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I have an exercise where I have to say if a vector $\vec{u} = \left(\begin{matrix}1 \\ 2\end{matrix}\right)$ is or not in a column space of a matrix $A = \left(\begin{matrix}1 & -2\\ -2 & 4\end{matrix}\right)$. I am quite newbie, and I am still not so comfortable with these concepts.

What I did was to create an augment matrix:

$$\left(\begin{array}{cc|c}1 & -2 & 1 \\ -2 & 4 & 2\end{array}\right)$$

If I try to reduce this I get the second row like $\left(\begin{array}{cc|c}0 & 0 & 4 \end{array}\right)$.

Which I think means that the $\vec{v}$ is not in $A$, because $0 +/- 0 \neq 4$.

Could you provide a more technical explanation of why? I know this might seem to easy, but just to understand better...

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You did perform the operation correctly.

The property you are taking advantage of with this method is the fact that the column space of a matrix is the same as the range of the corresponding matrix transformation (i.e. $x \mapsto A \vec{x}$). By definition of the range of a function, $\vec{u}$ is in the range if and only if there exists some $\vec{x}$ such that $A \vec{x} = \vec{u}$. So you attempted to solve that matrix equation, and determined that there was no solution (by producing an inconsistency).

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  • $\begingroup$ And if it's consistent, does that mean it's in the column space or the null space? $\endgroup$ – Aaron Franke Mar 30 at 3:31
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Your row-reduction is a general method, and you've done it correctly.

However, one easy way you can see that your vector is not in the column space of that specific matrix is to notice that the columns are scalar multiples of each other (multiply the first by $-2$), so the column space is a line in $\mathbb{R}^2$ with slope $-2$ passing through the origin, and the point $(1,2)$ is very clearly not on this line.

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To show something is in the span of a set of vectors, you want to show that it's a linear combination of those vectors. So technically what you're doing is you're looking for constants $c_1$ and $c_2$ such that $\left(\begin{matrix}1 \\ 2\end{matrix}\right)=c_1\left(\begin{matrix}1 \\ -2\end{matrix}\right)+c_2\left(\begin{matrix}-2\\4\end{matrix}\right)$, which is equivalent to the equation $A\vec{c}=\vec{u},$ where $\vec{c}=\left(\begin{matrix}c_1\\c_2\end{matrix}\right)$. And then this equation can be solved(or shown to be inconsistent) by the method you used.

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  • $\begingroup$ And if it's consistent, does that mean it's in the column space or the null space? $\endgroup$ – Aaron Franke Mar 30 at 3:31
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what you did is the correct procedure. it works no matter the size of the system. the key point was there was a pivot on the last column indicating an equation of the form $$0x_1 + 0x_2 + \cdots + 0 x_n = 1$$ which cannot be satisfied. that in turn means that the last column is not a linear combination of the previous columns.

the problem you have in your post deals with column vectors in a plane. there linear independence means one vector is a multiple of the other. you can see that second column is a multiple of the first column and the third is not. therefore, the third column cannot be in the column space of the first two columns.

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You could form the projection matrix, $P$ from matrix $A$:

$$P = A(A^TA)^{-1}A^T$$

If a vector $\vec{x}$ is in the column space of $A$, then

$$P\vec{x} = \vec{x}$$

i.e. the projection of $\vec{x}$ unto the column space of $A$ keeps $\vec{x}$ unchanged since $\vec{x}$ was already in the column space.

$\therefore$ check if $P\vec{u} \stackrel{?}{=} \vec{u}$

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