How can this be equal to the unit matrix?

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The discussion focuses on verifying the expression (C^{1/2})^{T}C^{-1}C^{1/2}=I, where C is a nonsingular covariance matrix and I is the unit matrix. The user seeks understanding rather than a rigorous proof, emphasizing the importance of comprehension over simply arriving at the solution. It is clarified that if C is decomposed into two invertible matrices A and B, then the inverse of C can be expressed as C^{-1} = B^{-1}A^{-1}. The user learns that the order of matrix multiplication does not need to change in this context, leading to a clearer understanding of the original expression. This exchange highlights the importance of grasping matrix properties in linear algebra.
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Homework Statement


At the lecture yesterday the teacher just ended up with a result I could not arrive at. So, how
can the below stated expression be verified?

\left(C^{1/2}\right)^{T}C^{-1}C^{1/2}=I
Here C is a nonsingular covariance matrix, obviously, and I is the unit matrix.

I will not make an attempt of a solution because then it feels like I would solve it but not understand. I hope that is ok. What I seek here is not an rigorous proof. I just want to understand.

Hope someone can help me!
 
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Suppose some invertible matrix n×n C is decomposed into the product of two n×n matrices A and B:

C = AB

Then A and B must themselves be invertible and the inverse of C is given by

C^{-1} = B^{-1}A^{-1}

The definition of the matrix square root of some matrix C is that

C=\left(C^{1/2}\right)^T\,C^{1/2}

Combine the above two and the result in the original post falls right out.
 
If I understand you correctly, then it is allowed to change the order in the matrix multiplication?
C^{-1}C^{1/2}=C^{1/2}C^{-1}
 
There was no need to change the order. Now I see. :) Thanks!
 
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