arthurhenry
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Why is the Trace of a projection is its Rank.
Thank you
Thank you
Why Rank is the Trace of a Projection
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Hi arthurhenry! 
A projection operator P satisfies P2=P. So it's only eigenvalues are 0 and 1. It is easy to see that the rank of P is the number of eigenvalues that are 1. Thus the sum of the eigenvalues is in this case the rank...
Now, take the Jordan normal form of P, then the diagonal contains all the eigenvalues. In particular, the trace is the sum of all the eigenvalues. And thus equals the rank of P.
A projection operator P satisfies P2=P. So it's only eigenvalues are 0 and 1. It is easy to see that the rank of P is the number of eigenvalues that are 1. Thus the sum of the eigenvalues is in this case the rank...
Now, take the Jordan normal form of P, then the diagonal contains all the eigenvalues. In particular, the trace is the sum of all the eigenvalues. And thus equals the rank of P.
arthurhenry
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Dear Micromass,
I thank you for your help --on two occasions now, as you answered another post of mine. Some of these questions come as I verify a comment or at times directly a trying to do an exercise. I am reading a book "Algebras of Linear Transformations" by Douglas Farenick, to teach myself some of that material. I do realize some of my questions are rather rudimentary, I apologize.
I thank you for your help --on two occasions now, as you answered another post of mine. Some of these questions come as I verify a comment or at times directly a trying to do an exercise. I am reading a book "Algebras of Linear Transformations" by Douglas Farenick, to teach myself some of that material. I do realize some of my questions are rather rudimentary, I apologize.
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Don't apologize! It's only by asking such a questions that you'll learn the material. Everybody has to go through it 
It's only by asking such a questions that you'll learn the material. Everybody has to go through it