A specific spanning property of a family of vectors

Question

Let $v_1, v_2, v_3, \dots$ be a family of vectors in $\mathbb R^n$ that span $\mathbb R^n$. Given this family, define the family of vectors \begin{align*} \begin{pmatrix} v_1 \\ \alpha_1 v_1 \end{pmatrix}, \begin{pmatrix} v_2 \\ \alpha_2 v_2 \end{pmatrix}, \begin{pmatrix} v_3 \\ \alpha_3 v_3 \end{pmatrix}, \dots \end{align*} in $\mathbb R^{2n}$. My question is under which conditions on $v_1, v_2, v_3, \dots$ and the $\alpha_1, \alpha_2, \alpha_3, \dots$ the family of vectors in $\mathbb R^{2n}$ is spanning. Partial solutions are also welcome.

Answer 1

Let's say there are $N$ indices $i$. Let $V$ be the $n \times N$ matrix whose columns are the $v_i$ and $A$ the $N \times N$ diagonal matrix with diagonal entries $\alpha_i$. Then what you're asking for is that the $2n \times N$ block matrix $$ \pmatrix{V\cr VA}$$ has rank $2n$. As Gerry commented, we need $N \ge 2n$. This is equivalent to there being a set of $2n$ indices such that the corresponding $2n \times 2n$ matrix $\pmatrix{W\cr W B}$ is nonsingular, where $W$ is the $n \times 2n$ submatrix of $V$ corresponding to those $2n$ columns, and $B$ the $2n \times 2n$ submatrix of $A$ corresponding to those $2n$ rows and columns.

For example, in the case $n=2$, $N=4$, the condition can be written as $$ \eqalign{&\left( v_{{1,3}}v_{{2,4}}-v_{{1,4}}v_{{2,3}} \right) \left( v_{{1,1} }v_{{2,2}}-v_{{1,2}}v_{{2,1}} \right) \alpha_{{1}}\alpha_{{2}}\cr -& \left( v_{{1,2}}v_{{2,4}}-v_{{1,4}}v_{{2,2}} \right) \left( v_{{1,1} }v_{{2,3}}-v_{{1,3}}v_{{2,1}} \right) \alpha_{{1}}\alpha_{{3}}\cr+& \left( v_{{1,2}}v_{{2,3}}-v_{{1,3}}v_{{2,2}} \right) \left( v_{{1,1} }v_{{2,4}}-v_{{1,4}}v_{{2,1}} \right) \alpha_{{1}}\alpha_{{4}}\cr+& \left( v_{{1,2}}v_{{2,3}}-v_{{1,3}}v_{{2,2}} \right) \left( v_{{1,1} }v_{{2,4}}-v_{{1,4}}v_{{2,1}} \right) \alpha_{{2}}\alpha_{{3}}\cr-& \left( v_{{1,2}}v_{{2,4}}-v_{{1,4}}v_{{2,2}} \right) \left( v_{{1,1} }v_{{2,3}}-v_{{1,3}}v_{{2,1}} \right) \alpha_{{2}}\alpha_{{4}}\cr+& \left( v_{{1,3}}v_{{2,4}}-v_{{1,4}}v_{{2,3}} \right) \left( v_{{1,1} }v_{{2,2}}-v_{{1,2}}v_{{2,1}} \right) \alpha_{{3}}\alpha_{{4}} \ne& 0}$$

Answer 2

As has been noted the number $N$ of vectors needs to be at least $2n$. Also the list $V=(v_1,\ldots,v_N)$ needs to contain a basis of $\mathbb{R}^n$, otherwise the matrix $V$ has rank less than $n$ and thus $$ \begin{pmatrix} V \\ VA \end{pmatrix}$$ has rank less than $2n$. Without loss of generality we may assume that the first $n$ of the list are linearly independent. Let's partition $V= (V_1,V_2)$ accordingly and do the same for the diagonal matrix. We are then looking at $$ \begin{pmatrix} V \\ VA \end{pmatrix}=\begin{pmatrix} V_1 & V_2 \\ V_1A_1 & V_2 A_2 \end{pmatrix}.$$ As $V_1$ is invertible we may eliminate the block $V_2$ without changing the rank. I.e. $$ \begin{pmatrix} V_1 & V_2 \\ V_1A_1 & V_2 A_2 \end{pmatrix}\begin{pmatrix} I_n & -V_1^{-1}V_2 \\ 0 & I_{N-m} \end{pmatrix}=\begin{pmatrix} V_1 & 0 \\ V_1A_1 & V_2 A_2 - V_1 A_1 V_1^{-1}V_2\end{pmatrix}.$$ The question is thus whether $V_2 A_2 - V_1 A_1 V_1^{-1}V_2$ has full rank, or equivalently whether $$ V_1^{-1}V_2 A_2 - A_1 V_1^{-1}V_2 $$ has full rank. This looks like a Sylvester equation in $X=V_1^{-1}V_2$. So for instance, if $V_2$ is also $n$-dimensional and the diagonal elements of $A_1$ and $A_2$ are distinct, you can take any invertible matrix $C$ and there exists a unique $X$ such that $$ X A_2 - A_1 X = C.$$ Then choose any invertible $V_1$ and let $V_2 = V_1 X$.