03-14-03

# 03-14-03 - Lecture 18 Andrei Antonenko 1 Theoretical facts...

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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 ) = 0 . This is the same as f ( a r +1 e r +1 + a r +2 e r +2 + · · · + a n e n ) = 0 . 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 = 0 .

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