a system of linear equations $x-y+z=0$

Question

Yall are probably gonna think me a noob. But I am working on this eigenvector problem and I reduced the matrix to $x-y+z=0$ . How do I describe this solution set. I know how to do it if it's just $x-y$ , but not with $3$ variables.

Answer 1

The mentioned constraint describes a $n-1$ deimensional subspace $V$ of $K^n$ if the equation is to be understood to live in $K^n$ for some field $K$.

The subspace is characterized by $$V=\{v\in K^3 \ |\ v\cdot(1,-1,1) = 0 \}\tag1$$ if we are in a 3-dimensinal space, and if your $x$, $y$, $z$ are labeling the 1st, 2nd and 3rd coordinate, respectively. This means that $V$ is the orthogonal complement of that vector: $$ V = \operatorname{span}((1,-1,1))^\bot $$ As $V$ is again a vector space, so it has a basis of 2 elements because $V$ is 2-dimensional then. The basis elements must satisfy the condition in $(1)$, say $b_1=(1,1,0)$ the second basis vector can then befuond vy the cross product $b_2 = b_1\times (1,-1,1)$ because this ensures $b_2$ is independent of $\{b_1, (1,-1,1)\}$. $V$ can then be reresented as $$V=\operatorname{span} (b_1,b_2) = \{v\in K^n\ |\ v=\alpha b_1 + \beta b_2\text{ where }\alpha,\beta\in K\}$$

Higher dimensional cases are similar, except that you will find $n-1$ basis vectors for $V$.