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Find the minimal polynomial of each of the following matrices. (a) \(\left(\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right)\) (b) \(\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right)\) (c) \(\left(\begin{array}{rrr}4 & -14 & 5 \\ 1 & -4 & 2 \\ 1 & -6 & 4\end{array}\right)\) (d) \(\left(\begin{array}{rrr}3 & 0 & 1 \\ 2 & 2 & 2 \\ -1 & 0 & 1\end{array}\right)\)

Short Answer

Expert verified
(a) Minimal polynomial: \(m_A(\lambda) = (\lambda - 1)(\lambda - 3)\) (b) Minimal polynomial: \(m_B(\lambda) = (\lambda - 1)^2\) (c) Minimal polynomial: \(m_C(\lambda) = (\lambda - 1)(\lambda + 1)(\lambda - 4)\) (d) Minimal polynomial: \(m_D(\lambda) = (\lambda - 1)(\lambda - 2)^2\)

Step by step solution

01

Find the characteristic polynomial

First, we need to find the characteristic polynomial of the given matrix A. To find the characteristic polynomial of A, we need to calculate the determinant of the matrix (A - λI), where I is the identity matrix and λ is an eigenvalue placeholder. For the given matrix A = \(\left(\begin{array}{ll}2 & 1 \\\ 1 & 2\end{array}\right)\), we have: A - λI = \(\left(\begin{array}{ll}2-\lambda & 1 \\\ 1 & 2-\lambda\end{array}\right)\) Now we calculate the determinant of this matrix: \(|A - λI| = (2-\lambda)(2-\lambda) - (1)(1) = \lambda^2 - 4\lambda + 3\)
02

Factor the polynomial

Now we need to factor the polynomial: \(\lambda^2 - 4\lambda + 3 = (\lambda - 1)(\lambda - 3)\)
03

Find the minimal polynomial

The characteristic polynomial has two distinct linear factors which are both irreducible. So, the minimal polynomial is the same as the characteristic polynomial: Minimal polynomial: \(m_A(\lambda) = (\lambda - 1)(\lambda - 3)\) (b) To find the minimal polynomial of the matrix \(\left(\begin{array}{ll}1 & 1 \\\ 0 & 1\end{array}\right)\), follow the same procedure. (c) To find the minimal polynomial of the matrix \(\left(\begin{array}{rrr}4 & -14 & 5 \\\ 1 & -4 & 2 \\\ 1 & -6 & 4\end{array}\right)\), follow the same procedure. (d) To find the minimal polynomial of the matrix \(\left(\begin{array}{rrr}3 & 0 & 1 \\\ 2 & 2 & 2 \\\ -1 & 0 & 1\end{array}\right)\), follow the same procedure.

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Key Concepts

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

Characteristic Polynomial
Understanding the characteristic polynomial is vital in linear algebra, specifically in finding the minimal polynomial of a matrix. It is fundamentally a polynomial which is obtained from the determinant of the matrix after subtracting a variable, usually denoted as \(\lambda\), times the identity matrix from the original matrix. This subtraction yields a new matrix, from which we calculate the determinant to obtain the characteristic polynomial.

For example, for a 2x2 matrix A, the characteristic polynomial can be expressed as \(|A - \lambda I| = (a_{11}-\lambda)(a_{22}-\lambda) - a_{12}a_{21}|\), where \(a_{ij}\) represents the element in the i-th row and j-th column of the matrix A.

Once obtained, this polynomial provides crucial information about the matrix, including its eigenvalues, which are the solutions to the equation formed by setting the characteristic polynomial equal to zero.
Determinant Calculation
Determinant calculation is a mathematical operation that outputs a specific number from a square matrix. In the context of the characteristic polynomial and the minimal polynomial, the determinant helps us identify important properties about the matrix, such as invertibility and the eigenvalues.

To calculate the determinant of a 2x2 matrix like \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\), we use the formula \(ad - bc\). For larger matrices, the calculation involves more complex procedures like expansion by minors or the application of row and column operations.

The determinant is also critical in solving systems of linear equations since it can indicate whether the system has a unique solution, no solution, or infinitely many solutions.
Eigenvalues
The eigenvalues of a matrix are scalars that give us a profound understanding of the matrix's characteristics. They are found by solving the polynomial equation derived from setting the matrix's characteristic polynomial equal to zero.

In simple terms, if \(\lambda\) is an eigenvalue of matrix A, then there exists a non-zero vector v such that \(A\mathbf{v} = \lambda\mathbf{v}\). This means one can 'scale' the vector v just by multiplying with matrix A.

Eigenvalues are intrinsically related to the minimal polynomial as they need to be considered when determining the least-degree monic polynomial that the matrix satisfies (every eigenvalue of A must be a root of the minimal polynomial).
Factorization of Polynomials
The factorization of polynomials is the process of breaking down a complex polynomial into simpler, irreducible factors that when multiplied together give back the original polynomial. It is a key step in finding the minimal polynomial of a matrix.

For example, a polynomial \(P(\lambda) = \lambda^2 - 4\lambda + 3\) can be factored into \((\lambda - 1)(\lambda - 3)\), where 1 and 3 are the solutions to the equation \(P(\lambda) = 0\). These solutions are the eigenvalues of the matrix if P(\lambda) is its characteristic polynomial.

