Eigenvalue Decomposition of a Symmetric Matrix

L. El Ghaoui

Eigenvalue Decomposition of a Symmetric Matrix

Let

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

We solve for the characteristic equation:

$0 = \det(\lambda I-A) = (\lambda -(3/2))^2 -(1/4) = (\lambda-1)(\lambda-2).$

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 $u= u_1$

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

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

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

Eigenvalue Decomposition of a Symmetric Matrix

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