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Let $x_1,\ldots,x_n$ be the variables satisfying the equations of motion $\ddot{x_i}=f_i(x_1,\ldots,x_n)$ for $i=1,\ldots,n$

We introduce a small perturbation such that $x_i(t)=x_i^0 +h_i(t)$ where $x_i^0$ is the equilibrium point.

Then in matrix form $\ddot{\vec{h}}=A\vec{h}$ where $A_{ij}=\left({\partial f_i\over \partial x_j}\right)$

Now, it is said that $A$ in general has different right $w_k$ and left $v_k$ eigenvectors such that $Aw_k=\lambda_k w_k$ and $v_k^TA=\lambda_kv_k^T$.

Why is it true that $v_a\cdot w_b=\delta_{ab}$?

If it is at all relevant, we are given that the system has only real eigenvalues.

Thank you.

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up vote 1 down vote accepted

Evaluate $v^t_aAw_b$ two different ways: it's $v^t_a\lambda_bw_b$ but it's also $\lambda_av_a^tw_b$. So $\lambda_av_a\cdot w_b=\lambda_bv_a\cdot w_b$.

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