Is support of an operator same as row space?

Question

Support of an operator is vector space that is orthogonal to its kernel.

Does this mean support is same as row space? How to calculate support for any matrix?

Answer 1

Yes, by Fundamental theorem of linear algebra.

Answer 2

Let's try to prove this statement in a slightly more rigorous way. We need to prove the following:

$\textbf{Claim 1.} \,\, \text{span}(r_1,r_2\ldots r_n) = \text{supp}(A)$ where $r_1,r_2 \ldots r_n$ are the rows of the given $n \times n$ operator $A$.

We can show this with the proving the following two claims:

$\textbf{Claim 1.1.} \,\,\text{span}(r_1,r_2\ldots r_n) \subseteq \text{supp}(A).$

$\textbf{Proof 1.1.} \,\, \langle \sum_i a_ir_i, x\rangle = 0$ for all combinations $\{a_i\}$ and $x \in \text{null}(A)$.

$\textbf{Claim 1.2.} \,\, \text{supp}(A) \subseteq \text{span}(r_1,r_2\ldots r_n).$

$\textbf{Proof 1.2.} \,\, $ Assume for contradiction that $\text{span}(r_1,r_2\ldots r_n) \subset \text{supp}(A)$. This implies $\text{dim}(\text{span}(r_1,r_2\ldots r_n)) < \text{dim}(\text{supp}(A))$. If $k \leq n$ is the rank of the operator $A$, then from the relation between dimensions of a subspace and its orthogonal complement along with rank-nullity theorem we get, $\text{dim}(\text{supp}(A)) = k$. Additionally, as the dimension of the row space is same as the operator rank $k$, we get a contradiction.