Let $T$ be a linear operator on a finite dimensional vector space $V$ s.t. $T^2=0$ Show that $$\dim V=\dim\left(\ker T+\ker T^{(t)}\right)$$ $t$ is transpose. Field is Real number

Is is true?

  • $\begingroup$ Do you mean $\dim V = \dim\ker T+\dim\ker T^t$? Are yo aware that $0^t=0$? $\endgroup$ – Hagen von Eitzen May 17 '14 at 16:06
  • $\begingroup$ No. I mean dimension of sum $\endgroup$ – user122794 May 17 '14 at 16:07
  • $\begingroup$ @HagenvonEitzen - Note that if the equation was as you wrote in the comment then $T=0$ would be a counterexample $\endgroup$ – Belgi May 17 '14 at 16:09
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    $\begingroup$ @PraphullaKoushik - this example seem to check out.. $\endgroup$ – Belgi May 17 '14 at 16:13
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    $\begingroup$ But $\ker T^t\subseteq V^*\ne V$, isn't it? Also, even if we pick a basis and identify $V$ with $V^*$ (in other words, consider $T$ and $T^t$ as matrices), what about the case $T=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}$? $\endgroup$ – Hagen von Eitzen May 17 '14 at 16:13

Take $V=\mathbb{R}^n$ with $(\cdot,\cdot)$ the usual inner product. Then a coordinate-free description of the transpose is $(Tv,w)=(v,T^t w)$.

For a subspace W of V, we let $W^\perp=\{v\in V\mid (v,w)=0\ \forall\ w\in W\}$. Then $V=W+W^\perp$ (actually a direct sum, but we don't need that). This is the point in the proof where we use the fact that the ground field is $\mathbb{R}$.

So it suffices to show that $(\ker T)^\perp\subset \ker(T^t)$.

Suppose $w\in (\ker T)^\perp$ and $v\in V$. Then $(v,T^t w)=(Tv,w)$. Since $T^2=0$, $Tv\in \ker T$ and therefore $(Tv,w)=0$. Thus we've shown that $(v,T^t w)=0$ for all $v$ and hence $T^tw=0$, implying $w\in \ker(T^t)$, as required.


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