Matrix - Orthonormally Diagonalizable

A square matrix $A$ is orthonormally diagonalizable if it can be written as

$$A=QD{Q}^T$$

Where:

• $D$ is a diagonal matrix of eigenvalues
• $Q$ is a orthonormal matrix

This is a special case of Matrix - Diagonalizable

Key Theorem

Spectral Theorem

Spectral Theorem

This theorem is made up of two parts

1. All real symmetric matrices are orthonormally diagonalizable.
2. All real symmetric matrices have real eigenvalues.

Proof for Theorem 2

Suppose $\lambda$ and $\vec{v}$ is an eigenvalue and eigenvector for the matrix $A$.

Then, we have that:

$$A\vec{v}=\lambda \vec{v}$$

Multiplying by the transpose of $\vec{v}$, we have that:

$${\vec{v}}^{T}A\vec{v}=\lambda {\vec{v}}^T\vec{v}=\lambda ||\vec{v}|{|}^2$$

Taking the complex conjugate of both sides, we have that:

$$\overline{{\vec{v}}^{T}A\vec{v}}=\overline{\lambda ||\vec{v}|{|}^2}=\overline{\lambda }||\vec{v}|{|}^2$$

But as $A$ is real and symmetric, we have $A=\overline{A}=A^T=\overline{A^T}$, and thus:

$${\vec{v}}^{TA\vec{v}}=\overline{{\vec{v}}^{T}A\vec{v}}$$

And finally

$$\lambda ||\vec{v}|{|}^2=\overline{\lambda }||\vec{v}|{|}^2$$

As $||\vec{v}|{|}^2\ne 0$, we have that $\lambda =\overline{\lambda }$, and thus $\lambda$ is real.

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Why This Matters

• No matrix inversion required for diagonalization
• Numerically stable (due to orthogonality of $Q$)
• Preseves geometry (lengths and angles)
• Useful in applications like Gram Schmidt Process, PCA, etc.