In Gilbert Strang's book in the SVD chapter, he states that $u$ and $v$ give the bases for the image and kernel of a matrix.
Say we have a matrix $A$ that has rank $r$.
Why is it that the orthonormal basis of $Im(A)$ can be written as $u_{1}, \dots, u_{r}$? and the orthonormal basis of $\ker(A)$ can be written as $v_{r+1}, \dots, v_{n}$?
If we decompose matrix $A$ in SVD form we will have $U \Sigma V^{T}$.
Edit: $A \in R^{n \times m}$, $U \in R^{n \times n}$, $\Sigma \in R^{n \times m}$, $V \in R^{m \times m}$ $$ A = U \Sigma V^{T} = \begin{pmatrix} | & & | \\ u_{1} & \dots & u_{n} \\ | & & | \end{pmatrix} \begin{pmatrix} \sigma_{1} & 0 & 0 \\ 0 & \ddots & 0 \\ 0& 0 & \sigma_{n} \end{pmatrix} \begin{pmatrix} - & \vec{v}_{1}^T& - \\ & \vdots & \\ -& \vec{v}_{m}^T & - \end{pmatrix} $$
If rank is r then the entries in the diagonal after $\sigma_{r}$ is $0$. So the Ker would be associated with the singular values $\sigma_{r+1}, \dots ,\sigma_{n}$?