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$\ T: \mathbf R^5 \rightarrow \mathbf R^5 , \ T^2 =0$ and let $U$ be a subspace of $\ T $ and $\dim U = 3 $

prove: $\ U \cap \ker T \not= \{0\} $

If I understand correct then $w \in \operatorname{im}T, \ v \in \mathbf R^5, \ T(v) = w,\ T(w) = T(T(v)) = 0 $

so $\operatorname{im}T \subseteq \ker T$ and therefore $\dim(\operatorname{im}T) \leq \dim(\ker T) $ (not sure if it is true?)

and because $\dim(\mathbf R^5) = 5 $ so $\dim(\ker T) \geq 3 $

$\dim(\ker T) + \dim(U) = 6 > \dim(\mathbf R^5) $ it must be that $\ U \cap\ker T \not = \{0 \}$

I feel like i'm way off here..

Michael Hardy
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bm1125
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1 Answers1

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Your reasoning looks good, but it should be $\dim(\ker T) + \dim(U) \ge 6 $, as you cannot guarantee equality.

If you'd like to justify your conclusion more carefully, consider proving that $ \dim(U) + \dim(V) = \dim(U + V) + \dim(U \cap V) $, for U and V subspaces of $ \mathbf R^n $ (or more generally, subspaces of some vector space W. EDIT: I should have said for some finite dimensional vector space W). If you have this fact, can you figure out how to conclude the result you want?

paul blart math cop
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  • I guess you meant $\ dim(U) + dim(V) = dim(U+V) - dim(U \cap V)$ ? also not sure what do you mean finite dimensional, is $\ \mathbf R^5 $ finite dimensional space?? – bm1125 Jun 22 '18 at 13:56
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    The equation I wrote is correct. You can ignore what I wrote in parentheses, it's something you'll cover if you study abstract vector spaces. But to answer your question, R^5 is a finite dimensional vector space over R. – paul blart math cop Jun 22 '18 at 16:38