Show that det(I-xy'T)= 1-y'T x

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Homework Help Overview

The discussion revolves around a problem in linear algebra involving determinants, specifically the expression det(I - xy'T) and its relation to 1 - y'T x, where I is the identity matrix and y'T is the transpose of vector y. The original poster expresses uncertainty about how to approach the determinant of a matrix involving subtraction.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the general properties of determinants and the implications of subtracting matrices. There is a focus on understanding the calculation of determinants in the context of the given expression.

Discussion Status

Some participants provide clarifications regarding the calculation of determinants and the nature of the matrices involved. There is an indication that the original poster feels more confident about proceeding with the problem after receiving some insights.

Contextual Notes

There is a noted confusion regarding the notation used for x and y, with some participants questioning the assumption that these are matrices rather than vectors. This highlights a potential misunderstanding in the problem setup.

kutaybulbul
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Need Help Please!

I am very much in need of your help. I have a question saying:

Let x,y E R. Show that det(I-xy'T)= 1-y'T x

y'T is transpose of y and I is identity matrix.

Actually I don't know how to solve something like det( A-B). What am I going to do when there is addition or subtraction in the determinant.

Thanks for your response.
 
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Well for some square matrix M, you know how to calculate [itex]\det(M)[/itex], right? And for two nxn square matrices A and B, A-B is an nxn matrix C, true again? Then calculating [itex]\det(A-B)[/itex] is no different than calculating [itex]\det(C)[/itex].
 
So is this so simple wov thank you very much I think I can handle the rest.
 
In future it might be a good idea not to say things like
"Let x,y E R" and then assert that they are matrices!
 

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