Question

The sum of eigenvalues of the matrix \( \begin{bmatrix} 4 & 3 & 2 \\ 0 & -1 & 2 \\ 0 & 0 & -3 \end{bmatrix} \) is ................... (in integer).

Hint:

If you are asked for eigenvalues or their sum for a triangular matrix, immediately look at the main diagonal. The diagonal elements are the eigenvalues, and their sum is the trace. This is a very common shortcut in exam questions.

Answer 1

Step 1: Understanding the Concept:
There are two key properties of matrices that relate to the sum of eigenvalues.
1. The sum of the eigenvalues of any square matrix is equal to the trace of the matrix (the sum of the elements on the main diagonal).
2. For a triangular matrix (either upper or lower), the eigenvalues are the elements on the main diagonal.
Step 2: Key Formula or Approach:
We can use either property. Using the second property is the most direct. The given matrix is: \[ A = \begin{bmatrix} 4 & 3 & 2 \\ 0 & -1 & 2 \\ 0 & 0 & -3 \end{bmatrix} \] This is an upper triangular matrix because all the elements below the main diagonal are zero.
Step 3: Detailed Calculation:
Method 1: Using the property of triangular matrices
Since the matrix is upper triangular, its eigenvalues (\(\lambda\)) are simply the elements on its main diagonal. So, the eigenvalues are: \[ \lambda_1 = 4 \] \[ \lambda_2 = -1 \] \[ \lambda_3 = -3 \] The sum of the eigenvalues is: \[ \text{Sum} = \lambda_1 + \lambda_2 + \lambda_3 = 4 + (-1) + (-3) = 4 - 1 - 3 = 0 \] Method 2: Using the trace property
The trace of a matrix is the sum of its diagonal elements. \[ \text{Tr}(A) = 4 + (-1) + (-3) = 4 - 4 = 0 \] Since the sum of eigenvalues is equal to the trace, the sum is 0.
Step 4: Final Answer:
The sum of the eigenvalues is 0.
Step 5: Why This is Correct:
Both methods, which are standard properties of linear algebra, yield the same result of 0. The calculation is correct.