Linear Transformations: M 2x2 to R^3 with Ker(L) and Rank Calculations

[kdizzle711]

Homework Statement

Let L : M 2x2 ---> R^3 be defined by L( a b ) = (a, b+c , d)
c d

a. Find ker(L), is L a 1-1 function?
b. What is the rank of L and does it map M 2x2 onto R^3

Homework Equations

The Attempt at a Solution

a. ker(L) = (0, 0, 0),
b. Rank= 3
c. 4= N + 3
N= 1 So it does map it




Can you guys see if this is right? It is an odd number in the book

 

[Mark44]

Homework Statement

Let L : M 2x2 ---> R^3 be defined by L( a b ) = (a, b+c , d)
c d

a. Find ker(L), is L a 1-1 function?
b. What is the rank of L and does it map M 2x2 onto R^3

Homework Equations

The Attempt at a Solution

a. ker(L) = (0, 0, 0),

This is not right, and in fact, not even close. The kernel of L consists of 2 x 2 matrices, not vectors in R³, as the vector you show is. The 2 x 2 zero matrix is included in the kernel of this problem, but there are many other matrices as well.

b. Rank= 3
c. 4= N + 3
N= 1 So it does map it

What does N represent here? Convince us that you understand what you are doing. Also, unless you have a theorem to invoke, you should show that the transformation is 1-to-1 (or not) by showing that if L(M₁) = L(M₂), then M₁ = M₂.

Can you guys see if this is right? It is an odd number in the book

 

[kdizzle711]

N represents the dimension of the nullspace, I am stumped
Am I supposed to prove it my T(v+w) = T(v) +T(w) and T(cV)=cT(v)?

 

[kdizzle711]

Also the nullspace is the solution set to Ax=O, so it being 1 means there is only 1 solution set

 

[HallsofIvy]

You don't appear to know any of the definitions. The "nullspace" and "kernel" are the same thing- just different names. You say at one point that the the kernel is just {(0,0,0)} and at another that the dimension of the nullspace is 1. Those can't both be true. The "subspace" {(0,0,0)}" has dimension 0, not 1. Learn the definitions!

In any case, since L is applied to "2 by 2 matrices" the kernel (nullspace) must be a subspace of matrices not ordered triples.

Here, L is defined as
L\left(\begin{bmatrix}a & b \\ c & d\end{bmatrix}\right)= (a, b+ c, d)

For a matrix to be in the kernel we must have
L\left(\begin{bmatrix}a & b \\ c & d\end{bmatrix}\right)= (a, b+ c, d)= (0, 0, 0)
or a= 0, b+ c= 0, d= 0. What numbers satisfy that? What matrices are in the kernel?

Answer that and we'll work on the rest of the problem.