# Dimension of solution space of system of linear equations

I have the system of linear equations

\begin{align} x_1+x_3=0 \\ -4x_1+2x_2=0 \\ -13x_1+4x_2-5x_3=0 \end{align}

and I have converted it to an augmented matrix and row reduced to

\begin{align} \begin{pmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \end{align}

which, as far as I know, tells me that the dimension of the solution space is $$2$$ since there are $$2$$ linearly independent rows. However, the general solution

$$\begin{pmatrix} -x_3 \\ -2x_3 \\ x_3 \end{pmatrix}$$ tells me that the dimension of the solution space is actually $$1$$.

Question: Could someone tell me where I've made a mistake?

• The dimension of the space of solutions is not the number of independent rows, but the number of variables minus the number of independent rows: $3-2=1$. Each independent row is a restriction, it cuts down a degree of freedom from the total space of values of the variables that is available for the the space of solutions. – MoonLightSyzygy Dec 30 '19 at 20:57

Dimension of the solution space is equal to number of basis vector in fundamental system of solutions: 1 in your case. In general case, from rank-nullity theorem for linear map $$L:V\rightarrow W$$ $$\dim(Ker(L))+\dim(Im(L))=\dim(V)$$. In your case $$L:\Bbb R ^3\rightarrow \Bbb R ^3$$ and $$\dim(Im(L))=2$$, so $$dim(Ker(L))=1$$ and dimension of the solution space is exactly 1.