I am looking at a the following proof:

If $A$ is a real symmetric matrix, then $A$ has real eigenvalues.

Suppose that $\lambda$ is an eigenvalue of $A$ and $x$ is a corresponding eigenvector, where we allow for the possibility that $\lambda$ is complex and $x \in \mathbb{C}^n$. Thus, $$Ax=\lambda x$$ where $x \ne 0$. Now multiply both sides of this equation by $\overline{x}^T$(...)

My question is: Why do we multiply by $\overline{x}^T$? I know that symmetric means $A=A^T$.


Let $X^{\text{H}}$ denote the conjugate transpose of $X$. In your case $\text{A}^{\text{H}} = \text{A}$ therefore what follows also works for Hermitian type matrices. \begin{equation} \begin{split} & Ax = \lambda x \\ \Rightarrow & \big( Ax \big) ^{\text{H}} = \big( \lambda x\big) ^{\text{H}} \\ \Rightarrow & x^{\text{H}}A^{\text{H}} = \lambda^{*}x^{\text{H}} \\ \Rightarrow & \big(x^{\text{H}}A^{\text{H}}\big)x = \big(\lambda^{*}x^{\text{H}}\big)x \\ \Rightarrow & \big(x^{\text{H}}A^{\text{H}}x\big) = \lambda^{*}\big(x^{\text{H}}x\big) \end{split} \end{equation} But $A^{\text{H}} = A$, so $\lambda^{*} = \frac{x^{\text{H}}Ax}{x^{H}x}$. Note that premultiplying the first equation by $x^{\text{H}}$ gives you $\lambda = \frac{x^{\text{H}}Ax}{x^{H}x}$, therefore $\lambda^{*} = \lambda$ and the proof is done.


We multiply both sides by $x^T$ because that's the matrix notation version of taking a dot product of vectors $\vec{x}\cdot \vec{x} = x^T x$, if $x$ is a column vector.

  • $\begingroup$ Yes I know that, but why do we want to use this fact? $\endgroup$ – Alim Teacher Aug 30 '16 at 13:28
  • 2
    $\begingroup$ In the particular case of symmetric matrices, that step is overkill. You can just complex conjugate $Ax = \lambda x$ and use the fact that every element of both $A$ and $x$ are real. When you have Hermitian matrices, where you work with $A^\dagger = (A^*)^T$ (the conjugate transpose) then the reason for doing it is to make the left hand side symmetric: $x^\dagger A x = \lambda x^\dagger x$ so you can complex conjugate both sides and show that $\lambda = \lambda^*$. $\endgroup$ – Sean Lake Aug 30 '16 at 13:56

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