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I have to prove that if $T$ is an idempotent ($T^2=T$) linear operator then space $V = \ker T\oplus\operatorname{im}T$.

My first try was to think about the basis of subspace $\ker T$.

Let say $(e_1,...,e_k)$ is the basis of $\ker T$ and $\dim V = n < \infty$.

Then, of course we can add $n-k$ vectors to $\ker T$ basis and get the basis of whole space. So every vector $v\in V$ can expressed as linear combination $v = a_1e_1 + \cdots +a_ke_k + a_{k+1}e_{k+1} + \cdots + a_ne_n$.

But then I stuck, because I don't know where to use this idempotency. Maybe there is another solution which do not consider this basis thing.

Thanks!

user26857
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janusz
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2 Answers2

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Here's a tricky but quick answer: note that every $v$ can be written in the form $$ v = (v - Tv) + Tv $$ and that $v - Tv$ is in the kernel, while $Tv$ is in the image.

To show that the subspaces have a trivial intersection, note that $T(Tv)$ can only be zero if $(Tv)$ was originally zero. So, no element of the image is also in the kernel. Then, note that $\ker T$ is the image of $I-T$.

Ben Grossmann
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Since $T^2=T$ so $T^2(v)=T(T(v))=T(v)\Rightarrow T^2(v)-T(v)=0\forall v$ now think a bit what happen if we apply $T$ on $v-T(v)$

so take any vector $v$ you can decompose it as $v-T(v)$ and $T(v)$ where $v-T(v)\in Ker(T)$ and $T(v)\in Im(T)$ this is true for all $v$

$v=(v-T(v))+T(v)$

so $V=ker T\oplus im(T)$

Myshkin
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