If M is a 3 × 3 matrix, where MTM = I and det (M) = 1, then prove that det (M – I) = 0.

To approach the problem, let's first break down the given conditions and what they imply. We have a 3 × 3 matrix \( M \) such that \( M^T M = I \), which indicates that \( M \) is an orthogonal matrix. Additionally, the determinant of \( M \) is given to be 1. Our goal is to prove that \( \text{det}(M - I) = 0 \).

Understanding Orthogonal Matrices

An orthogonal matrix has the property that its rows and columns are orthonormal vectors. This means that the dot product of any two different rows (or columns) is zero, and the dot product of a row (or column) with itself is one. The condition \( \text{det}(M) = 1 \) indicates that the matrix represents a rotation rather than a reflection.

Eigenvalues of Orthogonal Matrices

For orthogonal matrices, the eigenvalues have specific characteristics. Since \( M^T M = I \), we can infer that the eigenvalues \( \lambda \) of \( M \) must satisfy the equation \( |\lambda| = 1 \). This means that all eigenvalues lie on the unit circle in the complex plane. In a 3 × 3 orthogonal matrix with \( \text{det}(M) = 1 \), we can expect the eigenvalues to be of the form:

• One eigenvalue is 1,
• The other two eigenvalues are complex conjugates of each other, say \( e^{i\theta} \) and \( e^{-i\theta} \), where \( \theta \) is some angle.

Relating Eigenvalues to Determinants

The determinant of a matrix is equal to the product of its eigenvalues. For our matrix \( M \), this gives us:

\( \text{det}(M) = 1 \cdot e^{i\theta} \cdot e^{-i\theta} = 1 \).

This confirms our earlier assertion about the eigenvalues—specifically, that one of the eigenvalues is 1, which is crucial for our next step.

Examining \( M - I \)

Next, we consider the matrix \( M - I \). The eigenvalues of the matrix \( M - I \) can be derived from the eigenvalues of \( M \). If \( \lambda \) is an eigenvalue of \( M \), then the corresponding eigenvalue of \( M - I \) is \( \lambda - 1 \).

Given our earlier findings, the eigenvalues of \( M \) are \( 1 \), \( e^{i\theta} \), and \( e^{-i\theta} \). Thus, the eigenvalues of \( M - I \) are:

• \( 1 - 1 = 0 \)
• \( e^{i\theta} - 1 \)
• \( e^{-i\theta} - 1 \)

Determinant of \( M - I \)

Since one of the eigenvalues of \( M - I \) is \( 0 \), it follows that:

\( \text{det}(M - I) = (0) \cdot (e^{i\theta} - 1) \cdot (e^{-i\theta} - 1) = 0 \)

This result shows that the determinant of \( M - I \) is indeed zero, proving what we set out to demonstrate.

Summary of Findings

In summary, we established that because \( M \) is an orthogonal matrix with determinant 1, it must have an eigenvalue of 1. This leads us to conclude that \( M - I \) has a zero eigenvalue, thereby confirming that \( \text{det}(M - I) = 0 \). This aligns well with our understanding of the properties of orthogonal matrices and their eigenvalues.