Let $\mathbf{A}$ be a $n\times n$ matrix with Singular Value Decomposition (SVD) $\mathbf{A}=\mathbf{U}\mathbf{S}\mathbf{V}$ and $\mathbf{a}_1$ be the first row of $\mathbf{A}$. What can we say about the Singular Values and Vectors of $\mathbf{B}$ based on the SVD of $\mathbf{A}$? \begin{align} \mathbf{B}=\begin{bmatrix} \mathbf{A}\\\mathbf{a}_1 \end{bmatrix}. \end{align}
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$\begingroup$ $B$ is a rank-one update of $A$, so you can use the various results for that case, see for example stats.stackexchange.com/q/177007/338593 $\endgroup$– Carlo BeenakkerMar 6, 2022 at 16:46
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$\begingroup$ Yes. There is a numerical solution. Is there a closed form solution for this particular case? $\endgroup$– Math_YMar 6, 2022 at 21:26
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1 Answer
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The singular values of $\sigma_i(B)$ are given by $$\sigma_i(B)^2=\lambda_i(B^*B)=\lambda_i(BB^*).$$
We have that $$\begin{align} \mathbf{B}\mathbf{B}^*=\begin{bmatrix} \mathbf{A}\mathbf{A}^*&\mathbf{A}a_1^*\\\mathbf{a}_1\mathbf{A}^*&a_1a_1^* \end{bmatrix} \end{align}$$ is a $(n+1)\times (n+1)$ block matrix. The Cauchy's interlacing theorem gives the inequality $$\lambda_{n+1}(\mathbf{B}\mathbf{B}^*)\leq \lambda_n(\mathbf{A}\mathbf{A}^*)\leq \cdots \leq \lambda_2(\mathbf{B}\mathbf{B}^*)\leq \lambda_1(\mathbf{A}\mathbf{A}^*)\leq \lambda_1(\mathbf{B}\mathbf{B}^*).$$
Remark: You can find related results searching for "\(A^*A+a_1^*a_1\) with \(a_1\) a row matrix singular values" on SearchOnMath. Like this thread which gave me some glimpses on what I wrote above.
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$\begingroup$ In particular, $B$ has rank $\leq n$, so $\lambda_{n+1}(BB^*) = 0$. $\endgroup$ Feb 4, 2023 at 17:24