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If you have a linear transformation T:V→W and the dimension of V is greater than the dimension of W, then the kernel of T has to have a dimension at least as large as the difference.

For example, a transformation $T:\mathbb{R}^7\rightarrow \mathbb{R}^3$ has to have a kernel of dimension 4 or greater.

May I use this?

$\dim⁡ \ker(T)+ \dim⁡ R(T) = \dim ⁡V$

$\dim \ker (T) = \dim V - \dim R(T)$

But $R(T)$ belongs to $W$, so the dimension of $R(T)$ has to be equal or less than dimension of $W$ (How can I prove this?). So,

$\dim \ker (T) \geq \dim V - \dim W$

And $\dim V > \dim W$, so $\dim V - \dim W$ is positive.

Is this right?

May I use this property to say when V is infinite dimensional and W has a finite dimension, then the kernel has to be infinite dimensional?

Thank you very much for your help.

Yanko
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Harry
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  • This is correct indeed. The third line uses the fact that vector spaces can't have two bases of different size. If $\dim W = m$ then there are up to $m$ independent vectors in $W$ and therefore also in $Im(T)$. – Yanko Jan 11 '19 at 20:10
  • See this question it probably answers your last question : https://math.stackexchange.com/questions/752056/does-the-rank-nullity-theorem-hold-for-infinite-dimensional-v – Yanko Jan 11 '19 at 20:14
  • Thank you, Yanko! :) – Harry Jan 11 '19 at 20:15
  • You are saving my life! – Harry Jan 11 '19 at 20:15

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