# Question of Theorem Singular Value Decomposition

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). Sep 30, 2021 at 16:15
• @user: hence "German/English portmanteau". (Though technically it should be a compound since the words are not truncated...) Sep 30, 2021 at 16:40
• @ArturoMagidin Ah ok I've misintepreted the german/english as a dubitative statement! Thanks
– user
Sep 30, 2021 at 16:42

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

• Hi, so sorry but I didn't understand it right, there's a way to explain it better
– riva
Sep 30, 2021 at 19:30
• I know that null space have that dimension, but why these fact imply in that equation?
– riva
Sep 30, 2021 at 19:55
• Refer to Rank–nullity theorem
– user
Sep 30, 2021 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, 2021 at 20:38
• I had misinterpreted your doubt at first. Let me now if now it is clear.
– user
Oct 1, 2021 at 7:00