Orthogonal Eigenvector, Proof is bothering me

[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?

 

[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.

 

[HallsofIvy]

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?

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 A^T= 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.