# How To Prove Left Singular Vectors Are Orthogonal?

For SVD, for a matrix $A$ that is $m~\times~n$, we select the right singular vectors as

$$\mathbf{v_1} \equiv \text{arg max}_{\mathbf{v}:\vert\mathbf{v}\vert=1}\vert{A\mathbf{v}}\vert$$

$$\mathbf{v_i} \equiv \text{arg max}_{\mathbf{v}\perp\mathbf{v_1},...,\mathbf{v}\perp\mathbf{v_{i-1}},\vert\mathbf{v}\vert=1}\vert{A\mathbf{v}}\vert$$

And we select the singular values and left singular vectors as $$\sigma_{1}\equiv\text{max}_{\mathbf{v}:\vert\mathbf{v}\vert=1}\vert A\mathbf{v}\vert$$ $$\sigma_{i}\equiv\text{max}_{\mathbf{v}\perp\mathbf{v_1},...,\mathbf{v}\perp\mathbf{v_{i-1}},\vert\mathbf{v}\vert=1}\vert A\mathbf{v}\vert$$ $$\mathbf{u_i}\equiv\frac{A\mathbf{v_i}}{\sigma_{i}}$$ How can I prove that the $\mathbf{u_i}$'s, the left singular vectors, are orthogonal? I know of the proof by induction, but I was wondering if there is a simpler proof?

But that answer assumes that $A^T\mathbf{u_i}=\sigma_i\mathbf{v_i}$. But I'm not making that assumption. Or is there a way to derive that?