How to say that a projection mapping is digonalizable.further it can be represented in the diagonal matrix having a block of identity matrix of order r inside it.where r is the rank of the operator.
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1Welcome to MSE. When asking a question, it is helpful to provide what you have tried. Please also provide the definition of a projection mapping that you’re using. Thank you. – Benjamin Wang Jun 09 '20 at 07:47
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You may refer to https://math.stackexchange.com/questions/73862/diagonalization-of-a-projection – dust05 Jun 09 '20 at 09:06
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In my book a linear projection mapping is by definition one that is diagonalisable with no other eigenvalues than $0$ and $1$. It might be said in a slightly different way, for instance whenever $V$ and $W$ are complementary subspaces, there is a projection operator $p$ onto $V$ and parallel to $W$ such that whenever $u=v+w$ is a vector, $p(u)=v$. But that is just saying that $V$ is an eigenspace for the eigenvalue $1$ (since $p(v)=v$) and $W$ is an eigenspace for the eigenvalue $0$ (since $p(w)=0$), and since their direct sum fills the whole space, that is all there is. On any basis formed by combining a basis of $V$ and a basis of $W$, the matrix of $p$ will be a diagonal one, with diagonal entries $1$ for the initial "$V$-part", and remaining diagonal entries $0$.
Since the eigenspace for $0$ does not contribute to the rank, the rank of the operator is equal to the dimension of the eigenspace for $1$. The fact that a linear operator $\phi$ is a projection operator if and only if $\phi^2=\phi$ is something (easy) to prove, but I would think not the proper definition of a projection mapping.
Marc van Leeuwen
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