Theorem20.1
The rank of \(A\) is the number of nonzero singular values.
We returned to the topic of the SVD, and we finally completed the proof of its existence. After that, we proved two short results.
The rank of \(A\) is the number of nonzero singular values.
\(\range(A)=\Span(u_1, u_2, \dots, u_r),\) and \(\Null(A)=\Span(v_{r+1},\dots, v_n),\) where \(r\) is the number of nonzero singular values of \(A\text{.}\)
We proved in class that the SVD of any matrix exists. Carefully read the details in the Lecture 4 of the textbook that demonstrate that, if the matrix is square and the singular values distinct, then the left and right singular vectors are uniquely determined up to complex sign.
In class we proved that \(\range(A)=\Span(u_1, u_2, \dots, u_r),\) where \(r\) is the number of nonzero singular values of \(A\text{.}\) Now prove that
\begin{equation*} \Null(A)=\Span(v_{r+1},\dots, v_n). \end{equation*}