# The space spanned by the rows of the Kernel matrix.

Let's consider the linear model $y=Ax$ where $y$ is $m$ dimensional vector, $A$ is $m\times n$ matrix and $x$ is $n$ dimensional vector.

We have $A:\mathbb{C}^{n}\longrightarrow \text{Range}(A)$ and we know that $\mathbb{C}^n = \text{Range}(A^T) \oplus \text{Ker}(A)$ where $\text{dim}(Ker(A))=s$. Let $B=Null(A)$ be the $n\times s$ matrix that contains the $s$ vectors that span the subspace $\text{Ker}(A)$.

My question is can we find a subspace $L$ spanned by $s$ linearly independent rows from the $n$ rows of $B$ where the projection of $x$ onto $L$ is the null vector (0 everywhere). If so, how can we prove that.

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This question has been cross-posted on Computational Science Stack Exchange. See scicomp.stackexchange.com/q/10860/276. –  Geoff Oxberry Feb 24 at 19:26