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say for example I have two 3 dimensional vectors that are linearly independent then I would be be able to form a 2 dimensional subspace which is a plane in the 3 dimension. the null space in my understanding is the missing vectors that are needed to represent the whole 3 dimension and conventionally is it formed by finding the orthogonal complement for both vectors in my subspace which is for sure linearly independent. my question is, do the nullspace vectors need to be orthogonal, I am under the assumption that any vector that is linearly independent is good enough to represnt the whole dimension ? if that is not the case, why is it conventional for them to be orthogonal is it for simplicity?

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  • $\begingroup$ Can you please state your definition of null space very carefully? $\endgroup$
    – littleO
    Jan 1, 2016 at 13:11
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    $\begingroup$ This is a bit of a strange question. We don't speak of "a null space" or "a row space" in isolation -- these two terms are properties of a particular matrix, but you don't mention any matrix in your question at all, which makes me suspect you're confusing some terms. $\endgroup$ Jan 1, 2016 at 13:11
  • $\begingroup$ Certainly you can do the following: Take any two independent vectors; they span a plane. If you supplement your two original vectors by any vector outside the plane, you get a basis for all of $\mathbb R^3$, and $\mathbb R^3$ will be a direct sum of your plane and the line spanned by the additional vector. The two spaces will just not be "row space" and "null space" of anything that is relevant for this operation. $\endgroup$ Jan 1, 2016 at 13:15
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    $\begingroup$ @Raed and you still haven't mentioned a matrix. Are you aware that "row space" and "null space" are, by definition, properties of a matrix? $\endgroup$ Jan 1, 2016 at 13:29
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    $\begingroup$ @RaedTabani: There is no such thing as "the nullspace for a given vector space". Vector spaces (or subspaces) do not have nullspaces; only matrices (and perhaps linear transformations) do. $\endgroup$ Jan 1, 2016 at 13:42

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Your observations can be formalized as follows:

Let $A$ be a $m \times n$ matrix over a field $K$ (e.g. $K = \mathbb{R}$ for geometric intuition). For every $1 \leq i \leq m$ let $A_i = (A_{i1}, \dotsc, A_{in})^T \in K^n$ be the transposed of the $i$-th row of $A$ and let $$ R = \mathrm{span}_K\{A_1, \dotsc, A_m\} \subseteq K^n. $$ For $x,y \in K^n$ let $x \cdot y = \sum_{i=1}^n x_i y_i$ and call $x$ and $y$ orthogonal if $x \cdot y = 0$. For any subspace $U \subseteq K^n$ let $$ U^\perp = \{x \in K^n \mid \text{$x \cdot y = 0$ for every $y \in U$}\} $$ be the orthogonal complement of $U$.

Then $(A \cdot x)_i= \sum_{j=1}^n A_{ij} x_j$ for all $x \in K^n$ and $1 \leq i \leq m$. Therefore $x \in \ker A$ if and only if $\sum_{j=1}^n A_{ij} x_j = 0$ for every $1 \leq i \leq m$. Notice that $\sum_{j=1}^n A_{ij} x_j = A_i \cdot x$. Thus $x \in \ker A$ if and only if $A_i \cdot x = 0$ for every $1 \leq i \leq m$, i.e. if $x$ is orthogonal to each row of $A$. It then follows that $x$ is already orthogonal to every $y \in R$ because $$ \left( \sum_{i=1}^m \lambda_i A_i \right) \cdot x = \sum_{i=1}^m \lambda \underbrace{A_i \cdot x}_{=0} = 0. $$ So we have $\ker A = R^\perp$. In this way is makes sense to say that the nullspace $\ker A$ is precisely the orthogonal complement $R^\perp$ of the subspace $R$ spanned by the rows of $A$.

(These who know the dual space will recognize this as a special case of the equality $\mathrm{im}(f^*) = \ker(f)^\perp$ where $f \colon V \to W$ is some linear map and $f^* \colon W^* \to V^*$ is the induced linear map.)

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