I tried hard to find explicit Linear transformation but I couldn't find it. I am doing it from $2$ dimensions. Also by rank nullity theorem I see that dimension of $V$ must be even ($T: V \to V$)
Kindly suggest
Thanks
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5See When does the kernel of a function equal the image? – Weaam Aug 13 '16 at 16:37
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@Crostul it does not satisfy $T^2=0$ – Gathdi Aug 13 '16 at 16:41
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1@crostul I think you mean $T(x,y) = (y,0)$ – Adam Hughes Aug 13 '16 at 16:42
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If the range and null space of $T$ are equal (or even if the range is a subset of the null space),then it does satisfy $T^2-=0$. – Andreas Blass Aug 13 '16 at 17:21
2 Answers
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One example of a linear operator on an infinite-dimensional space, consider the map $T:\ell^\infty\to\ell^\infty$ ($\ell^\infty$ is the space of bounded real or complex sequences) given by $$ T(x_1,x_2,x_3,x_4,\ldots)=(0,x_1,0,x_3,0,\ldots). $$ Then we have $$N(T)=\{(x_i)_{i\in\mathbb{N}}\in\ell^\infty:x_k=0 \textrm{ for $k$ even\}}=R(T). $$
To get a $2$-dimensional example, consider the restriction of this example to the first two coordinates.
Aweygan
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Let $V$ be any vector space. Consider a projection $\pi$ followed by an inclusion $\iota$, say $V\oplus V\to V\hookrightarrow V\oplus V$ that kills $V$.
Then ${\rm im}\,{\iota \pi }={\rm im}\,\iota =V\oplus 0$ and $\ker \iota\pi =\ker \pi = V\oplus 0$.
Pedro
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