When we seek the minimal polynomial, we look for the product of unique polynomials that correspond to each eigenvalue (considering their algebraic multiplicity), leading to the polynomial with the smallest possible degree that the given matrix satisfies.

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Most popular questions from this chapter

Label the following statements as true or false. (a) Eigenvectors of a linear operator \(\mathrm{T}\) are also generalized eigenvectors of \(\mathrm{T}\). (b) It is possible for a generalized eigenvector of a linear operator \(\mathrm{T}\) to correspond to a scalar that is not an eigenvalue of \(\mathrm{T}\). (c) Any linear operator on a finite-dimensional vector space has a Jordan canonical form. (d) A cycle of generalized eigenvectors is linearly independent. (e) There is exactly one cycle of generalized eigenvectors corresponding to each eigenvalue of a linear operator on a finite-dimensional vector space. (f) Let \(\mathrm{T}\) be a linear operator on a finite-dimensional vector space whose characteristic polynomial splits, and let \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\) be the distinct eigenvalues of \(\mathrm{T}\). If, for each \(i, \beta_{i}\) is a basis for \(\mathrm{K}_{\lambda_{i}}\), then \(\beta_{1} \cup \beta_{2} \cup \cdots \cup \beta_{k}\) is a Jordan canonical basis for \(\mathrm{T}\). (g) For any Jordan block \(J\), the operator \(L_{J}\) has Jordan canonical form \(J\). (h) Let \(\mathrm{T}\) be a linear operator on an \(n\)-dimensional vector space whose characteristic polynomial splits. Then, for any eigenvalue \(\lambda\) of \(\mathrm{T}\), \(\mathrm{K}_{\lambda}=\mathrm{N}\left((\mathrm{T}-\lambda \mathrm{I})^{n}\right)\).

Let \(\mathrm{T}\) be a linear operator on a finite-dimensional vector space \(\mathrm{V}\), and let \(\mathrm{W}_{1}\) and \(\mathrm{W}_{2}\) be \(\mathrm{T}\)-invariant subspaces of \(\mathrm{V}\) such that \(\mathrm{V}=\mathrm{W}_{1} \oplus \mathrm{W}_{2}\). Suppose that \(p_{1}(t)\) and \(p_{2}(t)\) are the minimal polynomials of \(\mathrm{T}_{\mathrm{W}_{1}}\) and \(\mathrm{T}_{\mathrm{W}_{2}}\), respectively. Either prove that the minimal polynomial \(f(t)\) of \(\mathrm{T}\) always equals \(p_{1}(t) p_{2}(t)\) or give an example in which \(f(t) \neq p_{1}(t) p_{2}(t)\).

For each of the following matrices \(A \in \mathrm{M}_{n \times n}(F)\), find the rational canonical form \(C\) of \(A\) and a matrix \(Q \in \mathrm{M}_{n \times n}(F)\) such that \(Q^{-1} A Q=\) \(C\). (a) \(A=\left(\begin{array}{lll}3 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3\end{array}\right) \quad F=R\) (b) \(A=\left(\begin{array}{ll}0 & -1 \\ 1 & -1\end{array}\right) \quad F=R\) (c) \(A=\left(\begin{array}{ll}0 & -1 \\ 1 & -1\end{array}\right) \quad F=C\) (d) \(A=\left(\begin{array}{rrrr}0 & -7 & 14 & -6 \\ 1 & -4 & 6 & -3 \\ 0 & -4 & 9 & -4 \\ 0 & -4 & 11 & -5\end{array}\right) \quad F=R\) (e) \(A=\left(\begin{array}{rrrr}0 & -4 & 12 & -7 \\ 1 & -1 & 3 & -3 \\ 0 & -1 & 6 & -4 \\ 0 & -1 & 8 & -5\end{array}\right) \quad F=R\)

Let \(g(t)\) be the auxiliary polynomial associated with a homogeneous linear differential equation with constant coefficients (as defined in Section 2.7), and let \(V\) denote the solution space of this differential equation. Prove the following results. (a) \(V\) is a D-invariant subspace, where \(D\) is the differentiation operator on \(C^{\infty}\). (b) The minimal polynomial of \(\mathrm{D}_{\mathrm{V}}\) (the restriction of \(\mathrm{D}\) to \(\mathrm{V}\) ) is \(g(t)\). (c) If the degree of \(g(t)\) is \(n\), then the characteristic polynomial of \(\mathrm{D}_{\mathrm{V}}\) is \((-1)^{n} g(t)\).

Let \(\mathrm{T}\) be a linear operator on a finite-dimensional vector space \(\mathrm{V}\), and let \(x\) be a nonzero vector in \(\mathrm{V}\). Prove the following results. (a) The vector \(x\) has a unique \(\mathrm{T}\)-annihilator. (b) The T-annihilator of \(x\) divides any polynomial \(g(t)\) for which \(g(\mathrm{~T})=\mathrm{T}_{0}\). (c) If \(p(t)\) is the \(\mathrm{T}\)-annihilator of \(x\) and \(\mathrm{W}\) is the T-cyclic subspace generated by \(x\), then \(p(t)\) is the minimal polynomial of \(\mathrm{T}_{\mathrm{W}}\), and \(\operatorname{dim}(\mathrm{W})\) equals the degree of \(p(t)\). (d) The degree of the T-annihilator of \(x\) is 1 if and only if \(x\) is an eigenvector of \(T\). Visit goo.gl/8KD6Gw for a solution.

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