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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
- Context: Undergrad
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Discussion Overview
The discussion centers around the relationship between the trace of a projection operator and its rank, exploring the mathematical properties of projection operators in linear algebra.
Discussion Character
- Technical explanation, Conceptual clarification
Main Points Raised
- One participant questions why the trace of a projection is equal to its rank.
- Another participant explains that a projection operator P satisfies P² = P, leading to eigenvalues of only 0 and 1, and that the rank corresponds to the number of eigenvalues equal to 1.
- This participant further elaborates that the trace, being the sum of all eigenvalues, thus equals the rank of P.
- A third participant expresses gratitude for the assistance received and mentions their ongoing study of linear transformations, indicating a desire to verify concepts through questions.
- A subsequent reply encourages the questioning participant, emphasizing that asking questions is a vital part of the learning process.
Areas of Agreement / Disagreement
The discussion does not appear to have any explicit disagreements, but it reflects varying levels of understanding and engagement with the topic. There is no consensus on the foundational concepts as the initial question remains open for further exploration.
Contextual Notes
The discussion may be limited by the participants' varying levels of familiarity with linear algebra concepts, and the assumptions underlying the properties of projection operators are not fully explored.
Who May Find This Useful
Readers interested in linear algebra, particularly those studying projection operators and their properties, may find this discussion beneficial.
- #2
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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.
- #3
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.
- #4
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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 Similar threads
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