I'm self-studying a bit of introductory linear algebra, watching the lectures on MIT OCW given by Gilbert Strang. The course isn't too rigourous and gives many things without proof, most of which I can reason through and convince myself of so far, but I've run into one thing that I can't wrap my head around.

So he's discussing an algorithm to solve the system of linear equations $\boldsymbol{\mathbf{A}}\boldsymbol{\mathbf{x}} = \boldsymbol{\mathbf{b}}$. And basically he says that all we need to do is find a particular solution to the equation (after elimination) and then add it to any vector in the nullspace of the matrix. It's obvious why the sum of the particular solution and the nullspace vector is part of the solution set to $\boldsymbol{\mathbf{A}}\boldsymbol{\mathbf{x}} = \boldsymbol{\mathbf{b}}$. It's not obvious to me, however, why all solutions to the equation can be described as that type of sum. Could someone explain to me why this is true?

  • 2
    $\begingroup$ If $\mathbf{Ax} = \mathbf{b} = \mathbf{Ay}$ then $\mathbf{A}(\mathbf{x}-\mathbf{y}) = \mathbf{Ax} - \mathbf{Ay} = \mathbf{b}-\mathbf{b}= \mathbf{0}$, so $\mathbf{A}(\mathbf{x}-\mathbf{y}) = 0$ and hence $\mathbf{x}-\mathbf{y}$ is an element of the nullspace of $\mathbf{A}$. In words: any two solutions $\mathbf{x}$ and $\mathbf{y}$ differ by an element of the nullspace of $\mathbf{A}$. $\endgroup$
    – t.b.
    Jul 12, 2011 at 16:00
  • $\begingroup$ And this is analogous to the recipe to find the general solution of a linear differential equation: the "particular" solution plus the "general" solution of the homogeneous equation. $\endgroup$
    – leonbloy
    Jul 12, 2011 at 16:23

1 Answer 1


Sure, if $Ax_1 = b$ and $Ax_2 = b$ then by linearity, $A(x_1 - x_2) = 0$ hence $x_1-x_2$ is a nullspace vector.

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    $\begingroup$ Of course, can't believe I missed that, thanks @Gortaur $\endgroup$
    – Arpon
    Jul 12, 2011 at 16:03

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