Skip to main content

Intersection of hyperplanes in $\mathbb{R}^5$

Consider the $4$ hyperplanes in $\mathbb{R}^5$ given by the equations $$x_1-x_2+x_3+x_4+2x_5=0$$ $$2x_1-x_2+6x_3+2x_4+6x_5=0$$ $$3x_1-3x_2+3x_3+4x_4+7x_5=0$$ $$x_1+x_2+9x_3+x_4+6x_5=0$$ Let $V \leq \mathbb{R}^5$ be the intersection of these hyperplanes. Find a basis for $V$.

Comments

Popular posts from this blog

[Community Question] Calculus: Manifold with boundary - finding the boundary

One of our user asked: I have the manifold with boundary $M:= \lbrace (x_1,x_2,x_3) \in \mathbb R^3 : x_1\geq 0, x_1^2+x_2^2+x_3^2=1\rbrace \cup\lbrace (x_1,x_2,x_3) \in \mathbb R^3 : x_1= 0, x_1^2+x_2^2+x_3^2\leq1\rbrace$ and I need to find the boundary of this manifold. I think it is $\lbrace (x_1,x_2,x_3) \in \mathbb R^n : x_1= 0, x_2^2+x_3^2=1\rbrace$ , the other option is that the boundary is the empty set? I think the first is right? Am I wrong?

[Community Question] Linear-algebra: Are linear transformations between infinite dimensional vector spaces always differentiable?

One of our user asked: In class we saw that every linear transformation is differentiable (since there's always a linear approximation for them) and we also saw that a differentiable function must be continuous, so it must be true that all linear operators are continuous, however, I just read that between infinite dimensional vector spaces this is not necessarily true. I would like to know where's the flaw in my reasoning (I suspect that linear transformations between infinite dimensional vector spaces are not always differentiable).