# Symmetric Matrices

• Sep 23rd 2009, 07:27 AM
steiner
Symmetric Matrices
Show that every symmetric matrix A=A^T can be written as U=UDU^T where D is a diagonal matrix and UU^T=I (orthogonal:unitary)

i am pretty sure that this should be tried through induction.
• Sep 24th 2009, 12:28 PM
aliceinwonderland
Quote:

Originally Posted by steiner
Show that every symmetric matrix A=A^T can be written as U=UDU^T (A=UDU^T ?) where D is a diagonal matrix and UU^T=I (orthogonal:unitary)
i am pretty sure that this should be tried through induction.

Lemma 1. The eigenvalues of a symmetric matrix with real entries are real numbers.

Proof. If $\displaystyle \lambda$ is an eigenvalue and v a corresponding eigenvector of an $\displaystyle n \times n$ symmetric matrix A.
Then $\displaystyle Av = \lambda v$ . If we multiply each side of this equation on the left by $\displaystyle v^T$, then we obtain
$\displaystyle v^TAv = v^T(\lambda v) = \lambda v^T v = \lambda ||v||^2$ . Then,

$\displaystyle \lambda = \frac{v^TAv}{||v||^2}$ .

Thus, $\displaystyle \lambda$ is a real number. To further verify, you apply a conjugate transpose and

$\displaystyle ({v^TAv})^* = v^TA^T(v^T)^T = v^TAv$ , since A is a symmetric matrix.

Lemma 2. Let A be a real symmetric $\displaystyle n \times n$ matrix. There is an orthogonal matrix $\displaystyle U \in O_n(\mathbb{Re})$ such that $\displaystyle U^TAU$ is diagonal.

Proof. We first show that eigenvectors from different eigenspaces with respect to a symmetric matrix A are orthogonal.
Let $\displaystyle v_1$ and $\displaystyle v_2$ be eigenvectors corresponding to distinct real eigenvalues $\displaystyle \lambda_1$ and $\displaystyle \lambda_2$ of the matrix A. We shall show that $\displaystyle v_1 \cdot v_2 = 0$. Since A is a symmetric matrix, $\displaystyle Av_1 \cdot v_2 = v_1 \cdot A^Tv_2 = v_1 \cdot Av_2$ . This implies that $\displaystyle \lambda_1 v_1 \cdot v_2 = v_1 \cdot \lambda_2 v_2 \Longleftrightarrow (\lambda_1 - \lambda_2)(v_1 \cdot v_2) = 0.$ Since $\displaystyle \lambda_1$ and $\displaystyle \lambda_2$ are distinct by hypothesis, $\displaystyle v_1 \cdot v_2 = 0$ .

The above showed that eigenvectors from different eigenspaces are orthogonal. By applying a Gram-Schmit process, eigenvectors obtained within same eigenspaces become orthonormal. This ensures that there exists an orthogonal matrix $\displaystyle P \in O_n(R)$ whose columns are normalized eigenvectors of a symmetric matrix A.

Let $\displaystyle U= \begin{bmatrix}v_{11} & v_{12} & \cdots & v_{1n} \\ v_{21} &v_{22} & \cdots & v_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ v_{n1} & v_{n2}&v_{n3} & v_{n4}\end{bmatrix}$
be an orthogonal matrix whose columns are eigenvectors $\displaystyle v_k= \begin{bmatrix}v_{1k} \\ v_{2k}\\ \vdots \\v_{nk} \end{bmatrix}$ of A.

Since $\displaystyle Av_1 = \lambda_1 v_1, Av_2 = \lambda_2 v_2, ... ,Av_n = \lambda_n v_n$,

$\displaystyle AU=\begin{bmatrix}\lambda_1 v_{11} & \lambda_2 v_{12} & \cdots & \lambda_n v_{1n} \\ \lambda_1 v_{21} & \lambda_2 v_{22} & \cdots & \lambda_n v_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ \lambda_1 v_{n1} & \lambda_2 v_{n2} & v_{n3} & \lambda_n v_{n4}\end{bmatrix}$ = $\displaystyle \begin{bmatrix}v_{11} & v_{12} & \cdots & v_{1n} \\ v_{21} &v_{22} & \cdots & v_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ v_{n1} & v_{n2}&v_{n3} & v_{n4}\end{bmatrix} \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 &\lambda_2 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0& \cdots & \lambda_n \end{bmatrix}=UD$.

Thus, $\displaystyle D=U^TAU$ and $\displaystyle A=UDU^T$, since U is an orthogonal matrix

To apply an induction, assume $\displaystyle n \times n$ symmetric matrix A holds the above lemma and show $\displaystyle (n+1) \times (n+1)$ symmetric matrix holds the above lemma as well. The construction shown in the lemma 2 can be used without much modification.