Three presets drop the right angle. The frozen picture is the skewed basis, mathbfb1=(1,0.4) and mathbfb2=(−0.3,1), where the dashed basis grid is made of parallelograms rather than squares.The same mathbfv now reads (2.634,0.446), and the two decomposition legs meet at an oblique angle. Skewed basis, frozen
The basis grid is made of parallelograms rather than squares, and the two legs meet at an oblique angle. Still a valid basis: det = 1.12, so v has unique coordinates (2.634, 0.446).
Nothing here is invalid. The only requirement for a basis is that the two vectors be linearly independent — equivalently $det B
eq 0$ — and perpendicularity is a convenience, not a condition. Every vector still has exactly one pair of coordinates in this basis.
What is lost is the convenience. Without orthogonality, $B^{-1}
eq B^{mathsf{T}}$ and the inverse must be computed properly; the Pythagorean formula for length no longer applies to the coordinates; and projecting onto one axis is no longer independent of the other. Non-orthogonal bases are common in practice — crystal lattices and eigenbases are rarely perpendicular — which is why the tool insists they are legitimate.