Skip to main content

[Community Question] Linear-algebra: Number of possible zero entries in orthogonal matrices

One of our user asked:

It's easy to check that in orthogonal matrix dimension $2 \times 2$ if there is entry $0$ in the matrix necessary one additional zero must be present. Then the total number of zeros is $2$.

In an orthogonal matrix dim. $3 \times 3$ number of zeros can be (if they are present) , I suppose from observations, only $4$ or $6$ - once again we obtain an even number of possible zeros.

  • Can this observation be extended for other orthogonal matrices of greater dimensions? The number of zeros is always even? How to prove this?

  • Maybe, it is known the explicit formula for the number of possible zeros in orthogonal matrices of any dimension?


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).

Order of elements of the Prüfer groups $\mathbb{Z}(p^{\infty})$

Let $\mathbb{Z}(p^{\infty})$ be defined by $\mathbb{Z}(p^{\infty}) = \{ \overline{a/b} \in \mathbb{Q}/ \mathbb{Z} / a,b \in \mathbb{Z}, b=p^i$ $ with$ $ i \in \mathbb{N} \}$ , I wish show that any element in $\mathbb{Z}(p^{\infty})$ has order $p^n$ with $n \in \mathbb{N}$ . i try several ways but I have not been successful, some help ?? thank you