📂Matrix Algebra

Orthogonal Similarity and Orthogonal Diagonalization of Matrices

Definition¹

Orthogonal similarity

For two square matrices $A$ and $B$, if there exists an orthogonal matrix $P$ that satisfies the following, then $A$ and $B$ are said to be orthogonally similar.

$$ A = P^{\mathsf{T}}BP $$

Orthogonal diagonalization

A square matrix $A$ is said to be orthogonally diagonalizable (or $P$ is said to orthogonally diagonalize $A$) if it is orthogonally similar to some diagonal matrix $D$. In other words, $A$ is orthogonally diagonalizable if there exists an orthogonal matrix $P$ such that the following holds.

$$ A = P^{\mathsf{T}}DP $$

Remarks

Since an orthogonal matrix satisfies $P^{\mathsf{T}} = P^{-1}$, if $A$ and $B$ are orthogonally similar then $A$ and $B$ are similar.

Theorem

(a) If orthogonally diagonalizable then diagonalizable. The converse does not hold.

For a real square matrix $A$ the following conditions are equivalent.

(a) $A$ is orthogonally diagonalizable.

(b) $A$ has an orthonormal set consisting of $n$ eigenvectors.

(c) $A$ is a symmetric matrix.

---

As a corollary of (a) $\implies$ (b), if $A$ is orthogonally diagonalizable and $A = P^{\mathsf{T}} D P$, then each diagonal entry $d_{ii}$ of $D$ is an eigenvalue of $A$ corresponding to the $i$-th column vector of $P$.

Proof

(a) $\implies$ (b)

Assume the matrix $A$ in $n \times n$ is orthogonally diagonalizable. Then there exists an orthogonal matrix $P$ satisfying

$$ D = P^{-1} A P $$

Let the column vectors of $P$ be $\mathbf{p}_{i}$. Then $\mathbf{p}_{i}$ are linearly independent eigenvectors. Moreover, since $P$ is orthogonal, the column vectors form an orthonormal set.

■

---

From the above proof we obtain the following equation.

$$ PD = AP \implies \begin{bmatrix} d_{11} \mathbf{p}_{1} & \cdots & d_{nn} \mathbf{p}_{n} \end{bmatrix} = \begin{bmatrix} A \mathbf{p}_{1} & \cdots & A \mathbf{p}_{n} \end{bmatrix} $$

Therefore, the $i$-th diagonal entry $d_{ii}$ of $D$ is an eigenvalue of $A$ corresponding to the $i$-th column vector $\mathbf{p}_{i}$ of $P$.

(b) $\implies$ (c)

Suppose $A$ has an orthonormal set $\left\{ \mathbf{p}_{i} \right\}$ consisting of $n$ eigenvectors. The matrix $P = \begin{bmatrix} \mathbf{p}_{1} & \mathbf{p}_{2} & \cdots & \mathbf{p}_{n} \end{bmatrix}$ having these as column vectors diagonalizes $A$.

$$ A = P^{-1}DP $$

Since $\left\{ \mathbf{p}_{i} \right\}$ is an orthonormal set, $P$ is an orthogonal matrix.

$$ A = P^{\mathsf{T}}DP $$

Hence the following holds, and $A$ is a symmetric matrix.

$$ A^{\mathsf{T}} = (P^{\mathsf{T}}DP)^{\mathsf{T}} = P^{\mathsf{T}}D^{\mathsf{T}}P = P^{\mathsf{T}}DP = A $$

■

(c) $\implies$ (a)

If $A$ is symmetric then it is a normal matrix, and by the spectral theorem $A$ is unitarily diagonalizable. Unitarily diagonalizable implies orthogonally diagonalizable.

■