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seen Oct 19 at 10:57

Oct
19
comment What are the possible dimensions of all $n \times n$ matrices such that $A\vec{x} = \vec{0}$?
That matrix would not do a basis change. But then, you don't need to explicitly find a matrix that does it. You just have to answer the question: "Does there exist a basis for $\mathbb R^n$ that has $\vec x$ as basis vector? About the dimension: Is the dimension a property of the space, or just a property of the individual basis?
Oct
19
comment What are the possible dimensions of all $n \times n$ matrices such that $A\vec{x} = \vec{0}$?
No. $(1,1,0,\ldots,0)^T$ will also be sent to $\vec 0$ e.g. if the first column is all $1$ and the second column is all $-1$. To elaborate on my previous hint: Is it possible to change the basis to one having $\vec x$ as basis vector? And what happens to dimensions when you change the basis?
Oct
19
comment What are the possible dimensions of all $n \times n$ matrices such that $A\vec{x} = \vec{0}$?
The $n^2-n$ is right. The $n$ is not. Hint: What happens for a basis change?
Oct
19
comment What are the possible dimensions of all $n \times n$ matrices such that $A\vec{x} = \vec{0}$?
@JennParker: You're right about the form of the matrix, but not the dimension. Hint: How many matrix elements can you still choose freely?
Oct
19
comment What are the possible dimensions of all $n \times n$ matrices such that $A\vec{x} = \vec{0}$?
@JennParker: The dimension for $\vec x=0$ you've got right, but in the other cases you're wrong. Hint: What matrices will map the vector $(1,0,\dots,0)^T$ to $\vec 0$?
Oct
19
comment Showing that $e^{-2} < \ln 2$
It's not only allowed, if the answer is good you may even get a badge for it.
Oct
19
answered What are the possible dimensions of all $n \times n$ matrices such that $A\vec{x} = \vec{0}$?
Oct
18
comment Are there countably many infinities?
Ah, now I got it. Thank you!
Oct
18
comment Are there countably many infinities?
Well, I didn't yet agree that $\aleph_\alpha$ exists for all $\alpha<\epsilon_0$ exist. For that I first need an argument that all countable ordinals are reachable through the addition/limit process.
Oct
18
comment Are there countably many infinities?
Let's start with $\aleph_0$. Now with the $\alpha+1$ rule, I can generate $\aleph_1$, $\aleph_2$ and so on. Basically, everything $\aleph_n$, $n\in\mathbb N$. With the limit rule I then get $\aleph_\omega$. Then I get $\aleph_{\omega+1}$ etc., then again with the limit rule $\aleph_{2\omega}$, and indeed $\aleph_{m\omega+n}$. I even see how you can extend that to arbitrary polynomials of $\omega$. But those are still countably many ordinals, and thus countably many cardinals associated to them. So, how do you get beyond countability?
Oct
18
comment Are there countably many infinities?
But I don't see how you can get more than countably many cardinalities by that procedure.
Oct
18
comment Are there countably many infinities?
Is there an easy way to see that there is an $\aleph_\alpha$ for each ordinal $\alpha$?
Oct
18
comment Does $1.0000000000\cdots 1$ with an infinite number of $0$ in it exist?
@DavidRicherby: Maybe the cardinality of proper classes?
Oct
18
comment Does $1.0000000000\cdots 1$ with an infinite number of $0$ in it exist?
@AsafKaragila: The banner at the top of the site contains, besides a set of images, "StackExchange" and some numbers, as well as the words "review", "help" and "search", but not the word "Mathematics". ;-)
Oct
18
comment Does $1.0000000000\cdots 1$ with an infinite number of $0$ in it exist?
Actually, you don't need $10^{-\omega}$ to be well defined, since that doesn't occur in the answer. What you need is a well defined limit.
Oct
10
answered Addition of Vectors?
Oct
9
answered Is it possible to not have irrational numbers?
Oct
9
comment How is greater than defined for real numbers?
@Lano: Good point, see my edit.
Oct
9
revised How is greater than defined for real numbers?
Added how to define the topology without comparison operators
Oct
8
comment The “Empty Tuple” or “0-Tuple”: Its Definition and Properties
@SteveKass, Mathemanic: Thank you for explaining why my idea doesn't work.