Eigenvalues of symmetric matrices [closed]

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Why does a symmetric matrix with elements in $\mathbb R$ have all eigenvalues in $\mathbb R$? How do you show that?

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1 Answer

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Let $A$ be a real symmetric $n\times n$-matrix. Suppose that $Av=\lambda v$, i.e. $v$ is an eigenvector with eigenvalue $\lambda$. Consider the standard Hermitian product $\left\langle X,Y\right\rangle=\sum_{i=1}^nX_i\overline{Y_i}$ on $\mathbb{C}^n$. Then \begin{eqnarray} \lambda\left\langle v,v\right\rangle&=& \left\langle \lambda v,v\right\rangle \\ &=& \left\langle Av,v\right\rangle\\ &=& \left\langle v,A^tv\right\rangle \:\: (\mbox{ in general }\left\langle v,A^*v\right\rangle, \mbox{ where $A^*$ denotes $\overline{A}^t$})\\ &=& \left\langle v,Av\right\rangle\\ &=& \left\langle v,\lambda v\right\rangle\\ &=& \overline{\lambda}\left\langle v,v\right\rangle. \end{eqnarray}

Hence $(\lambda-\overline{\lambda})\left\langle v,v\right\rangle=0$. But $\left\langle v,v\right\rangle\neq 0$ as $v$ is an eigenvector hence $(\lambda-\overline{\lambda})=0$, in other words, $\lambda$ is real.

In a very similar fashion you can show that eigenvectors belonging to different eigenvalues of a symmetric matrix are orthogonal. Try that as an exercise.

Edit: As indicated in the comments below, we also prove that eigenvalues for $A$ exist. Consider the characteristic polynomial $p_A(X)$ of $A$. A priori $p_A(X)\in \mathbb{R}[X]$, but similarly as above (where we considered a complex inner-product rather than a real one), we have that $p_A(X)\in \mathbb{C}[X]$. By the fundamental theorem of algebra, $p_A(X)=\prod_{i=1}^n(X-\lambda_i)$ for certain not necessarily distinct $\lambda_i\in \mathbb{C}$. Then the $\lambda_i$ are complex eigenvalues and by the reasoning above they are real.

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