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The problem asks to show that for a nonzero projector \( P \in \mathbb{C}^{m \times m} \), the 2-norm \( \|P\|_2 \geq 1 \), with equality if and only if \( P \) is an orthogonal projector.

### Step 1: Basic properties of projectors
A **projector** \( P \) is a matrix that satisfies \( P^2 = P \), meaning it is idempotent. This property implies that applying \( P \) multiple times does not change the outcome after the first application.

### Step 2: 2-norm definition
The **2-norm** of a matrix, \( \|P\|_2 \), is defined as:
\[
\|P\|_2 = \sup_{\|x\|_2 = 1} \|Px\|_2
\]
where \( \|x\|_2 \) is the usual Euclidean norm of a vector.

### Step 3: Eigenvalues of a projector
Since \( P \) is a projector, its eigenvalues must be either 0 or 1. This follows from the fact that for any eigenvector \( v \) corresponding to an eigenvalue \( \lambda \), we have:
\[
P v = \lambda v
\]
and using \( P^2 = P \), we get:
\[
P^2 v = P v \implies \lambda^2 v = \lambda v \implies \lambda(\lambda - 1) v = 0
\]
Hence, \( \lambda = 0 \) or \( \lambda = 1 \). Therefore, the only possible eigenvalues of \( P \) are 0 and 1.

### Step 4: Lower bound for the 2-norm
The 2-norm of a matrix is the largest singular value, which for Hermitian matrices (including projectors) coincides with the largest absolute value of the eigenvalues. Since the eigenvalues of \( P \) are 0 and 1, we have:
\[
\|P\|_2 = 1
\]
if \( P \) has at least one nonzero eigenvalue (which it must, since \( P \) is nonzero by the problem's assumption).

Thus, \( \|P\|_2 \geq 1 \).

### Step 5: Equality condition
Equality \( \|P\|_2 = 1 \) holds if the 2-norm is exactly 1, which implies that the matrix has no singular values greater than 1. This occurs if and only if \( P \) is an **orthogonal projector**, meaning that \( P = P^* \), where \( P^* \) is the conjugate transpose (adjoint) of \( P \).

Thus, the condition \( \|P\|_2 = 1 \) holds if and only if \( P \) is an orthogonal projector.

### Conclusion:
We have shown that:
1. \( \|P\|_2 \geq 1 \) for any nonzero projector \( P \).
2. \( \|P\|_2 = 1 \) if and only if \( P \) is an orthogonal projector.

Would you like to go into more details about any step?

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Here are 5 related questions:
1. What are other types of matrix norms, and how do they compare to the 2-norm?
2. How are projectors used in solving linear systems?
3. What is the relationship between the spectral norm and the eigenvalues of Hermitian matrices?
4. How can we compute the 2-norm of a general matrix numerically?
5. What are the geometric interpretations of orthogonal projectors?

**Tip**: Orthogonal projectors are very useful in applications like least squares problems and QR factorization!

Let P ∈ C^m×m be a nonzero projector. Show that ||P||_2 ≥ 1, with equality if and only if P is an orthogonal projector.

This detailed explanation proves that for a nonzero projector P in C^m×m, ||P||_2 is greater than or equal to 1, with equality holding if and only if P is an orthogonal projector. Learn about the properties of projectors, matrix norms, and their eigenvalues, key concepts in linear algebra.

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