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$\def\Col{\mathop{\mathrm{Col}}}\def\Nul{\mathop{\mathrm{Nul}}}$In preperation for a linear algebra exam I have coming up I was asked to construct a $3×3$ matrix $A$ and nonzero vector $b$ such that $b$ is in $\Col A$ but $b$ is not the same as any of the columns in $A$. I came up with the following:

Let $$A = \begin{bmatrix}1&-1&-1\\4&-2&-2\\1&-2&0\end{bmatrix}$$

and let $$b = \begin{bmatrix}2\\1\\1\end{bmatrix}.$$

Would this be correct? If not, what is a good answer? Any pointers someone can give on how to answer questions like this?

Can I also say $b$ is in $\Nul A$?

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  • $\begingroup$ To check if $b$ is in nul$A$, check if $Ab=0$. $\endgroup$
    – take008
    Mar 22 '18 at 4:17
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So basically you have to find out nonzero vector $b$ which is in columns space of $A$ but not in columns of $A$. Take $b=pC_1+qC_2$, where $C_1$ and $C_2$ are independent columns of $A$ and $p$, $q$ are any non zero real numbers.

For your answer, consider $(2\; 1\; 1)=p(1\;4\;1)+q(-1\;-2\;-2)+r(-1\;-2\;0)$ and check that atleast two of $p,q$ and $r$ is non zero. If such $p,q$ and $r$ does not exist that it is not possible.

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Whenever any matrix is multiplied by a column vector (of compatible size) the resulting vector is in the column space of that matrix. And this process covers the whole of column space of the matrix. (If you end up getting zero vector as the answer the column vector used is in the null space).

So what you should do is take a random column vector $v$ (say all components non-zero) and multiply with the matrix given the result will be the desired one: $b =Av$

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