# If $(U,\Sigma,V)$ is a singular value decomposition of $A$, do the first $\text{rank}A$ columns of $V$ and $U$ form orthonormal bases?

Let

• $$m,n\in\mathbb N$$
• $$A\in\mathbb R^{m\times n}$$ and $$|A|:=\sqrt{A^TA}$$
• $$r:=\operatorname{rank}A$$
• $$\sigma_1>\cdots>\sigma_r>\sigma_{r+1}=\cdots=\sigma_n=0$$ denote the singular values of $$A$$

We say that $$(U,\Sigma,V)$$ is a singular value decomposition of $$A$$ if

1. $$U\in\mathbb R^{m\times n}$$ is a partial isometry;
2. $$\Sigma=\operatorname{diag}(\sigma_1,\ldots,\sigma_n)\in\mathbb R^{n\times n}$$;
3. $$V\in\mathbb R^{n\times n}$$ is orthogonal

and $$A=U\Sigma V^T\tag1.$$

Can we show that

1. The first $$r$$ columns $$(e_1,\ldots,e_r)$$ and $$(f_1,\ldots,f_r)$$ of $$V$$ and $$U$$ are orthonormal bases of $$\mathcal R(|A|)$$ and $$\mathcal R(A)$$, respectively?
2. $$\mathcal N(U)=\mathcal N(A)$$ (noting that $$\mathcal N(A)=\mathcal N(|A|)$$)?
• What have you tried? May 27, 2020 at 13:27
• @user251257 The claims are clear to me when $(U,\Sigma,V)$ is constructed by the polar decomposition and spectral theorem. I'm now trying to figure out whether they hold for any $(U,\Sigma,V)$ with the given assumptions in the questions. It's clear to me that the columns of $V$ are an orthonormal basis of $\mathbb R^n$ (since they are orthonormal). May 27, 2020 at 13:56
• Hint: let $y=Ax$ then $y = U\Sigma V^T x= \sum_{i=1}^r u_i \sigma_i v_i^T x$. May 27, 2020 at 14:01
• @user251257 I don't see why $\sigma_i$ pops up in your equation. We've got $A^TA=V\Sigma U^TU\Sigma V^T$. The question is: Why can we remove the $U^TU$? We know that it is an orthogonal projection onto $\mathcal N(U)^\perp$. May 28, 2020 at 14:06
• $U^TU$ is the identity, as $U$ is orthogonal by assumption. But I only used $A$ not $A^TA$. May 28, 2020 at 14:08

You have $$\tag1 AA^*=U\Sigma^2 U^*.$$ Taking the trace in $$(1)$$, $$\operatorname{Tr}(\Sigma^2)=\operatorname{Tr}(AA^*)=\operatorname{Tr}(U\Sigma^2U^*)=\operatorname{Tr}(\Sigma^2 U^*U).$$ Then $$0=\operatorname{Tr}(\Sigma^2\,(I-U^*U))=\operatorname{Tr}(\Sigma\,(I-U^*U)^2\,\Sigma)$$ As the trace is faithful, we get $$\Sigma(I-U^*U)^2\Sigma=0$$, and so $$(I-U^*U)\Sigma=0$$. So $$\tag2 \Sigma=U^*U\Sigma=\Sigma\,U^*U.$$ Now $$A^*A=V\Sigma U^*U\Sigma V^*=V\Sigma^2\,V^*.$$ Now $$\ker A=\ker A^*A=\ker V\Sigma^2 V^*=\ker \Sigma V^*.$$ So, taking orthogonals,
$$\operatorname{ran} A^*=\operatorname{ran}V\Sigma.$$ This shows that the first $$r$$ columns of $$V$$ span the range of $$A^*$$ (which is the same as the range of $$|A|$$). Going back to $$(1)$$, $$\ker A^*=\ker AA^*=\ker U\Sigma^2\,U^*=\ker \Sigma U^*,$$ so $$\operatorname{ran} A=\operatorname{ran} U\Sigma,$$ so the first $$r$$ columns of $$U$$ span the range of $$A$$.
It is not true in general that $$\ker A=\ker U$$. For instance take $$A=\begin{bmatrix} 0&0\\1&0\end{bmatrix} \,\begin{bmatrix} 1&0\\0&0\end{bmatrix} \,\begin{bmatrix} 0&1\\1&0\end{bmatrix} =\begin{bmatrix} 0&0\\0&1\end{bmatrix}.$$
• Thank you for your answer. How did you remove $U^\ast U$ in $A^*A=V\Sigma U^*U\Sigma V^*=V\Sigma^2\,V^*$? May 28, 2020 at 17:43
• If $\Sigma$ is trace class, I would think it should work. With a few closures of ranges thrown in. May 28, 2020 at 17:48
• I didn't obtain that, you gave it in your hypotheses, when you said that the diagonal of $\Sigma$ were the singular values of $A$. May 28, 2020 at 18:19