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Magazine Chapter 9: Problem 28
Find the minimal polynomial \(m(t)\) of each of the following matrices: (a) \(A=\left[\begin{array}{ll}5 & 1 \\ 3 & 7\end{array}\right]\) (b) \(B=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 0 & 3\end{array}\right]\) \((\mathrm{c}) \quad C=\left[\begin{array}{rr}4 & -1 \\ 1 & 2\end{array}\right]\)
Short Answer
Expert verified
The minimal polynomials for the given matrices are:
(a) \(m(t) = t^2 - 12t + 32\)
(b) \(m(t) = (t-1)(t-2)(t-3)\)
(c) \(m(t) = t^2 - 6t + 9\)
Step by step solution
01
Write down the minimal polynomial equation
$
m(t) = t^2 - (\mathrm{tr}(A))t + \mathrm{det}(A)
$
02
Choose a basis for the minimal polynomial
Since the minimal polynomial must have a degree less than or equal to the degree of the matrix (in this case 2), we can choose a basis for the minimal polynomial as \((1, t)\).
03
Verify that matrix A satisfies the minimal polynomial equation
Plug in matrix A into the minimal polynomial equation to get:
$$
m(A) = A^2 - (\mathrm{tr}(A))A + \mathrm{det}(A)I
$$
We need to verify whether this expression simplifies to the zero matrix. Calculate the following:
1. Trace of A: \( \mathrm{tr}(A) = 5 + 7 = 12 \)
2. Determinant of A: \( \mathrm{det}(A) = (5)(7) - (3)(1) = 35 - 3 = 32 \)
3. Compute \(m(A)\):
$$
m(A) = A^2 - 12A + 32I = \left[\begin{array}{ll}11 & 9 \\\ 9 & 35\end{array}\right] - 12\left[\begin{array}{ll}5 & 1 \\\ 3 & 7\end{array}\right] + 32\left[\begin{array}{ll}1 & 0\\\\ 0 & 1\end{array}\right]
$$
$$
m(A) = \left[\begin{array}{ll}11 & 9 \\\ 9 & 35\end{array}\right] - \left[\begin{array}{ll}60 & 12 \\\ 36 & 84\end{array}\right] + \left[\begin{array}{ll}32 & 0 \\\ 0 & 32\end{array}\right]
$$
$$
m(A) = \left[\begin{array}{ll}0 & 0 \\\ 0 & 0\end{array}\right]
$$
04
Determine the minimal polynomial using the chosen basis
Since matrix A satisfies the minimal polynomial equation and we have verified that the expression simplifies to the zero matrix, the minimal polynomial for matrix A is:
$$
m(t) = t^2 - 12t + 32
$$
(b) Find the minimal polynomial of matrix B:
$$
B=\left[\begin{array}{lll}1 & 2 & 3 \\\ 0 & 2 & 3 \\\ 0 & 0 &
3\end{array}\right]
$$
Notice that matrix B is already in upper triangular form. The minimal polynomial of an upper triangular matrix has the form \(m(t)=(t-\lambda_1)(t-\lambda_2)...(t-\lambda_n)\), \(\lambda_i\) are diagonal entries. Therefore, minimal polynomial for B:
$$
m(t) = (t-1)(t-2)(t-3)
$$
(c) Find the minimal polynomial of matrix C:
$$
C=\left[\begin{array}{rr}4 & -1 \\\ 1 &
2\end{array}\right]
$$
05
Write down the minimal polynomial equation
$
m(t) = t^2 - (\mathrm{tr}(C))t + \mathrm{det}(C)
$
06
Choose a basis for the minimal polynomial
Since the minimal polynomial must have a degree less than or equal to the degree of the matrix (in this case 2), we can choose a basis for the minimal polynomial as \((1, t)\).
