I should have also mentioned that similar matrices have the same rank, trace and determinant. These can be sometimes used to quickly refute that two matrices are not similar.
In this case, $A^2$ and $B^2$ have different rank.
Hey, I'm not sure if it's okay to request answers to questions here, but if someone has a great geometric understanding of Linear Algebra, I would REALLY appreciate an answer to this question: math.stackexchange.com/questions/3266970/…
If it's not okay to request answers (which makes sense, since this could get flooded), I won't post requests again, just a moderator let me know. I just really want to understand this, because I feel like it should be so intuitive, but its not (at least not to me)!!!!
I've been trying (without success) to graphically understand why, if $D$ is a diagonal matrix, and I've got two random vectors $\vec{a}$ and $\vec{b}$, then:
$$\vec{a}^TD\vec{b}=\vec{b}^TD\vec{A}$$
I know that I could algebraically prove the above by just taking the transpose of the entire thin...
@JoshuaRonis Is it just a typo the the first two equations $$\vec{a}^TD\vec{b}=\vec{b}^TD\vec{A}$$ and $$\vec{a}^TD\vec{b}=\vec{b}^TD^T\vec{A}=\vec{b}^TD\vec{A}$$ have $\vec A$ rather than $\vec a$?
Or is there some meaning in the distinction between lowercase and uppercase in the notation for vectors.
It's just I don't go on the chat like, ever, so I didn't know if when someone answers something on the chat website, I get a notification on the normal website or not.
If you weren't here up to 15 minutes after the message, you would get the notification in your inbox - as I can say from my experience and from the answers here: Receiving Stack Overflow's Chat Notification.
To clarify, if you log out from all chatrooms now and if somebody pings you (using @username) here or in Mathematics (or in some other room where you posted recently), then it will turn to notification in your inbox after 15 minutes. Unless you see it in the chat before that.
Understood! I like that, because I don't go on the chat often. Is it generally accepted to link questions here from the normal site? And are the chats usually active?
Well, you can see in the room info that this chatroom is not very active - it shows average 35 messages per week.
The main chatroom is much more active. (Of course, that also means that if you post something, other conversation will make your message visible for shorter time.)
@JoshuaRonis Regarding linking questions, you can have a look at Main Chatroom Guidelines. They are for the main chatroom. In the rooms with much lower activity, this should be not a problem. (Rooms like this are not in danger of getting flooded.)
Yea, I posted my question there too. Alright, thanks for all the info! I gotta run, but if you could repost my question under here in the pretty format you just did (idk how you did that) I'd really appreciate it - that way people can see it and hopefully someone answers!
> Be mindful that if you post a link to Wikipedia, YouTube, StackExchange, or an image, it will automatically balloon up to a large preview; consider linking to it instead using the markup [text](url). Images may be uploaded to imgur.
I've been trying (without success) to graphically understand why, if $D$ is a diagonal matrix, and I've got two random vectors $\vec{a}$ and $\vec{b}$, then:
$$\vec{a}^TD\vec{b}=\vec{b}^TD\vec{a}$$
I know that I could algebraically prove the above by just taking the transpose of the entire thin...
When the question is that some matrix does not change inner product and we want to express this in terms of the map $\vec x\mapsto \vec xA$, I'd guess that an easier problem would be to ask about the condition $\langle \vec x A, \vec y A \rangle = \langle \vec x, \vec y \rangle$ rather than $\langle \vec x A, \vec y \rangle = \langle \vec x, \vec y A\rangle$.
At least easier in the sense that it would be easier to get some geometric intuition for that condition. The difference is that here we apply the same map to both vectors.
But maybe I just do not know the right intuition for the problem you're asking about.
I've been trying (without success) to graphically understand why, if $D$ is a diagonal matrix, and I've got two random vectors $\vec{a}$ and $\vec{b}$, then:
$$\vec{a}^TD\vec{b}=\vec{b}^TD\vec{A}$$
I know that I could algebraically prove the above by just taking the transpose of the entire thin...
Linear & Abstract algebra
Linear & Abstract algebra