Linear Algebra: Notation of vectors

Question

Can I write vectors in nullspaces or solution sets of linear systems as row vectors instead of column vectors? Do they mean the same thing?

For example, when a homogeneous equation Ax=0 has a general solution x =s$ (-1,1,0,0,0)^T \ $ + t$(-1,0,-1,0,1)^T \ $, where s and t are arbitrary parameters. Then the basis for the nullspace of A is $ \{(-1,1,0,0,0)^T \ $, $(-1,0,-1,0,1)^T \} $ Can I also write that the basis for nullspace of A to be $ \{(-1,1,0,0,0)$, $(-1,0,-1,0,1) \} $ as row vectors instead? Do they mean the same thing? I'm quite confused about the notation.

Answer 1

Conventionally, $x$ is taken to be a column vector, and since the nullspace is the set of all $x$ such that $Ax=0$, so the nullspace also consists of column vectors.

Answer 2

You have to pay attention: at first sight, column/row notations seem to be quite equivalent. But you should prefer the column notation for vectors and row notation for their transpose. Clearly, I am reffering to the coordinates of a vector with respect to a certain basis...

I don't know your level of knowledge, but I'll try to explain why: given a finite-dimensional vector space $V$, you can define its dual $V^*$: this the set of all linear maps $f:V\to K$, where $K$ is the field of scalars. If your space $V$ is finite-dimensional and comes with an inner product $\langle\cdot,\cdot\rangle$, then, there is a bijection $V^*\leftrightarrow V$, hence, each $f\in V^*$ can be identified with a vector $v\in V$. We will then denote the functional in $V^*$ associated with $v\in V$ by $v^*$. The action of $v^*$ on $V$ is as follows: for each $u\in V$ we define $v^*(u)=\langle v,u\rangle=v^T\cdot u$. Here the row vector $v^T$ and the column vector $u$ are evident.

Now, your answer: a column vector is used for the vector $v\in V$, while $v^*\in V^*$ is written as a row vector. This gives an immediate way to calculate scalar products using the rows-times-columns product for matrices.

Given a basis of $V$, say $\{e_1,...,e_n\}$, a basis for $V^*$ (which is again finite-dimensional of the same dimension of $V$) is given by $e_i^*$ acting as $e^*_i(e_j)=\delta_{i,j}$.

Answer 3

Row vectors are $1\times n$ matrices (matrix dimensions are given in row $\times$ column format), so $Ax=0$ would not make sense if $x$ was a row vector, as the matrix dimensions don't match. You could, rewrite the equation as $x^TA^T=0$ to get around that issue, but then notice that you have to take the transpose of $A$ to make it work...better to stick with the (somewhat annoying) column vector notation to avoid more serious potential for confusion.