$\textbf{The Economy SVD}$:

Let $r:=rank(A)\leq \min(m,n)$ where $A\in\mathbb{R^{m\times n}}$. We know that the number of singular values of $A$ or the cardinality of the set $\Sigma(A):=\{\sigma_{i}|i=1,2,\cdots,m\}$ is $r$. Considering $AA^{T}$, we know that it is symmetric and thus by the Jordan-Schur's decomposition, we may write: $$ AA^{T}=QDQ^{T} $$ Here $D$ which is a diagonal matrix, houses the eigenvalues of $AA^{T}$ which are $r$ in total since $rank(A)=rank(AA^{T})$. Now this means that if we consider $D_{r}$ which is an $r\times r$ diagonal matrix that removes the $0$'s from the diagonal entries and only contains that $r$ eigenvalues $\Sigma_{r}^{2}$, as well as $Q_{r}$ excluding the zero orthonormal vectors we can write : $$ AA^{T}Q_{r}\Sigma_{r}^{-1}=Q_{r}\Sigma_{r} $$
Now let $U_{r}:=Q_{r}$, $S_{r}:=\Sigma_{r}$, and $V_{r}:=A^{T}Q_{r}\Sigma_{r}^{-1}=A^{T}U_{r}S_{r}^{-1}$, we may write: $$ AV_{r}=U_{r}S_{r} $$ where $U_{r}^{T}U_{r}=I_{r}$ and it can also be shown that $V_{r}^{T}V_{r}=I_{r}$ which lets us to rewrite the equation above as: $$ A=U_{r}S_{r}V_{r}^{T} $$

Request: I hope someone can explain the beginning of the proof as I didn't understand it. The proof below essentially derives the full scale SVD from the economy SVD

$\textbf{Theorem (Full Scale SVD)}$ : Consider the matrix $A\in\mathbb{R}^{m\times n}$ with $\operatorname{rank}(A)=r\leq p=\min(m,n)$. We claim that there exist two orthogonal matrices $U\in\mathbb{R}^{m\times m}$ and $V\in\mathbb{R}^{n\times n}$ and a diagonal matrix $S\in\mathbb{R}^{m\times n}$, such that : $$ AV=US\iff A=USV^{T}\iff S=U^{T}AV $$ where $S$ has the singular values of $A$ on its diagonal $\operatorname{diag}(S):=\begin{bmatrix}\sigma_{1}&\sigma_{2}&\cdots&\sigma_{p}\end{bmatrix}$. Furthermore, $\sigma_{1},\sigma_{2},\cdots,\sigma_{r}>0$ and the remaining $p-r$ singular values $\sigma_{r+1}=\sigma_{r+2}=\cdots=\sigma_{p}=0$. The columns of $U$ and the columns of $V$ are called the left-singular vectors and right-singular vectors of $A$, respectively.

$\textbf{Proof}$. Construct $V:=[V_{r}\;\;V_{n-r}]$ and $U:=[U_{r}\;\;U_{n-r}]$, by adding orthogonal vectors so that $V$ forms a basis for $\mathbb{R}^{n}$ and $U$ forms a basis for $\mathbb{R}^{m}$.

Recall that $Av_i \in \operatorname{Im}(A) = \{y \in \Bbb C^n \mid y = Ax; x \in \Bbb C^n\}$ with $Av_i = \sigma_i u_i$ for $i = 1,\dots,r$ and $\sigma_i > 0$. Moreover, $u_i = \frac 1{\sigma_i } Av_i$ form a basis for $\operatorname{Im}(A)$ for $i = 1,\dots, r$. Thus, $v_{r+1},\dots,v_n \in \ker(A)$ i.e. $AV_{n-r} = 0$, since otherwise $\operatorname{rank}(A) > r$ (proof by contradiction)

The proof then goes on...

  • $\begingroup$ I rewrote the red text because some of the expressions were going beyond the margins of the question, making things very difficult to read. Honestly, the coloring does not seem particularly necessary because the portion of the proof that you want help with is most of the portion that you have copied to your question. $\endgroup$ Jun 1, 2021 at 13:25
  • $\begingroup$ Could you go into detail about what exactly you didn't understand? In particular, do you understand that the vectors $u_1,\dots,u_r$ form a basis for $\operatorname{Im}(A)$? $\endgroup$ Jun 1, 2021 at 13:28
  • $\begingroup$ Hello, thank you very much for your suggestion. I did not understand how the fact that $Av_{i}\in\text{Im}(A)$ and $u_{i}\in\text{Im}(A)$ imply that $V_{n-r}\in\ker(A)$ it seems I got lost in this transition. @BenGrossmann $\endgroup$
    – Tesslaqwe
    Jun 1, 2021 at 13:35
  • 1
    $\begingroup$ I understand what you're having difficulty with now, thank you $\endgroup$ Jun 1, 2021 at 14:03
  • $\begingroup$ Alright thank you sir :) @BenGrossmann $\endgroup$
    – Tesslaqwe
    Jun 1, 2021 at 14:12

1 Answer 1


Thus, Thus, $v_{r+1},\dots,v_n \in \ker(A)$ i.e. $AV_{n-r} = 0$, since otherwise $\operatorname{rank}(A) > r$ (proof by contradiction).

Let's fill in the details of this proof. Suppose that for some $k > r$, $Av_k \neq 0$. If $Av_k$ is linearly independent from $Av_1,\dots,Av_r$, then we are forced to conclude that $\operatorname{Im}(A)$ has dimension greater than $r$, which is what is referenced by the parenthetical remark.

However, we still need to rule out the possibility that $Av_k$ is non-zero but the set $\{Av_1,\dots,Av_r,Av_k\}$ is linearly dependent. It is not clear how the author of the proof intended for the reader to deduce this, but here is one approach. We begin by supposing that $Av_k = c_1 Av_1 + \cdots + c_r Av_r$ and then concluding that we much have $c_i = 0$ for all $i = 1,\dots,r$. As the author notes, the vectors $u_i = \frac 1{\sigma_i} Av_i$ form an orthonormal basis for the image of $A$. It follows that for each $i = 1,\dots,r,$ $$ Av_k = c_1 Av_1 + \cdots + c_r Av_r \implies\\ Av_k = c_1 \sigma_1 u_1 + \cdots + c_r \sigma_r u_r \implies \\ \langle Av_k, u_i \rangle = c_i \sigma_i \implies\\ c_i = \frac{\langle Av_k, u_i\rangle}{\sigma_i} = \frac{\langle v_k, A^Tu_i\rangle}{\sigma_i} = \frac{\langle v_k, A^TA v_i\rangle}{\sigma_i^2} = \langle v_k,v_i\rangle. $$ However, the fact that $v_1,\dots,v_n$ are orthonormal implies that $\langle v_k,v_i\rangle = 0$. So indeed: if $Av_k$ is a linear combination of $Av_1,\dots,Av_r$, then it must hold that $Av_k = 0$.


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