Chapter 7 - Eigenvalues and Eigenvectors - 7.5 Orthogonal Diagonalization and Quadratic Forms - Problems - Page 474: 20

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The $2 \times 2$ real symmetric matrix $A$ has two distinct eigenvalues $\lambda_1$ and $λ_2$ If $v_1 = (1, 2)$ is an eigenvector of A corresponding to the eigenvalue $\lambda_1$, we obtain: $=0\\ \rightarrow =0\\ \rightarrow v_1+2v_2=0\\ \rightarrow v_1=-2v_2\\ \rightarrow v=(-2v_2,v_2)=v_2(-2,1)$ An eigenvector corresponding to $\lambda_2$ is $(-2,1)$.