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Let's break down and solve the problems in the image:

### Problem A2:
#### (a) **Show that the eigenvalues of an idempotent matrix are always either 0 or 1.**

An idempotent matrix \( A \) satisfies the property:
\[
A^2 = A
\]
Let \( \lambda \) be an eigenvalue of \( A \), and \( \mathbf{v} \) be the corresponding eigenvector, such that:
\[
A\mathbf{v} = \lambda\mathbf{v}
\]
Applying the idempotent property to this eigenvalue equation:
\[
A^2\mathbf{v} = A(A\mathbf{v}) = A(\lambda\mathbf{v}) = \lambda A\mathbf{v} = \lambda^2 \mathbf{v}
\]
But since \( A^2 = A \), we also have:
\[
A^2 \mathbf{v} = A \mathbf{v} = \lambda \mathbf{v}
\]
Thus, \( \lambda^2 \mathbf{v} = \lambda \mathbf{v} \), which gives:
\[
\lambda^2 = \lambda
\]
This implies that the eigenvalues \( \lambda \) must satisfy:
\[
\lambda(\lambda - 1) = 0
\]
Therefore, \( \lambda = 0 \) or \( \lambda = 1 \). Hence, the eigenvalues of an idempotent matrix are always 0 or 1.

#### (b) **Let \( A \) be a \( k \times k \) matrix and idempotent matrix of rank \( r \). Show that \( r(A) = \text{tr}(A) \).**

The rank \( r(A) \) of an idempotent matrix equals the number of nonzero eigenvalues (which are all 1, since the eigenvalues are either 0 or 1). The trace \( \text{tr}(A) \) of a matrix is the sum of its eigenvalues. For an idempotent matrix, the trace counts the number of 1s among the eigenvalues (since the trace is the sum of the diagonal entries, which are eigenvalues for a diagonalizable matrix). Therefore:
\[
r(A) = \text{tr}(A)
\]

---

### Problem A3:
#### (a) **Show that the matrix for the following quadratic form is positive definite:**
\[
3x_1^2 + 2x_2^2 - 2\sqrt{2}x_1x_2
\]

We can write the quadratic form as:
\[
\mathbf{x}^T Q \mathbf{x} = 3x_1^2 + 2x_2^2 - 2\sqrt{2}x_1x_2
\]
where \( Q \) is the matrix of the quadratic form. The matrix \( Q \) is:
\[
Q = \begin{pmatrix}
3 & -\sqrt{2} \\
-\sqrt{2} & 2
\end{pmatrix}
\]

To check if \( Q \) is positive definite, we need to ensure that:
1. The leading principal minors are positive.
2. The eigenvalues are positive.

- The first leading principal minor is \( 3 > 0 \).
- The determinant of \( Q \) is:
\[
\text{det}(Q) = 3(2) - (-\sqrt{2})(-\sqrt{2}) = 6 - 2 = 4 > 0
\]
Since both minors are positive, \( Q \) is positive definite.

#### (b) **Suppose \( p = 2 \) and \( n = 1 \), and consider the random vector \( \mathbf{X}' = [X_1, X_2] \). Let the discrete random variable \( X_1 \) have the following probability function:**

\[
\begin{array}{c|cc}
x_1 & 0 & 1 \\
\hline
p_1(x_1) & 0.8 & 0.2
\end{array}
\]
Find \( E(X_1) \).

The expected value \( E(X_1) \) is calculated as:
\[
E(X_1) = \sum_{i} x_1 p_1(x_1)
\]
Substitute the values from the table:
\[
E(X_1) = (0)(0.8) + (1)(0.2) = 0 + 0.2 = 0.2
\]
Thus, \( E(X_1) = 0.2 \).

---

Would you like more details on any of the steps or concepts? Here are 5 questions to expand on this topic:

1. What are the properties of idempotent matrices in higher dimensions?
2. How would you compute the eigenvalues of a specific \( k \times k \) idempotent matrix?
3. Can a matrix be positive definite if one of its eigenvalues is zero?
4. How can you determine if a matrix is positive semidefinite using its principal minors?
5. How does the concept of expectation apply to continuous random variables?

**Tip:** When checking if a matrix is positive definite, remember that positive eigenvalues ensure positive definiteness, while non-negative eigenvalues ensure positive semi-definiteness.

The eigenvalues of an idempotent matrix and a quadratic form problem related to positive definiteness

This page explores eigenvalues of idempotent matrices, showing they are either 0 or 1, and demonstrates how to check if a quadratic form matrix is positive definite. Also, it covers calculating the expected value of a discrete random variable.

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