Does there exist a linear transformation T : R2 → R4 such that Range(T ) ={(x1, x2, x3, x4) : x1 + x2 + x3 + x4 = 0
Question
Does there exist a linear transformation T : R2 → R4 such that Range(T ) ={(x1, x2, x3, x4) : x1 + x2 + x3 + x4 = 0
Solution
Yes, there does exist a linear transformation T : R2 → R4 such that Range(T) = {(x1, x2, x3, x4) : x1 + x2 + x3 + x4 = 0}.
Here's a step-by-step explanation:
-
Define the linear transformation T : R2 → R4 as T(x, y) = (x, y, -x, -y).
-
For any (x, y) in R2, T(x, y) is in the set {(x1, x2, x3, x4) : x1 + x2 + x3 + x4 = 0} because x + y - x - y = 0.
-
Therefore, the range of T is a subset of {(x1, x2, x3, x4) : x1 + x2 + x3 + x4 = 0}.
-
Conversely, for any (x1, x2, x3, x4) in the set {(x1, x2, x3, x4) : x1 + x2 + x3 + x4 = 0}, we can find (x, y) in R2 such that T(x, y) = (x1, x2, x3, x4) by setting x = x1 and y = x2.
-
Therefore, the set {(x1, x2, x3, x4) : x1 + x2 + x3 + x4 = 0} is a subset of the range of T.
-
Since the range of T and the set {(x1, x2, x3, x4) : x1 + x2 + x3 + x4 = 0} are subsets of each other, they are equal.
So, the linear transformation T : R2 → R4 such that Range(T) = {(x1, x2, x3, x4) : x1 + x2 + x3 + x4 = 0} does exist.
Similar Questions
Define the linear transformation T by T(x) = Ax. Find ker(T), nullity(T), range(T), and rank(T).A = 4 −2 1 11 −1(a) ker(T) (b) nullity(T)(c) range(T){(s, t, s − 3t): s, t are any real number}R3 {(s, t, 0): s, t are any real number}{(3s, 3t, s − t): s, t are any real number}R2(d) rank(T)
show that the transformation defined by T(x1,x2)=(2x1-3x2,x1+4,5x2) is not linear ChatGPT
Let T: R2 --> R2 be the linear transformation given by T(x,y)=(2x+y, x+y). Find T -1(x,y).Select one:a. (x-y, -x+2y)b. (x-y, x+2y)c. None of themd. (x+y, -x+2y)
Find the linear transformation T : ℝ2 → ℝ2 that has the values given below on the given basis.T 2 1 = 4 6, T 1 1 = 3 7(a) First write a general vector v as a linear combination of the basis vectors.v =v1 v2 = v1−v2 2 1 + 2v2−v1 1 1(b) This implies the following.T(v) = T v1 v2= 4v1+3v2 6v1+7v2 = v
This problem provides a useful way to construct invertible linear maps between subspaces X ⊂ Rn and Y ⊂ Rm.3 Let B = {u1, u2, · · · , uk} be a basis of X. (a) Let {v1, v2, · · · , vk} ⊂ Y . Show that there exists a unique linear transform T : X → Y such that T(uj) = vj for every 1 ≤ j ≤ k. To be specific, you need to 1) define a function4 T : X → Y and verify that it is linear, 2) show that if T0 : X → Y is a linear map such that T0(uj) = vj, then T = T0.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.