# For the matrices (a) find k such that Nul A

For the matrices (a) find k such that Nul A is a subspace of $$\displaystyle{\mathbb{{{R}}}}^{{k}}$$, and (b) find k such that Col A is a subspace of $$\displaystyle{\mathbb{{{R}}}}^{{k}}$$
$A=\begin{bmatrix}4&5&-2&6&0\\1&1&0&1&0\end{bmatrix}$

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Feas1981
a) In order to find Nul A, we must solve
$$\displaystyle{A}{x}={0}$$
Since A is $$\displaystyle{2}\times{5}$$ matrix, it can multyplied from left only by vector with 5 rows. That means x is in $$\displaystyle{\mathbb{{{R}}}}^{{5}}$$, i.e. k=5
b) As we saw in part a) $$\displaystyle{x}\in{\mathbb{{{R}}}}^{{5}}$$, so we know
Col $$\displaystyle{A}={\left\lbrace{y}:{y}={A}{x}\ \text{ for some }\ {x}\in{\mathbb{{{R}}}}^{{5}}\right\rbrace}$$
So we have a multiplication of $$\displaystyle{2}\times{5}$$ matrix A and $$\displaystyle{5}\times{1}$$ vector x. From definition of matrix muplication, y is $$\displaystyle{2}\times{1}$$ vector, i.e k=2
Result: a) 5, b) 2