Question

Show that an orthogonal matrix M with all real eigenvalues is symmetric. Hints: Method 1. When the eigenvalues are real, so are the eigenvectors, and the unitary matrix which diagonalizes M is orthogonal. Use (11.27). Method 2. From Problem 46, note that the only real eigenvalues of an orthogonal M are ±1. Thus show that $\mathit{M}\mathit{=}{\mathit{M}}^{\mathit{-}\mathit{1}}$ . Remember that M is orthogonal to show that $\mathit{M}\mathbf{=}{\mathit{M}}^{\mathbf{T}}$.

Short answer

An orthogonal matrix M with real eigen values is symmetric.

Step-by-step solution

Given Information

If an orthogonal matrix has real eigenvalues, then its eigenvectors are also real. Since, the unitary matrix that diagonalizes has eigenvectors as columns, the unitary matrix is orthogonal. There is a theorem that states that a matrix which has real eigenvalues can be diagonalized by an orthogonal similarity transformation if and only if it is symmetric.

Therefore, M is symmetric.

Orthogonal Matrix

If the transpose of a square matrix containing real numbers or components equals the inverse matrix, the matrix is said to be orthogonal.

Symmetric Matrix

An orthogonal matrix whose only real eigenvalues are only $\pm 1$. We can show that $M=M^{-1}$. This means that$M^2=M{M}^{-1}=l$

The determinant of this, the LHS gives

$$\begin{gathered} det{M}^2={\left(detM\right)}^2 \\ =det{D}^2 \\ ={\left(\pm 1\right)}^2 \\ =1 \end{gathered}$$

Used the fact that $detAB=detAdetB$, the fact that $detM=detD$, where, is the diagonal matrix with eigenvalues of M , and the fact that the eigenvalues are $\pm 1$and $e^{\pm i\theta }$.

Therefore, $M=M^{-1}$. Since, M is orthogonal,$M^{-1}=M^T$ , which gives $M=M^T$. Therefore, M is symmetric.