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Suppose the complete set of solutions to $Ax = b$ where $b = [2, 4, 2]^T$ is given by $$ x = [2, 0, 0]^T + c[1, 1, 0]^T + d[0, 0, 1]^T. $$

This is from a YouToube Video of Gilbert Strang's lecture on Linear Algebra. He said that the rank of the matrix $A$ was $1$. I understand that $b$ is a column vector so its rank is $1$. How did the students identify the dimension of the null space of $A$ to be $2$? This is something that baffles me.

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

Firstly, why is it that if $[0, 0, 1]^T$ is in the null space, the third column is a zero vector.

Assuming $c = 1$, $d = 1,$

and $A$ is the matrix as proposed, in the first row of $B$, $1\times 2 - 1 \times 1 + 1 \times 0$ gives $1$ instead of $2.$

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    Your question is about the matrix $A$. Would you write it down in your post instead of giving the video link? – 2017-01-03
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    @Jack this is what is mentioned. The A matrix is not given – 2017-01-03

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I wonder if the easiest way to look at this is a follows:

$[2,0,0]^\top$ is a particular solution to $Ax=b$. Why? Because it is true that it doesn't matter what $c$ and $d$ values we pick: for any $c$ and $d$ we get a vector, $x = [2, 0, 0]^\top + c[1, 1, 0]^\top + d[0, 0, 1]^\top$ that satisfies $Ax=b$. Therefore, we can pick $c=0$ and $d=0$, and we end up with $x = [2, 0, 0]^\top.$

This is tantamount to saying that $c[1, 1, 0]^\top + d[0, 0, 1]^\top$ are really non-contributory. Or in other words, that they are in the null space of $A.$ If a system of three equations with three unknowns can be solved in more than one way is because the system is under-determined, and there are free variables.

$c[1, 1, 0]^\top + d[0, 0, 1]^\top$ represents all linear combinations of two linearly independent vectors in the null space, and any vector space (or subspace) observes closure under addition and scalar multiplication. Hence, $c[1, 1, 0]^\top + d[0, 0, 1]^\top=0.$

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    Hi, Thanks for answer. I modified my question to further expand on this question. Could you kindly help me out with this question – 2017-01-04
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    Sorry i do not know how to type in maths notation form so i would use square brackets for illustrations – 2017-01-04
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    $[\color{blue}{0},\color{red}{0},1]^\top$ is in the null space of $A$ because if you take $\color{blue}{0}\times \begin{bmatrix}1\\2\\1\end{bmatrix}+\color{red}{0}\times \begin{bmatrix}-1\\-2\\-1\end{bmatrix}+1\times \begin{bmatrix}0\\0\\0\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$ – 2017-01-04
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    So you are saying that there are 3 vectors in the solution x? First being [1\2\1] Second being [−1\-2\-1] and third being [0\0\0]. And the next question would be why not the second vector be a [0\0\0] as well? – 2017-01-04
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    $Ax$ where in this case $x=[\color{blue}{0},\color{red}{0},1]^\top=\begin{bmatrix}\color{blue}{0}\\\color{red}{0}\\1\end{bmatrix}$ is calculated (and that was the idea behind using colors) as taking $\color{blue}{0}\times \text{ first column of A} + \color{red}{0}\times \text{ second column of A} +1\times \text{ third column of A}.$ Now the two color zeros would nullify whatever first and second columns you could place in $A$, and as for the third entry of $x$ (in this case $1$), it doesn't even matter - any entry times the $\bf 0$ vector will result in the $\bf 0$ vector. – 2017-01-04
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    Can a possible solution of A be? – 2017-01-04
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    I think you are very close to seeing it clearly, but you may want to go to the first lectures of Professor Strang, where he talks about column and row interpretations of $Ax$. – 2017-01-04
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    Look up how [matrix vector multiplication works](https://en.wikipedia.org/wiki/Matrix_multiplication#Illustration) and try multiplying $A$ by $[0,0,1]^T.$ Then do the same with $[1,1,0]^T.$ What do you get? – 2017-01-04
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    Thanks all. Now i got it. @PiotrBenedysiuk I think the comment on doing matrix vector mulitplication enlightened me. – 2017-01-04
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    Thanks @Antoni Parellada as well for your kind assistance – 2017-01-04
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From the rank-nullity theorem, the following relation satisfies for an $m \times n$ matrix.

$$\text{rank}(A)+\text{nullity}(A)=n$$

The dimension of the null space is called the nullity of A.

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    Yes I understand this but where is the rank or rather the nullity? How did he get the rank of the matrix A to be 1? – 2017-01-03
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    See [here](https://www.youtube.com/watch?v=abYAUqs_n6I). – 2017-01-03
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Watch the video from 10:38 again.

First of all he points out that the size of the matrix $A$ is $3\times 3$. Secondly, the assumption of the complete sets of solutions tells you that $$ Null(A)=2. $$ Now the rank-nullity theorem tells you the rank of $A$ is given by $$ Rank(A)=3-Null(A)=1. $$

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    at which min mark should I watch? I started the video and it loaded from the start. And what is this assumption of complete set of solutions – 2017-01-03
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    @aceminer: I have edited my answer. – 2017-01-03
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    At 10:53, he said that "and I give you the 'complete solutions'". – 2017-01-03
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The homogeneous equation $A x = 0$ has as set of solutions $\mathcal{O}$ the set of differences of the solutions of the original equation (as $A x = A y = b$ implies $A (x - y) = 0$, and conversely if $Ax = b$ and $A u = 0$ then $A (x+u) = b$), so this set is $$ \mathcal{O} = \{ a\,[1, 1, 0]^T + b\,[0, 0, 1]^T : a, b \in \mathbb{R}\}, $$ a $2$-dimensional space.

Therefore the nullity of $A$ is $2$ and the rank is $3 - 2 = 1$.

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    how did you derive the last set of a and b? It seems likes there's a huge leap for me – 2017-01-03
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    Just take two of your solutions, and their difference. – 2017-01-04