Question of Theorem Singular Value Decomposition

Question

Let $A$ a matrix of order $m \times n$ of rank $r$; so there are real numbers $\sigma_1 \geq \ldots \geq \sigma_r > 0$, a orthonormal basis $v_1, \ldots , v_n $ of $\mathbb{R}^n$ and a orthonormal basis $u_1, \ldots , u_m $ of $\mathbb{R}^m$ such that $$Av_i = \sigma_i u_i \ \ \ \text{for} \ \ \ i=1, \ldots r \ \ \ \text{and} \ \ \ Av_i =0 \ \ \text{for} \ \ \ i = r+1, \ldots ,n $$

$$A^Tu_i = \sigma_i v_i \ \ \ \text{for} \ \ \ i=1, \ldots r \ \ \ \text{and} \ \ \ A^Tu_i =0 \ \ \ \text{for} \ \ \ i = r+1, \ldots ,n .$$

The vectors $v_1, \ldots , v_n $ are autovectors of $A^TA$, $u_1, \ldots u_m$ are autovectors of $AA^T$ and $\sigma_1^2, \ldots, \sigma_1^r$ are nonzero autovalues of $A^TA$ and $AA^T$.

Proof: Let $v_1, \ldots, v_n$ a orthonormal basis of $\mathbb{R}^n$ formed by autovalues of $A^TA$ and let $\lambda_1, \ldots, \lambda_n$ the autovalues associated. We have that every autovalue of $A^TA$ is not negative.

Assume $v_1, \ldots , v_n$ ordered so that $\lambda_1 \geq \ldots \geq \lambda_n$.

How $r=rank(A)=rank(A^TA)$, we have that $$ \lambda_1 >0, \ldots ,\lambda_r>0 \ \ \ \text{and } \ \ \ \lambda_{r+1}= \ldots = \lambda_n=0.$$

For $i=1, \ldots r$, we define $$ \sigma_i = \Vert Av_i \Vert_2 \ \ \ \text{and} \ \ \ u_i = \frac{1}{\sigma_i}Av_i.$$ These conditions imply that

$$ Av_i = \sigma_i u_i, \ \ \ \text{and} \ \ \ \Vert u_i \Vert_2 = 1 \ \ \ \text{and} \ \ \ \sigma_i^2= \lambda_i, $$ for $i=1, \ldots, r$.

Now my question: Why the next equality is true?

If $\lambda_i = 0$, for $i= r+1, \ldots, n$, so

$$Av_i = \lambda_i v_i = 0v_i = 0.$$

I am sorry for my english.

In English, they are usually called eigenvectors and eigenvalues (though that is a German/English portmanteau). — Arturo Magidin, Sep 30 '21 at 16:15
@user: hence "German/English portmanteau". (Though technically it should be a compound since the words are not truncated...) — Arturo Magidin, Sep 30 '21 at 16:40
@ArturoMagidin Ah ok I've misintepreted the german/english as a dubitative statement! Thanks — user, Sep 30 '21 at 16:42

user · Answer 1 · 2021-10-01T06:59:07.690

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Since rank of $A$ is $r$, we have that null space of $A$ has dimension $n-r$, that is

$$Av_i = 0 \implies A^TAv_i=0$$

and since $A^TA$ has also rank $r$, we have that exactly $n-r$ zero eigenvalues $\lambda_i=0$ exist (counting multiplicity) such that

$$A^TAv_i = \lambda_i v_i = 0$$

Refer also to

Prove $\operatorname{rank}A^TA=\operatorname{rank}A$ for any $A\in M_{m \times n}$

edited Oct 01 '21 at 06:59

answered Sep 30 '21 at 16:15

user

162,563

Hi, so sorry but I didn't understand it right, there's a way to explain it better – riva Sep 30 '21 at 19:30
I know that null space have that dimension, but why these fact imply in that equation? – riva Sep 30 '21 at 19:55
Refer to Rank–nullity theorem – user Sep 30 '21 at 19:58
Yes, I agree. But $\lambda_i$ is a eigenvalue of $A^TA$, why is it an eigenvalue of A? why $Av_i=\lambda_i v_i$ ? – riva Sep 30 '21 at 20:38
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I had misinterpreted your doubt at first. Let me now if now it is clear. – user Oct 01 '21 at 07:00

Question of Theorem Singular Value Decomposition

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