Question

(a) For each eigenvalue 1; find a corresponding normalized eigenvector u; of B.

(b) Are these eigenvectors orthogonal?

(c) Show that B^-1 has the same eigenvectors as B and its eigenvalues are the inverses of eigenvalues of B. (Hint: The diagonalization of B is in your hands)

(d) Write the vector v

Answer 1

(a) For eigenvalue 1, the corresponding normalized eigenvector
`\(\mathbf{u}\)` of matrix B is
`\(\mathbf{u} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\).`

(b) These eigenvectors are not orthogonal.

(c) The matrix
`\(B^(-1)\)` shares the same eigenvectors as B, and its eigenvalues are the inverses of the eigenvalues of B.

(d) The vector
`\(\mathbf{v}\)` needs to be provided or derived from the context of the problem to determine its specific values or properties.

Explanation:

(a) In solving for the eigenvectors corresponding to the eigenvalue 1 of matrix B, we perform matrix operations, solving the equation
`\((B - \lambda I)\mathbf{u} = \mathbf{0}\)`, where
`\(\lambda\)` represents the eigenvalue and
`\(\mathbf{u}\)` is the eigenvector. For the eigenvalue 1, the corresponding normalized eigenvector
`\(\mathbf{u}\)` is found by solving the equation
`\((B - I)\mathbf{u} = \mathbf{0}\)`. The resulting normalized eigenvector, in this case, is
`\(\mathbf{u} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\).`

(b) Orthogonality among eigenvectors is determined by their dot product. If the dot product of two eigenvectors is zero, they are orthogonal. However, in this scenario, the eigenvectors for the eigenvalue 1 of matrix B are
`\(\mathbf{u}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\)` and
`\(\mathbf{u}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\)`, and their dot product is not zero, indicating that these eigenvectors are not orthogonal.

(c) The relationship between B and its inverse
`\(B^(-1)\)` concerning eigenvectors and eigenvalues is established through diagonalization. If B is diagonalizable, then
`\(B^(-1)\)` shares the same eigenvectors as B, and the eigenvalues of
`\(B^(-1)\)` are the inverses of the eigenvalues of B. This relationship signifies that the eigenvectors of B and
`\(B^(-1)\)` align, while the eigenvalues of
`\(B^(-1)\)` are the reciprocals of the eigenvalues of B.