To determine if the given $2 \times 2$ matrix

$A = \begin{bmatrix} 0 & 5 \\ 1 & 0 \end{bmatrix}$

is diagonalizable, we proceed as follows:

Step 1: Compute the eigenvalues

The eigenvalues of $A$ are found by solving the characteristic equation:

$\det(A - \lambda I) = 0$

$\det \begin{bmatrix} -\lambda & 5 \\ 1 & -\lambda \end{bmatrix} = 0$

$\lambda^2 - (0) + 5 = 0 \implies \lambda^2 = -5 \implies \lambda = \pm i\sqrt{5}$

Thus, the eigenvalues are $\lambda_1 = i\sqrt{5}$ and $\lambda_2 = -i\sqrt{5}$, both of which are complex.

Step 2: Diagonalizability

A matrix is diagonalizable if:

1. It has $n$ linearly independent eigenvectors (where $n$ is the size of the matrix).
2. Its eigenvalues need not be real, but the algebraic multiplicity must equal the geometric multiplicity for each eigenvalue.

For this matrix, the eigenvalues are distinct ($i\sqrt{5} \neq -i\sqrt{5}$), and distinct eigenvalues guarantee $n$ linearly independent eigenvectors.

Conclusion

The matrix $A$ is diagonalizable. The correct answer is "Yes".

Would you like a step-by-step calculation of eigenvectors or further clarification?

Related Questions:

1. How do you calculate eigenvectors for a given matrix?
2. What does it mean for a matrix to have distinct eigenvalues?
3. Can a matrix with complex eigenvalues be diagonalized in the real number space?
4. What is the significance of algebraic and geometric multiplicities in diagonalizability?
5. How does diagonalizability relate to the properties of symmetric matrices?

Tip:

For $2 \times 2$ matrices, distinct eigenvalues (real or complex) always imply diagonalizability.