Let
$A=\begin{bmatrix}
1 & 2 & 3 &4 \\
0 & 0 & 0 & 1
\end{bmatrix}$
What's dim(Null(A))+dim(Col(A))?
Is that the most reduced form it could take or is there more i can do to the matrix?
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$\begingroup$This is the most reduced form you can have. Here, $dim(Col(A))=2$, in fact the first and the last columns are linearly independent, while the second and the third are both multiples of the first one. Moreover, a known theorem of linear algebra asserts that $dim(Col(A))+dim(Null(A))=$number of columns, i.e. the dimension of the domain of the map represented by the matrix $A$. So, $dim(Null(A))=2$. Try to find a basis for it!
$\endgroup$ $\begingroup$\begin{align*} \dim(\operatorname{Col}(A)) &= \operatorname{Rank}(A) = \text{# of pivot columns} \\ \dim(\operatorname{Nul}(A)) &= \text{# of columns that aren't pivot columns} \\ \dim(\operatorname{Col}(A)) + \dim(\operatorname{Nul}(A)) &= \text{# of columns} \end{align*}
Wish my teacher explained these terms like this, because span and subspace are way too vague for me. Hope this helps.
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