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10:52 PM
I'm busy enough without extra chatrooms ... What's up?
 
Excuse me Ted, will you have some time to review a question about determinant?
 
Why don't you just ask in the main room?
 
Oh sorry, i think could be in private.
Really i want your opinion. you're too good.
 
What's the question, but I really don't see the need for privacy ...
 
If you want, i can ask in global chat.
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Q: Prove about determinant of a block matrix.

Bvss12Let $X \in Mat_p(\mathbb{K})$, $C \in Mat_q(\mathbb{K})$, $D\in Mat_{p,q}(\mathbb{K})$ where $p+q=n$. Consider the function $\alpha: \mathbb{K}^p\times...\times\mathbb{K}^p\rightarrow\mathbb{K}$ Defined by $\alpha(X_1,...,X_p)=det\begin{bmatrix} X_{p\times p} & D_{p\times q} \\ 0_{q\...

 
10:58 PM
Whether $X$ is invertible or not, because of the 0s on the lower left, the determinant is always $(\det X)(\det C)$.
You can see this by doing row operations to make each of $X$ and $C$ upper triangular.
 
Uhhh... Okay
 
This is an exercise in most linear algebra books :P
 
Yeah, i was suppose X is invertible
 
You don't need that. But because invertible matrices are dense in the space of all matrices, you can use a continuity argument to deduce it holds for all $X$ if you prove it for invertible $X$.
 
uhmm... okay, thanks for that
i go to think the exercise without that.
Thanks for all Ted!
 
11:04 PM
OK, you're welcome.
 

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