Product of matrix and vectors in orthogonal subspaces Let $p < n$, and let $A \in \mathbb{R}^{n \times n}$ be rank $p$. Let $V \in \mathbb{R}^{n \times n}$ be an orthogonal matrix with columns $V = \begin{bmatrix} v_1 & \cdots & v_n \end{bmatrix}$.
Suppose that for any linear combination of the first $p$ columns of $V$, that the product $A(c_1 v_1 + \cdots + c_p v_p) \ne 0$. Is it true that $A v_i = 0$ for all $i > p$?
I believe the answer to this is yes. By rank-nullity theorem, $A$ must have nullity $n-p$. Since $Aw \ne 0$ for any $w$ spanned by $v_1, \cdots, v_p$, and since the subspace ${\rm span}(v_1, \cdots ,v_p)$ is orthogonal to the subspace ${\rm span}(v_{p+1}, \cdots ,v_n)$, it must follow that $A v_i = 0$ for all $i > p$.
Any corrections or a more detailed proof would be much appreciated!
 A: I suppose that you mean that $A(c_1 v_1 + \cdots + c_p v_p) \ne 0$ whenever at least one coefficient $c_i$ is non-zero.
The answer is no. For example, suppose that $V$ is a $4\times 4$ identity matrix, and take
$$
A = \pmatrix{1&0&1&0\\0&1&0&1\\0&0&0&0\\0&0&0&0}.
$$
In this case, we have $Av_1 = Av_3 = v_1$ and $Av_2 = Av_4 = v_2$.
This still fails in the case where $A$ is symmetric. As an example, take $V$ to be the identity matrix and
$$
A = \pmatrix{1&1\\1&1}.
$$
It is indeed the case that $A(c_1 v_1) = 0 \iff c_1 = 0$, but $Av_2 \neq 0$.
A similar statement applies in the case where
$$
A = \pmatrix{1&0&1&0\\0&1&0&1\\1&0&1&0\\0&1&0&1}
$$
and $V$ is, again, the identity matrix.
A: $A=\begin{pmatrix}1&0&1\\0&1&0\\0&0&0\end{pmatrix}$
$\operatorname{Rank} A=p=2$
$V=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$
Then $\begin{align}A(c_1v_1+c_2v_2) &=\begin{pmatrix}1&0&1\\0&1&0\\0&0&0\end{pmatrix}\begin{pmatrix}c_1\\c_2\\0\end{pmatrix}\\  &=\begin{pmatrix}c_1\\c_2\\0\end{pmatrix}\neq 0 [\text{ as $c\neq 0$ }]\end{align}$
$\begin{align}Av_3&=\begin{pmatrix}1&0&1\\0&1&0\\0&0&0\end{pmatrix}\begin{pmatrix}0\\0\\1\end{pmatrix}\\  &=\begin{pmatrix}1\\0\\0\end{pmatrix}\neq 0\end{align}$
A: The previous answers are sufficient answers to the question as they provide counterexamples. To understand the question geometrically, consider the inner product space $\mathbb{R}^n$ with the standard Euclidean inner product.
The condition $A(c_1 v_1 + \cdots + c_p v_p) \ne 0$ for any $c_1, \ldots, c_p$ not all equalling $0$ implies in particular that
$$A v_i \neq 0$$
for $i = 1, \ldots, p$. This means that every $v_i$ for $i \leq p$ has a non-zero projection onto the row space of $A$. In the first example of Ben Grossmann, $v_1$ and $v_2$ form a $45$ degree angle with their orthogonal projections onto the row space of A. It just so turns out that $v_3$ and $v_4$ do as well.
Geometrically in $\mathbb{R}^2,$ you can see very easily that a nonzero projection of  $v_1 = (1, 0)$ onto a 1-dimensional subspace (a line, the row space of $A$) does not mean that $v_2 = (0, 1)$ is orthogonal to that line. In fact, this happens if and only if the 1-dimensional subspace is $\{(\alpha, 0) : \alpha \in \mathbb{R}\}$! This is the crux of your question in probably the simplest example.
A similar question to yours that yields a positive answer is this. Let $p < n$, and let $A \in \mathbb{R}^{n \times n}$ have rank $p$. Let $V \in \mathbb{R}^{n \times n}$ be an orthogonal matrix such that $A(c_{p+1} v_{p+1} + \cdots + c_n v_n) = 0$ for any selection of coefficients in $c_{p+1}, \ldots, c_n  \in \mathbb{R}$. Then $A(c_1 v_1 + \cdots + c_p v_p) \ne 0$ for any $c_1, \ldots, c_p$ not all equalling $0$.
