Mathematics Asked by V. Elizabeth on November 27, 2020

Let $V$ be a real vector space. If $T: V rightarrow V$ is an idempotent linear map, and if $lambda$ is an eigenvalue of $T$, then it may be proven of $lambda$, that it is either $0$ or $1$. If $V$ were a complex vector space, one need only evoke the Fundamental Theorem of Algebra to conclude of the existence of this $lambda$. Yet, if $V$ is real, how may we be certain such a $lambda$ exists?

There is no need for space decomposition. Let $vneq0$ be an eigenvector of $E$ for eignevalue $lambda$, namely $$ Ev=lambda v. $$ So $$ E^2v=lambda Ev=lambda^2v. $$ Since $E^2=E$, one has $$ lambda v=lambda^2v $$ which gives $lambda=lambda^2$. So $lambda=0$ or $1$, namely $E$ only has eigenvalues $0$ and $1$.

Answered by xpaul on November 27, 2020

Suppose $V$ is a vector space over any field $mathbb{F}$. Suppose that $E:Vto V$ is an idempotent linear map, that is $E^2=E$. Then $V=ker E oplus ker (I-E)$, the direct sum of what is clearly the $0$-eigenspace with the $1$-eigenspace.

[The decomposition is trivial to prove: for $vin V$ we have $v=(I-E)v +Ev$, and $E(I-E)v=(I-E)Ev=0$; and if $vinker E cap ker (I-E)$ then $Ev=0, (I-E)v=0$ so add and get $v=Iv=0$.]

Answered by ancientmathematician on November 27, 2020

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