linear algebramatricessvd
I have to find a reduced SVD, A = UrDVrT, for the matrix A = BQT.
They give me the matrix B and the null-space of $A$. I've already found the singular values for A and the matrix Vr (with an orthonormal basis of row space of A) but I'm not so sure about how to find a basis for Col(A).
Is saying that Col(A) is the same as Col(B) correct? If so, why?
Let's start out from the standard basis $e_1,..,e_n$. Let $a_1,..,a_k$ be the column vectors of $A$.
Check that the step on rows $r_i':=r_i+\lambda\,r_j$ corresponds to the basis transformation $e_j':=e_j-\lambda\,e_i$, that is, for a vector $v$ we have $$v=\sum_i\alpha_ie_i=\sum_i\alpha_i'e_i'$$ where the row transformation is made for the coordinate vector $\pmatrix{\alpha_1\\ \alpha_2\\ \vdots} \leadsto \pmatrix{\alpha_1'\\ \alpha_2'\\ \vdots}$.
So, in this interpretation the column vectors all "stay" where they are in the $n$ dimensional space, but we keep on changing the basis. Of course, the vectors stay (in-)dependent. The crucial thing here is that we get another basis when applying (the inverse of) each step of the row transformation.
The topic of low rank approximation is sprinkled throughout Math SE:
Low-rank Approximation with SVD on a Kernel Matrix
Matrix values increasing after SVD, singular value decomposition
The singular value spectrum may span several orders of magnitude. It seems natural that the contributions from the larger values are more important. Numerically, it is difficult to tell whether small singular values are valid or simply machine noise in computing a $0$ singular value. This requires a threshhold to determine which singular values are discarded.
Let's look at the SVD in detail.
Singular Value Decomposition
Every matrix $$ \mathbf{A} \in \mathbb{C}^{m\times n}_{\rho} $$ has a singular value decomposition of the form $$ \begin{align} \mathbf{A} &= \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\ % &= % U \left[ \begin{array}{cc} \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}} \end{array} \right] % Sigma \left[ \begin{array}{cccc|cc} \sigma_{1} & 0 & \dots & & & \dots & 0 \\ 0 & \sigma_{2} \\ \vdots && \ddots \\ & & & \sigma_{\rho} \\\hline & & & & 0 & \\ \vdots &&&&&\ddots \\ 0 & & & & & & 0 \\ \end{array} \right] % V \left[ \begin{array}{c} \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ \color{red}{\mathbf{V}_{\mathcal{N}}}^{*} \end{array} \right] \\ % & = % U \left[ \begin{array}{cccccccc} \color{blue}{u_{1}} & \dots & \color{blue}{u_{\rho}} & \color{red}{u_{\rho+1}} & \dots & \color{red}{u_{n}} \end{array} \right] % Sigma \left[ \begin{array}{cc} \mathbf{S}_{\rho\times \rho} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right] % V \left[ \begin{array}{c} \color{blue}{v_{1}^{*}} \\ \vdots \\ \color{blue}{v_{\rho}^{*}} \\ \color{red}{v_{\rho+1}^{*}} \\ \vdots \\ \color{red}{v_{n}^{*}} \end{array} \right] % \end{align} $$
The connection to the row and column spaces follows: $$ \begin{align} % R A \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} &= \text{span} \left\{ \color{blue}{u_{1}}, \dots , \color{blue}{u_{\rho}} \right\} \\ % R A* \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} &= \text{span} \left\{ \color{blue}{v_{1}}, \dots , \color{blue}{v_{\rho}} \right\} \\ % N A* \color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)} &= \text{span} \left\{ \color{red}{u_{\rho+1}}, \dots , \color{red}{u_{m}} \right\} \\ % N A \color{red}{\mathcal{N} \left( \mathbf{A} \right)} &= \text{span} \left\{ \color{red}{v_{\rho+1}}, \dots , \color{red}{v_{n}} \right\} \\ % \end{align} $$ You are using is $\mathbf{S} \, \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*}$. This ignores the null space contributions in red.
Best Answer
In general, we have ${\cal R} (MN) \subset {\cal R}M$.
Hence we have ${\cal R} (B Q^TQ) \subset {\cal R} (B Q^T) \subset{\cal R} B$ and since $Q^TQ=I$ we have the desired result.