# Range of $A$ and null space of $A$ in singular value decomposition?

This is from Trefethen and Bau, page 33, theorem 5.2. SVD of $$A=U.\sum.V^*$$ where $$U,V$$ are unitary and $$\sum$$ is diagonal. It states that range$$(A) =\langle u_1,...,u_r \rangle$$ and null $$(A) = \langle v_{r+1},...,v_n \rangle$$.

$$r$$ is the rank of $$A$$.

The proof then states that this theorem is the consequence of range $$(\sum) = \langle e_1,...,e_r \rangle$$ and null $$(\sum) = \langle e_{r+1},...,e_n \rangle$$.

I get the range and null of $$\sum$$, intuitively. I don’t understand how the theorem follows from that statement.

Hint: Note that the range of $$\Sigma$$ is equal to the range of $$\Sigma V^*$$. Then, note that if $$x$$ is in the range of $$\Sigma V^*$$, then $$Ux$$ must be in the range of $$U\Sigma V^* = A$$. What is $$Ux$$ if $$x = e_k$$?
Similarly, note that the null space of $$\Sigma$$ is equal to the null space of $$U \Sigma$$. Then, note that if $$x$$ is in the null space of $$U \Sigma$$, then $$Vx$$ will be in the null space of $$U \Sigma V^*$$.
Regarding the range of $$\Sigma V^*$$: note that every vector $$x$$ in the range of $$\Sigma$$ can be expressed as $$x = \Sigma w$$ for some vector $$w$$. However, we can write $$\Sigma w = \Sigma(V^*V)w = (\Sigma V^*) (Vw).$$ Thus, $$\Sigma w$$ is also an element of the range of $$\Sigma V^*$$. By a similar argument, we can show that any element in the range of $$\Sigma V^*$$ is also an element of the range of $$\Sigma$$.
• Okay let me reason it out. Range of $\sum$ is equal to the range of $\sum.V^*$ because, $\sum$ is a diagonal matrix and hence, when it is multiplied with any vector, it’s effectively just a similar vector, with each component of the vector scaled by a factor equal to the corresponding diagonal element of $\sum$. I understood the rest. May 11, 2021 at 16:45
• Or if you could edit the answer and show why range of $\sum$ equals range of $\sum. V^*$? May 11, 2021 at 16:49
• So, similarly any element $y$ in the range of $\sum.V^*$ can be expressed as $y=\sum.V^*.w$ for some $w$. But, $\sum.V^*.w = (\sum.V^*).w$, so $y$ is also in the range of $\sum. V^*$. Does that work for the second part of the proof? May 11, 2021 at 18:29
• @user733666 You started with $y$ is in the range of $\Sigma V^*$ and ended with the same statement. You need to conclude that $w$ is in the range of $\Sigma$. May 11, 2021 at 18:31