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Lecture 16
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 letās consider kernel. Let f : V ā U be a linear function, and let its kernel be Ker f ā 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 ā 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 Ā· 0 = 0 ,
thus kv belongs to Ker f.
So, we proved the following theorem:
Theorem 1.1. The kernel of linear function f : V ā U is a vector subspace in V.
Example 1.2. Consider the projection function f (x, y, z) = (x, y, 0). Itās 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 letās consider the image. Let f : V ā U be a linear function, and itās image Im f is the set of all vectors from U where we can get by applying a function to vectors from V. Weāll state some properties of it.
Existence of zero. The zero vector is in Im f since by taking f ( 0 ) we can get to 0 : f ( 0 ) = 0.
Addition. Let u 1 and u 2 be elements from the image of f , so there exist v 1 and v 2 from V such that f (v 1 ) = u 1 and f (v 2 ) = u 2. Now we can consider the element v 1 + v 2 from V. We have: f (v 1 + v 2 ) = f (v 1 ) + f (v 2 ) = u 1 + u 2 , and thus u 1 + u 2 belongs to Im f.
Multiplication by a scalar. Let u be a vector from Im f. Then there exists a vector v from V such that f (v) = u. So, letās consider an element kv for any constant k. We have:
f (kv) = kf (v) = ku,
thus ku belongs to Im f.
As for the kernel, we proved the following theorem:
Theorem 1.3. The image of a linear function f : V ā U is a vector subspace in U.
Example 1.4. Consider the projection function f (x, y, z) = (x, y, 0). Itās image consists of vectors of the form (x, y, 0) for all x, y ā R. Geometrically speaking, this is an xy-plane in the 3-dimensional space. This is a vector subspace.
In order to continue studies of image and kernel, we would like to know more about linear functions.
2 Matrix of a linear function
When we studied linear function for the first time we considered the following example. If A is an m Ć n matrix, then we can define a linear function FA : Rn^ ā Rm^ by the following formula: FA(x) = Ax for any vector x ā Rn. In this part we can see, that it is one of the general cases of linear functions. Letās consider any linear function f : V ā W. Let vectors e 1 , e 2 ,... , en form a basis in the space V. And let we know the values f (e 1 ), f (e 2 ),... , f (en). Then we can compute the function f for any vector from V using only these given values. To show it letās note, that is eiās form a basis, then any vector v from V can be represented as a linear combination of them:
v = a 1 e 1 + a 2 e 2 + Ā· Ā· Ā· + anen.
Now letās show how to compute the value f (v):
f (v) = f (a 1 e 1 + a 2 e 2 + Ā· Ā· Ā· + anen) = f (a 1 e 1 ) + f (a 2 e 2 ) + Ā· Ā· Ā· + f (anen) = a 1 f (e 1 ) + a 2 f (e 2 ) + Ā· Ā· Ā· + anf (en).
Example 2.3. Now letās consider the function of taking a derivative in the space P 2 : D(at^2 + bt + c) = 2at + b. Letās take the standard basis in the space of polynomials P 2 and compute values of function on basis vectors:
- e 1 = t^2 , and D(e 1 ) = D(t^2 ) = 2t. The coordinates are (0, 2 , 0) since it is equal to 0 t^2 + 2t + 0.
- e 2 = t, and D(e 2 ) = D(t) = 1. The coordinates are (0, 0 , 1) since it is equal to 0 t^2 +0t+1.
- e 3 = 1, and D(e 3 ) = D(1) = 0. The coordinates are (0, 0 , 0) since it is equal to 0 t^2 +0t+0.
So, the matrix is
AD =
For example, letās take a derivative ofļ£« 3 t^2 +5t+7. Weāll write this polynomial as a column-vector
ļ£¬ļ£
ļ£·ļ£ø, and multiply AD by it:
ļ£« ļ£¬ļ£
So, the derivative of this polynomial 6 t + 5.
2.1 Proof
Letās prove that if a matrix Af is constructed using the method provided here, then
f (x) = Af x.
Proof. Letās take any vector x = (x 1 , x 2 ,... , xn) = x 1 e 1 + x 2 e 2 + Ā· Ā· Ā· + xnen. Since we have that
f (ej ) = (a 1 j , a 2 j ,... , amj )
ā j-th column of the matrix A, then
f (x) = x 1 f (e 1 ) + x 2 f (e 2 ) + Ā· Ā· Ā· + xnf (en) = x 1 (a 11 , a 21 ,... , am 1 ) + Ā· Ā· Ā· + xn(a 1 n, a 2 n,... , amn) = (a 11 x 1 + a 12 x 2 + Ā· Ā· Ā· + a 1 nxn,... , am 1 x 1 + am 2 x 2 + Ā· Ā· Ā· + amnxn) = (
j
a 1 j xj ,
j
a 2 j xj ,... ,
j
amj xj ).
Comparing this with the formal definition of matrix multiplication, we get that
f (x) = Af x.
