From S.L Linear Algebra:
Find the dimension of the space of solutions of the following systems of equations. Also find a basis for this space of solutions.
(a) $4x + 7y - \pi z = 0$, $2x - y +z = 0$.
We see that matrix associated with homogeneous equations (a) is:
$$A = \begin{pmatrix} 4 & 7 & - \pi \\ 2 & -1 & 1 \end{pmatrix}$$
It's therefore obvious that rank of the matrix is at most $2$. To get the precise result I simply ensure that there is a linear independence between first two rows:
$$a(4, 7, -\pi) + b(2, -1, 1) = (0, 0, 0)$$
Thus $4a + 2b =0$, $7a - b = 0$ and $-\pi a + b=0$. Considering that $a = -\frac{b}{2}$, $7(-\frac{b}{2}) - b = -\frac{7b-2b}{2}=- \frac{9b}{2}=b=0$.
Therefore it can be concluded that rank of the matrix is $2$. Then I consider the Rank-Nullity theorem:
$$\textrm{dim} \, \, \mathbb{R}^3 = \textrm{dim} \, \, \textrm{Im}(A) + \textrm{dim} \, \, \textrm{Ker}(A)$$
$$3 = 2 + \textrm{dim} \, \, \textrm{Ker}(A)$$
$$\textrm{dim} \, \, \textrm{Ker}(F)= 1$$
Thus the dimension of the space of solutions for the system of equations (a) is precisely $1$.
We know that dimension of the space of solutions is a $1$-dimensional space orthogonal to $A$. In order to find basis for solution space, we can extend $K=(4, 7, -\pi)$ and $B = (2, -1, 1)$ to basis of $\mathbb{R}^3$. Which can be done by selecting a vector $C$ such that $K, B, C$ is linearly independent, in this case, let $C=(0, 0, 1)$.
Final step is to ensure orthogonality between $K B ,C $. Using Gram-Schmidt:
$B' = (4, 7, -\pi) - \frac{\langle (2, -1, 1), \langle (4, 7, -\pi) \rangle}{\langle (2, -1, 1), \langle (2, -1, 1) \rangle} (2, -1, 1)=...$
$C' = B' - \frac{B', \langle (0, 0, 1) \rangle}{\langle (0, 0, 1), \langle (0, 0, 1) \rangle} (0, 0, 1)$=...
Therefore $S = \{C'\}$ is a basis of $\textrm{Ker}(A)$.
Is this a correct way to find basis without using Gaussian elimination method? I believe there might be few mistakes, especially during the orthogonalization step. Maybe $K, B$ should be treated as a whole matrix $A$? or maybe matrix $A$ must be flattened to a vector in $\mathbb{R}^{2 \times 3}$?
Thank you!