Proof with symmetric matrix (Direct and Contraposition) $S$ is a symmetric matrix. If $Sx=\lambda x$ and $Sy = \mu y$, where $x$ and $y$ are non-zero vectors and $\lambda , \mu \in \mathbb{R}$. Prove the following: (a) If $\mu \neq \lambda$, then $x^{t}y=0$. Prove by contraposition. (b) $(S^4 + S^2 + S)$ has an eigenvalue $\lambda^4+\lambda^2+\lambda$. Give a direct proof. Help on this would be much appreciated - I'm very new to proofs!

## Part a

$$\begin{array}{rcl} \lambda x^T y &=& (x^T S^T) y \\\ &=& (x^T S) y \\\ &=& x^T (S y) \\\ &=& x^T (\mu y) \\\ \end{array}$$

Therefore $(\lambda-\mu)(x^T y) = 0$

## Part b

$$\begin{array}{rcl} S x &=& \lambda x \\\ S^2 x &=& \lambda^2 x \\\ S^3 x &=& \lambda^3 x \\\ S^4 x &=& \lambda^4 x \\\ S^4 x + S^2 x + S x &=& \lambda^4 x + \lambda^2 x + \lambda x \\\ (S^4 + S^2 + S) x &=& (\lambda^4 + \lambda^2 + \lambda) x \\\ \end{array}$$