = A_T^3 - 2A_T = \begin{pmatrix} 0 & 1 & 0 \\ 2 & 0 & 2 \\ 0 & 2 & 0 \end{pmatrix}^3 - 2\begin{pmatrix} 0 & 1 & 0 \\ 2 & 0 & 2 \\ 0 & 2 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\]
Thus, the minimal polynomial of T is \(p(x) = x^3 - 2x\).
05
(c) Matrix representation of T
For this problem, we have the same basis as in problem (b). Now, compute the effect of T on those basis elements.
\[\mathrm{T}(1) = 2\]
\[\mathrm{T}(x) = 1 - x\]
\[\mathrm{T}(x^2) = 2x - x^2\]
So the matrix representation of T is:
\[A_T = \begin{pmatrix} 0 & 1 & 0 \\ 2 & -1 & 2 \\ 0 & 1 & -1 \end{pmatrix}\]
06
(c) Finding the minimal polynomial
The vector space is of dimension 3. The polynomial p(x) must have degree at most 3. We find \(p(x) = (x-1)^3 - 2(x-1)\):
\[p(A_T) = (A_T-I)^3 - 2(A_T - I) = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 0 & 2 \\ 0 & -1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 2 & 0 \\ 0 & 0 & 2 \\ 0 & -1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 2 & 0 \\ 0 & 0 & 2 \\ 0 & -1 & 0 \end{pmatrix} - 2\begin{pmatrix} 1 & 2 & 0 \\ 0 & 0 & 2 \\ 0 & -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\]
Thus, the minimal polynomial of T is \(p(x) = (x - 1)^3 - 2(x - 1)\).
07
(d) Matrix representation of T
For this problem, we have an \(n \times n\) matrix, and the standard basis consists of matrices with a 1 at a single position and 0 everywhere else. The effect of T on each basis element is a transpose, so every element remains unchanged under T. Therefore, the matrix representation of T is the identity matrix.
08
(d) Using the hint to find the minimal polynomial
Given the hint \(T^2 = 1\), we find that the minimal polynomial is \(p(x) = x^2 - 1\). Since \(T^2 - I = 0\), our result is verified. Thus, the minimal polynomial of T is \(p(x) = x^2 - 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Operator
Understanding a linear operator is key to grasping many complex ideas in linear algebra. A linear operator is a rule or function that maps elements from one vector space to another (or the same space), while preserving the operations of vector addition and scalar multiplication.
In our example, the linear operator is defined by specific rules for transformation, such as \(T(a, b)=(a+b, a-b)\). To better comprehend this, think of a linear operator as a kind of machine: you feed a vector into it, and it spits out another vector after applying certain linear changes to the input.
For an operator to be considered linear, it must satisfy two conditions for all vectors \(u\) and \(v\) in the vector space and any scalar \(c\):
In our example, the linear operator is defined by specific rules for transformation, such as \(T(a, b)=(a+b, a-b)\). To better comprehend this, think of a linear operator as a kind of machine: you feed a vector into it, and it spits out another vector after applying certain linear changes to the input.
For an operator to be considered linear, it must satisfy two conditions for all vectors \(u\) and \(v\) in the vector space and any scalar \(c\):
- Additivity: \(T(u + v) = T(u) + T(v)\)
- Homogeneity: \(T(cu) = cT(u)\)
Matrix Representation
The matrix representation of a linear operator provides a concrete way to handle abstract linear transformations numerically. It is a powerful tool to simplify computation, especially when dealing with large vector spaces.
The matrix of a linear operator \(T\) with respect to a basis is constructed by applying \(T\) to each basis vector and recording the outputs as columns in a matrix. This is what we've seen in steps 1, 3, 5, and 7 of the solution, where the effect of \(T\) is calculated on each basis element. The resulting matrix provides a blueprint for how \(T\) acts on the entire space.
For instance, with our first example, the matrix \[A_T = \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix}\] reflects the rules of our linear operator in a way that鈥檚 easy to work with using standard matrix operations. Having this representation enables us to solve various problems, like finding eigenvalues or the minimal polynomial, by utilizing matrices instead of dealing directly with abstract linear transformations.
The matrix of a linear operator \(T\) with respect to a basis is constructed by applying \(T\) to each basis vector and recording the outputs as columns in a matrix. This is what we've seen in steps 1, 3, 5, and 7 of the solution, where the effect of \(T\) is calculated on each basis element. The resulting matrix provides a blueprint for how \(T\) acts on the entire space.
For instance, with our first example, the matrix \[A_T = \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix}\] reflects the rules of our linear operator in a way that鈥檚 easy to work with using standard matrix operations. Having this representation enables us to solve various problems, like finding eigenvalues or the minimal polynomial, by utilizing matrices instead of dealing directly with abstract linear transformations.
Eigenvalues
Let鈥檚 delve into eigenvalues, a fascinating and deeply revealing concept in linear algebra. Eigenvalues are scalars associated with a given linear operator or matrix, indicating if there is a non-zero vector that remains on the same line after the transformation 鈥� essentially, the vector is only scaled by the linear operator and not knocked off course.
The term \(\lambda\) often denotes an eigenvalue, and it is typically discovered by solving the characteristic equation \(det(A - \lambda I) = 0\), where \(A\) is a matrix and \(I\) is the identity matrix. In our solution steps, finding the eigenvalues was an intermediate step before we could ascertain the minimal polynomial.
Eigenvalues are central to many areas of linear algebra and beyond, such as in stability analysis, where they can determine the dynamics of a system. In our example, the attempt to use linear polynomials derived from eigenvalues did not pan out, thus a quadratic polynomial was needed to determine the minimal polynomial.
The term \(\lambda\) often denotes an eigenvalue, and it is typically discovered by solving the characteristic equation \(det(A - \lambda I) = 0\), where \(A\) is a matrix and \(I\) is the identity matrix. In our solution steps, finding the eigenvalues was an intermediate step before we could ascertain the minimal polynomial.
Eigenvalues are central to many areas of linear algebra and beyond, such as in stability analysis, where they can determine the dynamics of a system. In our example, the attempt to use linear polynomials derived from eigenvalues did not pan out, thus a quadratic polynomial was needed to determine the minimal polynomial.
