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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.

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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 $\Sigma$.

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    $\begingroup$ 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. $\endgroup$
    – user733666
    May 11, 2021 at 16:45
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    $\begingroup$ Or if you could edit the answer and show why range of $\sum$ equals range of $\sum. V^*$? $\endgroup$
    – user733666
    May 11, 2021 at 16:49
  • $\begingroup$ @user733666 I don't understand what you're trying to say with you're first comment so I'm not sure if you have the right idea. In any case, I've edited my answer to address your question. $\endgroup$
    – Ben Grossmann
    May 11, 2021 at 18:11
  • $\begingroup$ 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? $\endgroup$
    – user733666
    May 11, 2021 at 18:29
  • $\begingroup$ @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 $\Sigma$. $\endgroup$
    – Ben Grossmann
    May 11, 2021 at 18:31

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