GATE CS 2011 Q.40 (CS Q.40): Linear Algebra

Consider the matrix as given below.

\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 7 \\ 0 & 0 & 3 \end{bmatrix}

Which one of the following options provides the CORRECT values of the eigenvalues of the matrix?

1, 4, 3

3, 7, 3

7, 3, 2

1, 2, 3

Official answer and solution

Solution

Use the property that the eigenvalues of any triangular matrix are simply its principal diagonal elements.

The given matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 7 \\ 0 & 0 & 3 \end{bmatrix}$ is an upper triangular matrix (all entries below the main diagonal are zero).

\det \begin{bmatrix} 1 - \lambda & 2 & 3 \\ 0 & 4 - \lambda & 7 \\ 0 & 0 & 3 - \lambda \end{bmatrix} = (1 - \lambda)(4 - \lambda)(3 - \lambda) = 0

The roots of the characteristic polynomial are the diagonal entries $\lambda_1 = 1$ , $\lambda_2 = 4$ , and $\lambda_3 = 3$ .

Final answer

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