Lets consider the symmetric matrix [tex]A[/tex], and the two eigenvectors [tex]\vec{v}_i[/tex] and [tex]\vec{v}_j[/tex] such as:
[tex]A \vec{v} _i = \lambda_i \vec{v} _i[/tex]
[tex]A \vec{v} _j = \lambda_j \vec{v} _j[/tex]
with
[tex]\lambda_i \ne \lambda_j[/tex].
The dot product between [tex]\vec{v}_i[/tex] and [tex]\vec{v}_j[/tex] can be obtained with:
[tex]\vec{v}_i \cdot \vec{v}_j = (\vec{v}_i )^t \vec{v}_j[/tex]
Using the first eigenvector equation we can find:
[tex] \vec{v}_i = \frac{1}{\lambda_i} A \vec{v} _i [/tex]
Lets transpose it
[tex] (\vec{v}_i)^t = (\frac{1}{\lambda_i} A \vec{v} _i)^t [/tex]
[tex] (\vec{v}_i)^t = (\vec{v} _i)^t A^t ((\frac{1}{\lambda_i})^t [/tex]
as [tex]\lambda_i[/tex] is an scalar
[tex] (\vec{v}_i)^t = (\vec{v} _i)^t A^t (\frac{1}{\lambda_i}) [/tex]
Now, as A is symmetric:
[tex]A^t = A[/tex]
so
[tex] (\vec{v}_i)^t = (\vec{v} _i)^t A (\frac{1}{\lambda_i}) [/tex]
Lets take the dot product again:
[tex]\vec{v}_i \cdot \vec{v}_j = (\vec{v}_i )^t \vec{v}_j[/tex]
but this is :
[tex]\vec{v}_i \cdot \vec{v}_j = (\vec{v} _i)^t A (\frac{1}{\lambda_i}) \vec{v}_j[/tex]
[tex]\vec{v}_i \cdot \vec{v}_j = (\frac{1}{\lambda_i}) (\vec{v} _i)^t A \vec{v}_j[/tex]
[tex]\vec{v}_i \cdot \vec{v}_j = (\frac{1}{\lambda_i}) (\vec{v} _i)^t ( A \vec{v}_j )[/tex]
But, the parenthesis is equal to
[tex]A \vec{v} _j = \lambda_j \vec{v} _j[/tex]
so
[tex]\vec{v}_i \cdot \vec{v}_j = (\frac{1}{\lambda_i}) (\vec{v} _i)^t \lambda_j \vec{v} _j[/tex]
[tex]\vec{v}_i \cdot \vec{v}_j = (\frac{\lambda_j }{\lambda_i}) (\vec{v} _i)^t \vec{v}_j[/tex]
Now, subtracting the dot product
[tex]\vec{v}_i \cdot \vec{v}_j - \vec{v}_i \cdot \vec{v}_j = (\frac{\lambda_j }{\lambda_i}) (\vec{v} _i)^t \vec{v}_j - (\vec{v} _i)^t \vec{v}_j = 0[/tex]
[tex] ( \frac{\lambda_j }{\lambda_i} - 1 ) (\vec{v} _i)^t \vec{v}_j= 0[/tex]
As the eigenvalues are distinct, [tex] \frac{\lambda_j }{\lambda_i} [/tex] can't be 1, so
[tex] ( \frac{\lambda_j }{\lambda_i} - 1 ) \ne 0[/tex]
this implies
[tex] (\vec{v} _i)^t \vec{v}_j= 0[/tex]
so the eigenvectors are orthogonal.
