I am trying to better understand SVD, and I have been struggling for a week now trying to understand this. Given the SVD of an $m \times n$ matrix $A = U \Sigma V^T$ where $m>n$, how do I show that the silent columns of $V$ span the null space of $A$. If $m=n$ or $n>m$ will the proof be different. Everything I look up states that facts about the null space and range of $A$ given the SVD, but the proofs are not presented.

  • $\begingroup$ In your notation, is $U \in \mathbb{R}^{m\times m}$ and $V \in \mathbb{R}^{n \times n}$? $\endgroup$ – PSPACEhard Nov 17 '16 at 14:56
  • $\begingroup$ Yes. That is correct. $\endgroup$ – Ralff Nov 17 '16 at 14:56

Let the columns of $V$ be $v_1$, $v_2$, $\cdots$, $v_n$. Without loss of generality, we assume that $v_{k+1}$, $v_{k+2}$, $\cdots$ and $v_n$ are the silent columns of $V$. Therefore, $A$ can be written as $$ A = U_k\Sigma_k V_k^T $$ where $\Sigma_k$ contains only non-zero singular values of $A$ and $V_k$ contains $v_1$, $v_2$, $\cdots$ and $v_k$ as columns. That is, $\mathsf{rank}(A) = k$. Moreover, we have $$ Av_i = U_k\Sigma_kV_k^Tv_i = 0 $$ for each $k < i \leq n$ becaue $v_i^Tv_j = 0$ for $i \neq j$. Therefore, $\mathsf{nullity}(A) \geq n - k$.

By rank-nullity theorem, we have $$ \mathsf{rank}(A) + \mathsf{nullity}(A) = n $$ we conclude that $\mathsf{nullity}(A) = n - k$ exactly. That is, $v_{k+1}$, $v_{k+2}$, $\cdots$ and $v_n$ in fact spans the null space of $A$.

  • $\begingroup$ Thanks! This was very clear and concise proof. I can now understand why these columns span the null space instead of just believing. $\endgroup$ – Ralff Nov 17 '16 at 15:33

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