Solution to $Ax=b$ is $x=\begin{bmatrix} 1\\ 0\\ 1\\ 0 \end{bmatrix}+\alpha_{1}\begin{bmatrix} 1\\ 1\\ -1\\ 0 \end{bmatrix}+\alpha_{2}\begin{bmatrix} 1\\ 0\\ 1\\ 1\end{bmatrix}$. For $b=(1,2,1)^T$, find $A$.

Somehow, Gilbert Strang says that it is obvious that first and third column should add up to $(1,2,1)^T$, and then he says that second column is third minus first and fourth is -(first + third). I don't see how he concluded all of this, so any clarification is very welcome.

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Note that the $\alpha_i$ given in the solution for $x$ are arbitrary. This means, that regardless of what values of $\alpha_i$ you have, the resulting $x$ is a solution for $Ax = b$.

Now, if we take $\alpha_1 = \alpha_2 = 0$, then we get that $x = (1,0,1,0)$ is a solution to $Ax = b$. If we look at how we multiply a matrix with a vector , then $(Ax)_i = \sum_{j} A_{ij}x_j$. Therefore, since $x_2=x_4 = 0$ and $x_1=x_3 = 1$, we get that $(Ax)_i = A_{i1} + A_{i3}$, for all $i$.

But this is equal to $b_i$, since $Ax = b$. Therefore, $b_i = A_{i1} + A_{i3}$, for all $i$.

Now, if you look at the first column vector $A_{j1}$ and the third column vector $A_{j3}$, the fact that $b_i = A_{i1} + A_{i3}$ for all $i$, is a restatement of the fact that $b$ is the sum of the first and third columns of $A$.


Similarly, use the fact that if $x_1$ and $x_2$ satisfy $Ax_1 = Ax_2 = b$ then $A(x_1 - x_2) = 0$ (zero vector). Now, see if you can take two different values of $\alpha_{i}$ for $x_1$ and $x_2$. Get a vector $x_1 - x_2 = y$ such that $Ay = 0$. Now, use the entries of $y$ , as I did above, to conclude certain relations about the columns of $A$. See what $y$ will give you the relations that Gilbert and Strang refer to.

You have that $$ A \begin{bmatrix} 1+\alpha_1+\alpha_2 \\ \alpha_1 \\ 1-\alpha_1+\alpha_2 \\ \alpha_2 \end{bmatrix}=\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} $$ If $A=[a_1\ a_2\ a_3\ a_4]$, this means that $$ (1+\alpha_1+\alpha_2)a_1 +\alpha_1a_2 +(1-\alpha_1+\alpha_2)a_3 +\alpha_2 a_4 =\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}=b $$ for every $\alpha_1$ and $\alpha_2$.

In particular you have $$ \begin{cases} a_1+a_3=b & (\alpha_1=0, \alpha_2=0) \\[4px] 2a_1+a_2+a_3=b & (\alpha_1=1, \alpha_2=0) \\[4px] 2a_1+a_3+a_4=b & (\alpha_1=0, \alpha_2=1) \end{cases} $$ This is quite similar to a linear system, isn't it? $$ \begin{bmatrix} 1 & 0 & 1 & 0 & b \\ 2 & 1 & 1 & 0 & b \\ 2 & 0 & 1 & 1 & b \\ \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 1 & 0 & b \\ 0 & 1 & -1 & 0 & -b \\ 0 & 0 & -1 & 1 & -b \\ \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 1 & 0 & b \\ 0 & 1 & -1 & 0 & -b \\ 0 & 0 & -1 & 1 & -b \\ \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 1 & 0 & b \\ 0 & 1 & -1 & 0 & -b \\ 0 & 0 & 1 & -1 & b \\ \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 1 & -1 & b \\ \end{bmatrix} $$ Thus $a_1=-a_4$, $a_2=a_4$, $a_3=a_4+b$. The fourth column can be anything; so the matrix $A$ is $$ \begin{bmatrix} -x & x & x+b & x \end{bmatrix} $$ where $x$ is an arbitrary $3\times 1$ column.

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