Let V be an n-dimensional (real) vector space and let T be a linear trans- formation from V into V  

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May 14th, 2015

We can prove this by using a Theorem known as the Kernel Rank Theorem which states that the dimension of V is equal to the sum of the dimension of the range of T and the dimension of the null space of T, where T is a linear transformation.

Since the range null space of T are identical, they each have a dimension of say k, where k is a natural number. This means the sum of their dimensions is 2k which the Kernel Rank Theorem says is equal to n. 

Therefore n = 2k and so is an even integer.

May 14th, 2015
i got it now.   thanks a lot
May 14th, 2015

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