Let v1,v2,v3....vn ∈V are the eigenvectors of T with respect to the eigenvalues λ1,λ2....λn
Which are distinct and non-zero. The vectors T(v1),T(v2)....T(vn) are linearly independent.
In fact, if c1,c2,c3.....cn∈K
Then c1T(v1)+c2T(v2)+c3T(v3).....cnT(vn)=0v
Hence, c1λ1v1+c2λ2v2+....cnλnvn=0 v
So, we can write it as c1λ1=c2λ2=c3λ3.....=cnλn=0
From the above, we can conclude that v1,v2,v3...vn are linearly independent, so c1=c2=c3.....cn=0
But, λi is non zero, So n=dimSpan(T(v1),...,T(vn))≤dimImageT=k
So, we can conclude that T has not k distinct non-zero eigenvalues, Hence only possible eigenvalues for T is zero, So we can say that T has k+1 distinct eigenvalues.
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