Question

Let \(A = [a_{ij}]\) be a \(3 \times 3\) real matrix such that
\[ A \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} = 2 \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}, \quad A \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = 2 \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, \quad \text{and} \quad A \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} = 4 \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}. \]
If \(m\) is the degree of the minimal polynomial of \(A\), then \(a_{11} + a_{21} + a_{31} + m\) equals ..................

Hint:

When a matrix's action on a set of basis vectors (in this case, eigenvectors) is known, you can find its action on any vector by expressing that vector as a linear combination of the basis vectors. This avoids having to explicitly construct the matrix A.

Answer 1

Step 1: Understanding the Concept:
This problem involves finding eigenvalues and eigenvectors, determining the minimal polynomial of a matrix, and calculating specific elements of the matrix. The given equations are in the form \(Av = \lambda v\), which define the eigenvalues and eigenvectors of A.
Step 2: Key Formula or Approach:
1. Identify the eigenvalues and eigenvectors from the given equations.
2. Determine the minimal polynomial. The minimal polynomial has the same roots as the characteristic polynomial. If the geometric multiplicity of each eigenvalue equals its algebraic multiplicity, the matrix is diagonalizable, and the minimal polynomial has distinct linear factors.
3. The first column of A is the vector \(A e_1\), where \(e_1 = [1, 0, 0]^T\).
4. Express \(e_1\) as a linear combination of the eigenvectors.
5. Use the property \(A(\sum c_i v_i) = \sum c_i A v_i = \sum c_i \lambda_i v_i\) to find \(A e_1\).
6. Sum the components of \(A e_1\) and add the degree of the minimal polynomial.
Step 3: Detailed Explanation or Calculation:
Finding Eigenvalues and Minimal Polynomial (m):
From \(A v_1 = 2 v_1\) with \(v_1 = [1, 2, 1]^T\), we have an eigenvalue \(\lambda_1 = 2\).
From \(A v_2 = 2 v_2\) with \(v_2 = [0, 1, 1]^T\), we have the same eigenvalue \(\lambda_2 = 2\).
From \(A v_3 = 4 v_3\) with \(v_3 = [-1, 1, 0]^T\), we have an eigenvalue \(\lambda_3 = 4\).
The eigenvalues are 2, 2, 4. The algebraic multiplicity of \(\lambda=2\) is 2.
The eigenvectors \(v_1\) and \(v_2\) corresponding to \(\lambda=2\) are linearly independent. Thus, the geometric multiplicity of \(\lambda=2\) is 2.
Since the geometric multiplicity equals the algebraic multiplicity for all eigenvalues, the matrix A is diagonalizable.
For a diagonalizable matrix, the minimal polynomial has distinct linear factors corresponding to the distinct eigenvalues.
Minimal polynomial \(m(x) = (x-2)(x-4)\).
The degree of the minimal polynomial is \(m = 2\).
Finding \(a_{11}, a_{21}, a_{31}\):
The first column of A is \([a_{11}, a_{21}, a_{31}]^T = A e_1\), where \(e_1 = [1, 0, 0]^T\).
We express \(e_1\) as a linear combination of the eigenvectors: \(e_1 = c_1 v_1 + c_2 v_2 + c_3 v_3\).
\[ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = c_1 \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} + c_3 \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} \]
This gives the system of equations:
\[ \begin{aligned} c_1 - c_3 &= 1 \quad (1) \\ 2c_1 + c_2 + c_3 &= 0 \quad (2) \\ c_1 + c_2 &= 0 \quad (3) \end{aligned} \]
From (3), \(c_2 = -c_1\). Substitute into (2): \(2c_1 - c_1 + c_3 = 0 \implies c_1 + c_3 = 0 \implies c_3 = -c_1\).
Substitute \(c_3 = -c_1\) into (1): \(c_1 - (-c_1) = 1 \implies 2c_1 = 1 \implies c_1 = 1/2\).
Then \(c_2 = -1/2\) and \(c_3 = -1/2\).
So, \(e_1 = \frac{1}{2}v_1 - \frac{1}{2}v_2 - \frac{1}{2}v_3\).
Now we find \(A e_1\):
\[ A e_1 = A \left( \frac{1}{2}v_1 - \frac{1}{2}v_2 - \frac{1}{2}v_3 \right) = \frac{1}{2}Av_1 - \frac{1}{2}Av_2 - \frac{1}{2}Av_3 \]
\[ A e_1 = \frac{1}{2}(2v_1) - \frac{1}{2}(2v_2) - \frac{1}{2}(4v_3) = v_1 - v_2 - 2v_3 \]
\[ A e_1 = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} - \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} - 2 \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \\ 0 \end{bmatrix} \]
Thus, \(a_{11} = 3\), \(a_{21} = -1\), \(a_{31} = 0\).
Step 4: Final Answer:
The question asks for \(a_{11} + a_{21} + a_{31} + m\).
\[ 3 + (-1) + 0 + 2 = 4 \]
Step 5: Why This is Correct:
The eigenvalues and eigenvectors were correctly identified. The concept of diagonalizability was correctly used to determine the minimal polynomial and its degree. The first column of the matrix A was found by expressing the standard basis vector \(e_1\) in terms of the eigenvectors and applying the linear transformation A. The final sum is calculated correctly.