A square matrix [math]A[/math] is called a [b]symmetric matrix[/b] if [math]A^T=A[/math].[br][br]It has many nice properties that can be summarized into the following theorem:[br][br][u]Spectral Theorem for symmetric matrices[/u]: For any [math]n\times n[/math] symmetric matrix [math]A[/math], we have[br][list=1][*][math]A[/math] has [math]n[/math] real eigenvalues (counting multiplicities).[/*][br][*] Any two eigenvectors corresponding to two distinct eigenvalues are orthogonal.[/*][br][*] The dimension of each eigenspace equals the multiplicity of the corresponding eigenvalue.[/*][br][*] [math]A[/math] is diagonalizable. Moreover, we can choose an orthogonal matrix [math]P[/math] such that [math]A=PDP^{-1}=PDP^T[/math], where [math]D[/math] is a diagonal matrix.[/*][br][/list]The proof of spectral theorem is beyond the scope of this course. We can only prove (2) as follows:[br]Suppose [math]u[/math] and [math]v[/math] are eigenvectors of [math]A[/math] corresponding to eigenvalues [math]\lambda[/math] and [math]\mu[/math] respectively. And we assume [math]\lambda\not=\mu[/math]. Then we have the following:[br][br][math]\lambda (u\cdot v)=\lambda u\cdot v =(Au)\cdot v=(Au)^Tv=u^TA^Tv=u^T(Av)=u^T(\mu v)=\mu u\cdot v=\mu(u\cdot v)[/math][br][math]\Rightarrow (\lambda-\mu)(u\cdot v)=0[/math][br]Since [math]\lambda\not=\mu[/math], [math]u\cdot v=0[/math], which implies that [math]u[/math] and [math]v[/math] are orthogonal.[br] [br][br]

In the applet below, you can visualize the diagonalization of a [math]2\times 2[/math] symmetric matrix.