To determine the number of distinct eigenvalues of the matrix $P = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$, let's proceed with the following steps:

1. Identify the Eigenvalues: For a matrix, the eigenvalues are found by solving the characteristic polynomial, $\det(P - \lambda I) = 0$, where $I$ is the identity matrix and $\lambda$ is the eigenvalue.

2. Formulate the Characteristic Polynomial: $P - \lambda I = \begin{bmatrix} 1 - \lambda & 1 & 0 \\ 0 & 1 - \lambda & 1 \\ 0 & 0 & 1 - \lambda \end{bmatrix}$ The determinant of this matrix is given by: $\det(P - \lambda I) = (1 - \lambda)^3$

3. Solve for Eigenvalues: The characteristic polynomial $(1 - \lambda)^3 = 0$ implies that $\lambda = 1$ is the only eigenvalue of the matrix $P$.

4. Determine Distinct Eigenvalues: Since $\lambda = 1$ is the only solution, the matrix $P$ has only one distinct eigenvalue.

Conclusion

The number of distinct eigenvalues of $P$ is $\boxed{1}$.

So, the correct answer is option (c) 1.

Would you like further details or have any questions?

Here are 5 related questions to explore further:

1. What is an eigenvalue, and how is it generally calculated?
2. How can we interpret eigenvalues in terms of transformations?
3. What does it mean for a matrix to have repeated eigenvalues?
4. Can a matrix with repeated eigenvalues be diagonalizable?
5. How does the number of distinct eigenvalues affect the properties of a matrix?

Tip: For diagonal matrices, the eigenvalues are simply the diagonal entries.