# Linear combination of the fundamental solutions a solution to a homogeneous linear system?

I'm reading a textbook about linear algebra and have trouble understanding how the author arrives at the notion of the fundamental solutions of a linear system:

Consider the reduced coefficient matrix A

$\begin{bmatrix} 0 & 1 & 2 & 0 & 3& -1\\0&0&0&1&2&0\end{bmatrix}$ of the homogeneous linear system $Ax = 0$.

Let $x=(x_1,x_2,x_3,x_4,x_5,x_6)^T$ be the general solutions vector. Solving for the corner variables $x_2 = −2x_3 −3x_5 + x_6$ and $x_4 = -2x_5$ gives a new expression $x=(x_1,−2x_3 −3x_5 + x_6,x_3,-2x_5,x_5,x_6)^T$ of the general solutions vector involving only the free variables.

At this point, the author computes the vector coefficients of each one of the free variables in $x$. These he calls the fundamental solutions $f_1,...,f_4$. Then $x$ can be written as $x=x_1 f_1 + x_3f_2+x_4f_3 + x_5f_4$

where $f_1=\begin{bmatrix} 1\\0\\0\\0\\0\\0\end{bmatrix}, f_2=\begin{bmatrix} 0\\0\\-2\\1\\0\\0\end{bmatrix}, f_3=\begin{bmatrix} 0\\-3\\0\\-2\\1\\0\end{bmatrix}, f_4=\begin{bmatrix} 0\\-1\\0\\0\\0\\1\end{bmatrix}$

such that every solution of $Ax=0$ is a linear combination of the fundamnetal solutions $f_1,...,f_4$.

I have trouble understanding this last step. I've searched online but have only found material dealing with fundamental solutions in the context of differential equations, and vector coefficients being presented as coordinate differences between two points? I would appreciate any help.

He simply calculated each "fundamental solution" by assigning 1 to one of the free variables and 0 to the others. Then by expressing x as the linear combination of $f_1, f_2, f_3, f_4$, one arrives at the general solution.

Note that this is just to say that the vector space of solutions to the linear system is of dimension $4$ and ${f_1, f_2, f_3, f_4}$ is its basis.

• Welcome to MSE. Please use MathJax. Aug 4, 2017 at 14:45
• Thanks for your answer, could you elaborate on the assignments to the variables? I can't see what you mean?
– user379802
Aug 4, 2017 at 14:52
• for example, $f_3$ was calculated by assigning $x_5 = 1 , x_1 = x_3 = x_6 = 0$. There seems to be a mistake in $f_2$, the nonzero indices need to be in positions 2 and 3. Aug 4, 2017 at 15:03
• Ah, I can see it. But aren't there then more mistakes in the presentation?-1 in $f_4$ must be 1, and the weights $x_4$ and $x_5$ in the linear combination must be replaced with $x_5$ and $x_6$ respectively, if I understand correctly.
– user379802
Aug 4, 2017 at 15:27
• Correct, there seem to be numerous mistakes Aug 4, 2017 at 16:34

Those $f_1,f_2,f_3,f_4$ as you get generates the null space of $A$ and also linearly independent. Hence the null space has dimension $4$.Observe that each of these $f_i$ are perpendicular to given row space of the matrix.So their linear combination also perpendicular to row vectors also hence belongs to null space of $A$.row vectors and the fundamental set forms the hole vector space $R^6$.