The problem from the image is asking about an orthogonal projection matrix onto a subspace $V$ of $\mathbb{R}^n$, and it specifically requires proving that the eigenvalues of this projection matrix $P$ are either 0 or 1.

Problem Breakdown:

1. Given:

   • $V$ is a subspace of $\mathbb{R}^n$.
   • $\{ \mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k \}$ is an orthonormal basis for $V$.
   • $P = \mathbf{V}\mathbf{V}^T$, where $\mathbf{V}$ is the matrix formed by the orthonormal basis vectors as columns.

2. To Prove:

   • The eigenvalues of $P$ are either 0 or 1.

3. Hint Provided:

   • You can write any $x \in \mathbb{R}^n$ as $\mathbf{x} = \mathbf{v} + \mathbf{v}^\perp$, where $\mathbf{v} \in V$ and $\mathbf{v}^\perp \in V^\perp$.

Solution Outline:

1. Orthogonal Projection Matrix Properties:

   • The projection matrix $P$ maps any vector $\mathbf{x}$ onto the subspace $V$.
   • If $\mathbf{x} \in V$, then $P\mathbf{x} = \mathbf{x}$, meaning $P \mathbf{x} = 1 \cdot \mathbf{x}$.
   • If $\mathbf{x} \in V^\perp$, the orthogonal complement of $V$, then $P \mathbf{x} = 0$, meaning $P \mathbf{x} = 0 \cdot \mathbf{x}$.

2. Eigenvalue Analysis:

   • For a vector $\mathbf{x} \in V$, $P \mathbf{x} = \mathbf{x}$. So, 1 is an eigenvalue of $P$ with multiplicity equal to the dimension of $V$.
   • For a vector $\mathbf{x} \in V^\perp$, $P \mathbf{x} = 0$. So, 0 is an eigenvalue of $P$ with multiplicity equal to the dimension of $V^\perp$.

3. Conclusion:

   • Since any vector in $\mathbb{R}^n$ can be decomposed into a part in $V$ and a part in $V^\perp$, and the action of $P$ on these parts yields eigenvalues 1 and 0 respectively, the only possible eigenvalues for $P$ are 0 and 1.

Would you like a more detailed step-by-step solution? Here are some related questions:

1. How do you find the orthonormal basis of a given subspace $V$?
2. What are the properties of orthogonal projection matrices?
3. How does the rank of the projection matrix $P$ relate to the dimension of $V$?
4. Can you explain the spectral theorem and its relation to projection matrices?
5. How do you generalize this concept to other linear transformations?

Tip: Always remember that the sum of the eigenvalues (considering multiplicity) of a matrix equals its trace, and the product of the eigenvalues equals its determinant.