Question

Find a (3 x 3 ) matrix with exactly one (real) eigenvalue -3 , such that the 3 -eigenspace is a line (Think geometrically.) Note: in this accour

Answer 1

The matrix that satisfies the conditions is:

`\[ \begin{bmatrix} -3 & 0 & 0 \\ 0 & -3 & 1 \\ 0 & 0 & -3 \end{bmatrix} \]`

Step-by-step explanation:

To create a matrix with a single eigenvalue of -3 and a 3-eigenspace forming a line, construct a matrix with the eigenvalue -3 along the main diagonal and a non-zero entry in the third row of the second column. This creates a matrix where the eigenvectors corresponding to the eigenvalue -3 form a line. The resulting matrix meets the criteria specified in the question.

Consider a
`\(3 * 3\)` matrix that should have an eigenvalue of -3 with a 3-eigenspace forming a line. For this purpose, let's construct a matrix in which the eigenvalue -3 appears thrice along the main diagonal to ensure that -3 is an eigenvalue with algebraic multiplicity 3. Thus, the matrix becomes:

`\[ \begin{bmatrix} -3 & 0 & 0 \\ 0 & -3 & 1 \\ 0 & 0 & -3 \end{bmatrix} \]`

The diagonal entries indicate the eigenvalues of the matrix. Here, -3 is the only eigenvalue, satisfying the requirement. The non-zero entry in the third row and second column breaks the diagonal symmetry, ensuring that the 3-eigenspace is not the entire plane but rather a line. This arrangement forces the eigenvectors corresponding to -3 to form a line, meeting the geometric condition specified. Therefore, this matrix fulfills the given criteria, having a single real eigenvalue of -3 while maintaining a 3-eigenspace that forms a line.