Eigenvalue decomposition of a symmetric matrix

Alicia Y. Tsai; Authors: Laurent EI Ghaoui; Contributors: Phan Hoang, Tony Tin, Ha To Tam, Nguyen Van Kep, Le Dinh Nam, Juliette Decugis, Nguyen Quoc Cuong.; Giuseppe C. Calafiore

Eigenvalue decomposition of a symmetric matrix

Let

$\begin{align*} A:=\left(\begin{array}{cc} 3/2 & -1/2 \\ -1/2 & 3/2 \end{array}\right). \end{align*}$

We solve for the characteristic equation:

$\begin{align*} 0 &= \det(\lambda I-A) = (\lambda -(3/2))^2 -(1/4) = (\lambda-1)(\lambda-2). \end{align*}$

Hence the eigenvalues are $\lambda_1 =1$ , $\lambda_2 =2$ . For each eigenvalue $\lambda$ , we look for a unit-norm vector $u$ such that $Au = \lambda u$ . For $\lambda =\lambda_1$ , we obtain the equation in $u= u_1$

$\begin{align*} 0 &= (A-\lambda_1) u_1 = \left(\begin{array}{cc} 1/2 & -1/2 \\ -1/2 & 1/2 \end{array}\right) u_1 \end{align*}$

which leads to (after normalization) an eigenvector $u_1: := (1/\sqrt{2})[1,1]$ . Similarly for $\lambda_2$ we obtain the eigenvector $u_2: := (1/\sqrt{2})[1,-1]$ . Hence, $A$ admits the SED (Symmetric Eigenvalue Decomposition)

$\begin{align*} A=&\left(\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right)\right)^T\left(\begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right)\left(\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right)\right). \end{align*}$

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