MATH 110: LINEAR ALGEBRA FALL 2007/08 PROBLEM SET 7 SOLUTIONS Let V be a vector space. The identity transformation on V is denoted by I V, ie.I V: V !V and I V (u) = u for all u 2V.

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On the other hand, T(d,l) F 2d contains F2, yielding the result. Example 1 When k = 2l, we can recover from Remark 2 the well-known result dim ker… dim ker f 4 3 1 by the fundamental theorem Now x x 1 x 2 is in ker f since dim from MATH 236 at McGill University dim(ker(A))+dim(im(A)) = m There are ncolumns. dim(ker(A)) is the number of columns without leading 1, dim(im(A)) is the number of columns with leading 1. 5 If A is an invertible n× n matrix, then the dimension of the image is n and that the dim(ker)(A) = 0.

Ker(f) = 〈(λ,−1)〉. 3. v2 = 0 und es gibt λ ∈ K mit v1 = λv2: Analog zum vorherigen Fall.

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Both of these are vector spaces. ker(T) is a sub- space of V , and q.e.d.. Definition 3. The dimensions of the kernel and Recall that a function f is said to be.

2. Trong toán học, một phép biến đổi tuyến tính (còn được gọi là toán tử tuyến tính hoặc là ánh xạ tuyến tính) là một ánh xạ → giữa hai mô đun (cụ thể, hai không gian vectơ) mà bảo toàn được các thao tác cộng và nhân vô hướng vectơ. MATH 110: LINEAR ALGEBRA FALL 2007/08 PROBLEM SET 7 SOLUTIONS Let V be a vector space. The identity transformation on V is denoted by I V, ie.I V: V !V and I V (u) = u for all u 2V. dim(Im f) = dim(Im (fe))+dim(ker(fe)) soit : rg (f) = rg (f2)+dim(kerf∩Im f) , ce qui donne bien la formule demandée. 2- Par le théorème du rang, n−dim(kerf) = n−dim(kerf2)+dim(kerf∩Im f) et donc dim(kerf2) = dim(kerf)+dim(kerf∩Im f) en n dim(kerf∩Im f) ≤ dim(kerf) puisque kerf∩Im f⊂ kerf donc dim(kerf2) ≤ 2dim(kerf) The rank–nullity theorem is a theorem in linear algebra, which asserts that the dimension of the domain of a linear map is the sum of its rank (the dimension of its image) and its nullity (the dimension of its kernel). The dimension would be: dim (Ker (ƒ)) = 4 - 2 = 2 To determine the base of Ker (ƒ), from what I understood, you have to take the rows and set 2 (because the dimension of Ker (ƒ) is 2) of x, y, z and t, equal to a scalar (a and b), put them into a system and then solve it until you get the two columns of the base of Ker (ƒ).