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let $V$ be the vector space consisting of polynomial with real coefficients in the variable $t$ of degree $≤ 9$.let $D:V→V$ be the linear operator by $D(f)=df/dt$.then $0$ is an eigenvalue of D.

is the above statement is true.i think it is true as rankD is 8 and it is less than 9. am i correct?

2 Answers 2

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Yes. Note that for $1\in V$,

$ D(1)=0=0\cdot 1 $

which implies that $0$ is an eigenvalue of $D$.

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Try writing the matrix of $D$ in the standard basis for $V$. It will be of a particular form, and matrices of this form have a special relationship with their eigenvalues.

I'm not sure what degree $\leq 9$ has to do with it, this holds for finite degree in general.