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Diagonalizable Matrices, Image, Kernels and Direct Sums

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Prove that if T Є L(V) is diagonalizable then V = im(T) + ker(T) (+ = direct sum)
(Hint: Use a basis of eigenvectors. The eigenvectors of the eigenvalue zero are a basis for the null space, and the remaining eigenvectors are a basis for the image)

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Proof:
suppose dim(V)=n. Since T is in L(V), T is diagonalizable, then we
can find a basis B={v1,v2,...,vn} of V, such that the representative
matrix of T under this basis B is a diagonal matrix D=diag(t1,t2,...,tn).
Then we have Tv1=t1v1, ...

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