07
Verify that matrix C satisfies the minimal polynomial equation
Plug in matrix C into the minimal polynomial equation to get:
$$
m(C) = C^2 - (\mathrm{tr}(C))C + \mathrm{det}(C)I
$$
We need to verify whether this expression simplifies to the zero matrix. Calculate the following:
1. Trace of C: \( \mathrm{tr}(C) = 4 + 2 = 6 \)
2. Determinant of C: \( \mathrm{det}(C) = (4)(2) - (1)(-1) = 8 + 1 = 9 \)
3. Compute \(m(C)\):
$$
m(C) = C^2 - 6C + 9I = \left[\begin{array}{ll}11 & -3 \\\ 3 & 5\end{array}\right] - 6\left[\begin{array}{ll}4 & -1 \\\ 1 & 2\end{array}\right] + 9\left[\begin{array}{ll}1 & 0\\\\ 0 & 1\end{array}\right]
$$
$$
m(C) = \left[\begin{array}{ll}11 & -3 \\\ 3 & 5\end{array}\right] - \left[\begin{array}{ll}24 & -6 \\\ 6 & 12\end{array}\right] + \left[\begin{array}{ll}9 & 0 \\\ 0 & 9\end{array}\right]
$$
$$
m(C) = \left[\begin{array}{ll}0 & 0 \\\ 0 & 0\end{array}\right]
$$
08
Determine the minimal polynomial using the chosen basis
Since matrix C satisfies the minimal polynomial equation and we have verified that the expression simplifies to the zero matrix, the minimal polynomial for matrix C is:
$$
m(t) = t^2 - 6t + 9
$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trace of a Matrix
The trace of a matrix is a straightforward yet powerful concept in linear algebra. It is simply the sum of the diagonal elements of a square matrix. For example, if we have a matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), then the trace is \( \mathrm{tr}(A) = a + d \). This sum provides insight into certain properties of matrices.
Here are some important applications and properties:
Here are some important applications and properties:
- **Relation to Eigenvalues:** The trace of a matrix is equal to the sum of its eigenvalues. This makes the trace useful in determining the characteristic polynomial of a matrix.
- **Invariance Under Cyclic Permutations:** If matrices have the same dimensions, then the trace of their product remains the same under cyclic permutations. For instance, \( \mathrm{tr}(AB) = \mathrm{tr}(BA) \).
- **Applications in Physics:** The trace is used in various fields, including quantum mechanics and statistics, where it often represents measurable quantities.
Determinant of a Matrix
The determinant of a matrix is a value that can be computed from its elements and is used frequently in linear algebra and calculus. For a 2x2 matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant is calculated as \( \mathrm{det}(A) = ad - bc \).
Some crucial aspects of determinants include:
Some crucial aspects of determinants include:
- **Existence of Inverse:** A square matrix has an inverse if and only if its determinant is non-zero. If the determinant is zero, the matrix is singular, meaning it does not have an inverse.
- **Relation to Area/Volume:** In two or three dimensions, the absolute value of the determinant of a matrix formed by vectors corresponds to the area or volume of the parallelepiped spanned by those vectors. It reflects how the matrix transforms space.
- **Change Under Elementary Row Operations:** Row operations alter the determinant in predictable ways:
- Swapping two rows changes the sign of the determinant.
- Multiplying a row by a scalar multiplies the determinant by that scalar.
- Adding a multiple of one row to another does not change the determinant.
Upper Triangular Matrix
An upper triangular matrix is a special kind of square matrix where all elements below the main diagonal are zero. For a 3x3 matrix \( \begin{pmatrix} a & b & c \ 0 & e & f \ 0 & 0 & g \end{pmatrix} \), it is upper triangular if all elements below the diagonal (those in positions (2,1) and (3,1), (3,2)) are zeros.
Key features and applications include:
Key features and applications include:
- **Simplified Calculations:** It is easier to compute the determinant of an upper triangular matrix because it is simply the product of its diagonal elements \( a \cdot e \cdot g \). This simplifies many calculations in linear algebra.
- **Gaussian Elimination:** Many matrices can be transformed into an upper triangular form using Gaussian elimination, which helps in solving systems of linear equations.
- **Eigenvalues:** The eigenvalues of an upper triangular matrix are directly the entries on its main diagonal. This property expedites the process of finding eigenvalues.
- **Stability in Numerical Methods:** Algorithms that require stability often prefer matrices in upper triangular form due to their predictable nature and reduced computational complexity.
