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April 7th, 2018, 04:26 AM  #1 
Newbie Joined: Apr 2018 From: Cracow Posts: 1 Thanks: 0  Kerf and Imf, basis and dimensions of f:R3→R2:
How to find Kerf and Imf, basis and dimension of f:R3→R2: f(1,1,1)=(1,−2),f(1,2,0)=(3,−4),f(0,1,0)=(1,3) ?
Last edited by aris182; April 7th, 2018 at 04:31 AM. 
April 7th, 2018, 11:04 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,089 Thanks: 846 
It helps to know what "ker(f)" and "Im(f)" (writing "kerf" and "imf" as single words is confusing) mean! The "kernel of f", "ker(f)", is the set of all vectors, u, in the domain space, such that f(u)= 0. The "image of ", "I'm(f)", is the set of all vectors, v, in the range space, such that there exist u, in the image space, such that f(u)= v. I assume you knew that. Let (x, y, z) be a vector in $\displaystyle R^3$, the domain space of f. We want to be able to write (x, y, z) as a linear combination of the given vectors in the domain space. That is, we want to find number, a, b, c, such that a(1, 1, 1)+ b(1, 2, 0)+ c(0, 1, 0)= (a+ b, a+ 2b+ c, a)= (x, y, z). We must solve a+ b= x, a+ 2b+ c= y, and a= z, for a, b, and c. Obviously, the third equation gives a= z. Putting that into the first equation, z+ b= x so b= x z. Putting those into the second equation, z+ 2(x z)+ c= 2x z+ c= y so c= 2x+ y+ z. (x, y, z)= z(1, 1, 1)+ (x z)(1, 2, 0)+ (2x+ y+ z)(0, 1, 0). Now, f(x, y, z)= f(z(1, 1, 1)+ (x z)(1, 2, 0)+ (2x+ y+ z)(0, 1, 0))= zf(1, 1, 1)+ (x z)f(1, 2, 0)+ (2x+ y+ z)f(0, 1, 0)= z(1, 2)+ (x z)(3, 4)+ (2x+ y+ z)(1, 3)= (z+ 3(x z)+ (2x+ y+ z), 2z+ 4(x z) 3(2x+ y+ z)= (x+ y z, 10x 3y 9x). (x, y, z) is in the kernel of f if and only if x+ y z= 0 and 10x 3y 9x= 0. That's two equations in three unknowns so we can solve for two of the unknowns in terms of the other one. The kernel is one dimensional. Specifically, we multiply the first equation by 2 and add that to the second equation, we get 13x 11z= 0. x= (11/13)z. Putting that into the first equation, (11/13)z+ y z= (2/13)z+ y= 0, y= (2/13)z. Taking z= 13, x= 11 and y= 2. Any vector in the kernel is a multiple of (11, 2, 13) so that a basis for the kernel is the set {(11, 2, 13)}. Now, to find the image we look at the span of the three vectors (1, 2), (3, 4), and (1, 3): a(1, 2)+ b(3, 4)+ c(1, 3)= (a 3b+ c, 2a 4b 3c)= (x, y). What x, y satisfy a 3b+ c= x and 2a 4b 3c=y for some values of a and b? It should be no surprise that, since there are only two equations in three unknowns, we can get infinitely many solutions for all x and y: if we multiply the first equation by 2 and add that to the second equation, 10b c= 2x+ y. In particular, taking b= 0, c= 2x y. Then the first equation becomes a 2x y= x so a= 3x+ y. That is one solution and is enough to show that the image is all of R^2. Of course that is of dimension 2 and a basis is (1, 0), (0, 1). We didn't really need to do that last calculation. Once we know that the kernel had dimension 1, we could use the "rank nullity theorem" (the dimension of Im(f) is the "rank" of f, the dimension of ker(f) is the "nullity" of f): if f is a linear function from a vector space of dimension n to any vector space, the kernel has dimension p and the image has dimension q, then p+ q= n. Here f is from R^3 which has dimension 3. Knowing that the kernel has dimension 1 (so the "nullity" is 1), the image must have dimension 3 1= 2 (the "rank" is 2). Last edited by Country Boy; April 7th, 2018 at 11:12 AM. 

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basis, dimensions, fr3→r2, imf, kerf 
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