03-10-03

# 03-10-03 - Lecture 16 Andrei Antonenko 1 Image and kernel...

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Andrei Antonenko March 10, 2003 1 Image and kernel Last lecture we studied image and kernel of a linear function. Now we will prove one of the properties of image and kernel. First let’s consider kernel. Let f : V U be a linear function, and let its kernel be Ker f — set of all elements v from V which map to 0 . Then we can state the following properties of it. Existence of zero. The zero vector 0 belongs to kernel of f , since f ( 0 ) = 0 — maps to 0 , so 0 is in kernel. Summation. Let vectors v and u belong to kernel, so, f ( v ) = 0 and f ( u ) = 0 . Then f ( v + u ) = f ( v ) + f ( u ) = 0 , and thus u + v belongs to Ker f . Multiplication by a scalar. Let vector v belongs to the kernel of f . Then we know that f ( v ) = 0 . Now for any constant k we have: f ( kv ) = kf ( v ) = k · 0 = 0 , thus kv belongs to Ker f . So, we proved the following theorem: Theorem 1.1. The kernel of linear function f : V U is a vector subspace in V . Example 1.2. Consider the projection function f ( x,y,z ) = ( x,y, 0) . It’s kernel consists of vectors of the form (0 , 0 ,c ) for any constant c . Geometrically speaking, this is a z -axis in the 3-dimensional space. This is a vector subspace. Now let’s consider the image. Let f : V U be a linear function, and it’s image Im f is the set of all vectors from U where we can get by applying a function to vectors from V . We’ll state some properties of it. 1

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03-10-03 - Lecture 16 Andrei Antonenko 1 Image and kernel...

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