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Unformatted text preview: Lecture 18 Andrei Antonenko March 14, 2003 1 Theoretical facts about image Now we’ll develop some theory about the image and its dimension and basis. Let f be a linear function from V to U , and dim V = n . Let’s consider the kernel of f . We can find the basis of the kernel. Let it consist of r vectors, e 1 ,e 2 ,...,e r . Now we can use the procedure of extending to the basis to find the basis of the whole V which contains the vectors e 1 ,e 2 ,...,e r . Let we added vectors from e r +1 to e n , so that the basis of V is basis of V z } { e 1 , e 2 , ..., e r  {z } basis of Ker f , e r +1 , ..., e n We know that the image of f is spanned by f ( e 1 ) ,f ( e 2 ) ,...,f ( e n ). But f ( e 1 ) , f ( e 2 ) , ..., f ( e r ) are all equal to 0, because they belong to the kernel. So, the image is spanned by f ( e r +1 ), f ( e r +2 ), ... , f ( e n ). Now let’s prove that they are linearly independent, and then it will be proved that they form a basis for the image. To check their linear independence, let’s write the linear combination which is equal to 0: a r +1 f ( e r +1 ) + a r +2 f ( e r +2 ) + ··· + a n f ( e n ) = . This is the same as f ( a r +1 e r +1 + a r +2 e r +2 + ··· + a n e n ) = . So, it follows that a r +1 e r +1 + a r +2 e r +2 + ··· + a n e n belongs to the kernel of f . But the kernel is spanned by e 1 ,e 2 ,...,e n , so, a r +1 e r +1 + a r +2 e r +2 + ··· + a n e n = b 1 e 1 + ··· + b r e r or a r +1 e r +1 + a r +2 e r +2 + ··· + a n e n b 1 e 1 ···  b r e r = ....
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This note was uploaded on 05/28/2011 for the course AMS 2010 taught by Professor Andant during the Spring '03 term at SUNY Stony Brook.
 Spring '03
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