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Show that {T $\in$ $L(R^5, R^4)$: dim null T > 2} is not a subspace of $L(R^5, R^4)$

I have no idea how to show this isn't a subspace the farthest I have gotten is to show that

dim range T < 3

But how would I go from there?

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Consider $(x_1, x_2, x_3, x_4, x_5)\mapsto (x_1, x_2, 0,0)$ and $(x_1, x_2, x_3, x_4, x_5)\mapsto (0, 0, x_3, x_4)$. – Git Gud Jul 23 '14 at 1:35
up vote 1 down vote accepted

This question does not lend itself to an application of the rank nullity theorem.

Hint: what does it mean for a set to be a subspace of $L(R^5,R^4)$? How can a subset fail to be a subspace?

Note that $$ \pmatrix{ 1&0&0&0&0\\ 0&1&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0 } + \pmatrix{ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0 } = \pmatrix{ 1&0&0&0&0\\ 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0 } $$

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So it must be closed under addition and scalar multiplication and be non empty, I don't understand what this is example is saying? Sorry would you mind explaining it to me? – Soaps Jul 23 '14 at 1:43
@Soaps the example shows it's not closed under addition – Omnomnomnom Jul 23 '14 at 2:48

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