# Dimension of solution space of homogeneous system of linear equations

I have the homogeneous system of linear equations

$$3x_1 + 3x_2 + 15x_3 + 11x_4 = 0,$$ $$x_1 − 3x_2 + x_3 + x_4 = 0,$$ $$2x_1 + 3x_2 + 11x_3 + 8x_4 = 0.$$

I have converted to a augmented matrix and row reduced to $$\begin{bmatrix}1 & 0 & 4 & -3 & 0\\0 & 1 & 1 & 2/3 & 0\\0 & 0 & 0 & 0 & 0\end{bmatrix}$$

And came up with the general solution: $$\begin{bmatrix}x_1 \\x_2\\x_3\\x_4\end{bmatrix} = s\begin{bmatrix}4 \\-1\\1\\0\end{bmatrix}+ t\begin{bmatrix}-3\\-2/3\\0\\1\end{bmatrix}$$

I know that the basis is:

$$\left\{\begin{bmatrix}4 \\-1\\1\\0\end{bmatrix},\begin{bmatrix}-3\\-2/3\\0\\1\end{bmatrix}\right\}$$

But how do I determine the dimension of the solution space?

• Why did you bother to augment the coefficient matrix? The last column will remain all zero regardless of the row operations that you perform. – amd May 16 '16 at 3:04

You don't have to write an augmented matrix since you have a homogeneous system. Just compute the rank using Gauß' pivot method; \begin{align*}\begin{bmatrix}3&3&15&11\\ 1&-2&1&1\\ 1&3&4&8\end{bmatrix}&\rightsquigarrow\begin{bmatrix} 1&-2&1&1\\3&3&15&11\\ 1&3&4&8\end{bmatrix}\rightsquigarrow \begin{bmatrix} 1&-2&1&1\\0&12&12&8\\ 0&9&9&6\end{bmatrix}\\\begin{bmatrix} 1&-2&1&1\\0&3&3&2\\ 0&3&3&2\end{bmatrix}&\rightsquigarrow\begin{bmatrix} 1&-2&1&1\\0&3&3&2\\ 0&0&0&0\end{bmatrix}\end{align*} Hence the rank of the matrix is $2$. This rank is the codimension of the system $S$ of solutions, i.e. $4-\dim S$. Thus $\dim S=2$.
• So if I understand correctly, the solution space lies in $\mathbb{R}^4$ since 4 variables, the rank is 2 meaning a subspace of dimension 2 is null space and the rest is the solution space? So is dim $S$ = dim image($A$) ?? – Aditya P Nov 19 '17 at 17:39
• In the present case, yes. That $\dim S=\dim(\operatorname{Im}A)$ results from the rank-nullity theorem, it's not the general situtaion. – Bernard Nov 19 '17 at 19:03