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

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SUMMARY

The discussion centers on proving the determinant identity det(I - xy'T) = 1 - y'T x, where I is the identity matrix and y'T represents the transpose of vector y. The user expresses confusion about calculating determinants involving subtraction, specifically det(A - B). A participant clarifies that calculating det(A - B) is equivalent to calculating det(C) for the resulting matrix C, thus simplifying the user's understanding of the determinant operation. The conversation emphasizes the importance of clear notation in mathematical expressions.

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  • Understanding of matrix operations, specifically determinants
  • Familiarity with matrix transpose notation
  • Knowledge of identity matrices
  • Basic linear algebra concepts
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  • Learn about matrix subtraction and its implications on determinants
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Students and professionals in mathematics, particularly those studying linear algebra, as well as anyone needing clarification on determinant calculations involving matrix operations.

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 \det(M), right? And for two nxn square matrices A and B, A-B is an nxn matrix C, true again? Then calculating \det(A-B) is no different than calculating \det(C).
 
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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