Structural dynamics: Why are modeshape-vectors referred to as orthogonal?

When dealing with an undamped MDOF dynamic system, one can find the natural frequency (eigenvalue) and corresponding modeshape-vector (eigenvector) for mode $n$ by solving the eigenvalue problem:

$$ [K] { phi }_n = omega_n^2 [M] { phi }_n $$

Furthermore, it is stated in a lot of litterature, that for any two modes, $n$ and $r$, their eigenvectors are orthogonal, because they can be shown to fulfill the following:

$$ { phi }_n [M] { phi }_r = 0 ;;;;;;;;;;;;;;;;;;;; { phi }_n [K] { phi }_r = 0 $$

Two vectors are, however generally, defined to be orthogonal if: $;;; { phi }_n^T { phi }_r = 0 ;;;$, and this is actually not true for the modeshape-vectors you find by solving the eigenvalue problem for an undamped dynamic system!

Just to be completely clear, i will give an example using a 2DOF system:

For such a system, the stiffness and mass matrices will be the following:

$$K =
begin{bmatrix}
k_1 + k_2 & -k_2
-k_2 & k_2
end{bmatrix} ;;;;;;;;;;;;;;;;;;;; M =
begin{bmatrix}
m_1 & 0
0 & m_2
end{bmatrix}
$$

I will just select some arbitrary values:

$$m_1 = 50 ;;;;;;;;;;;;;;;;;;;; m_2 = 20 ;;;;;;;;;;;;;;;;;;;; k_1 = 90 ;;;;;;;;;;;;;;;;;;;; k_2 = 50$$

which gives:

$$K =
begin{bmatrix}
140 & -50
-50 & 50
end{bmatrix} ;;;;;;;;;;;;;;;;;;;; M =
begin{bmatrix}
50 & 0
0 & 20
end{bmatrix}
$$

By solving the eigenvalue problem using these matrices, i get the following eigenvectors:

$$
{ phi }_1 =
begin{Bmatrix}
0.575
1.000
end{Bmatrix} ;;;;;;;;;;;;;;;;;;;; { phi }_2 =
begin{Bmatrix}
-0.695
1.000
end{Bmatrix}
$$

which gives:

$$
{ phi }_1^T { phi }_2 = 0.6 neq 0
$$

By using the mass or stiffness matrix, sure enough we get:

$$ { phi }_1 [M] { phi }_2 = 0 ;;;;;;;;;;;;;;;;;;;; { phi }_1 [K] { phi }_2 = 0 $$

but still ${ phi }_1^T { phi }_2 neq 0$ so they are not orthogonal! So why do we call them orthogonal?