# Orthogonal Eigenvector, Proof is bothering me

1. Jun 29, 2008

### rocomath

Suppose

$$A\overrightarrow{x}=\lambda_1\overrightarrow{x}$$
$$A\overrightarrow{y}=\lambda_2\overrightarrow{y}$$
$$A=A^T$$

Take dot products of the first equation with $$\overrightarrow{y}$$ and second with $$\overrightarrow{x}$$

ME 1) $$(A\overrightarrow{x})\cdot \overrightarrow{y}=(\lambda_1\overrightarrow{x})\cdot\overrightarrow{y}$$

BOOK ... skipped steps but only shows this 1) $$(\lambda_1\overrightarrow{x})^T\overrightarrow{y}=(A\overrightarrow{x})^T\overrightarrow{y}=\overrightarrow{x}^TA^T\overrightarrow{y}=\overrightarrow{x}^TA\overrightarrow{y}=\overrightarrow{x}^T\lambda_2\overrightarrow{y}$$

Now it looks like I have to transpose my first step, but if I do so, do I assume that $$y=y^T$$?

Last edited: Jun 29, 2008
2. Jun 30, 2008

### CompuChip

Note that the dot product of two vectors is just a matrix multiplication. E.g. if we take $\vec x, \vec y$ to be (n x 1) column vectors then the dot product is just
$$\vec x \cdot \vec y = \vec x^T \vec y$$
where the right hand side is just matrix multiplication of an (1 x n) with an (n x 1) matrix. Then the solution easily follows. Try to write it out and indicate what is being done in each step.

3. Jun 30, 2008

### HallsofIvy

Staff Emeritus
You don't, they are not the same.

What you are doing is thinking of the dot product $\vec{u}\cdot\vec{v}$ as the matrix product $\vec{u}^T\vec{v}$ where $\vec{u}$ and $\vec{v}$ are "column" matrices and $\vec{u}^T$ is the "row" matrix corresponding to $\vec{v}$. It Then follows that $(A\vec{x}\cdot\vec{y}= (A\vec{x})^T\vec{y}= \vec{x}^TA^T\vec{y}$ which, because AT= A is the same as $\vec{x}^TA\vec{y}= \vec{x}\cdot A\vec{y}$. It is not that $\vec{y}^T= \vec{y}$, you never have "$\vec{y}^T$.