1
$\begingroup$

I am currently proving a theorem where $L: \mathbb R^n \to \mathbb R^n$ and $ M:\mathbb R^n\to \mathbb R^n$ are linear transformations if:

  1. $L+M:\mathbb R^n\to \mathbb R^n$ defined by $(L+M)(v)=L(v)+M(v)$ is a linear transformation.

  2. $-L:\mathbb R^n \to \mathbb R^n$ defined by $(-L)(v)=-(L(v))$ is a linear transformation

  3. $cL:\mathbb R^n \to \mathbb R^n$ defined by $(cL)(\vec v)=c(L(\vec v))$ is a linear transformation (Where c is some scalar in the reals)

  4. $Z:\mathbb R^n \to \mathbb R^n$ defined by $ Z(\vec v)= \vec 0$ (where $Z=\vec 0$) is a linear transformation.

I believe I have proved the first two but am tripping up on the simple proofs for numbers 3 and 4.

3: $L+M(-\vec v) = L(-\vec v) + M (-\vec v) = -L + -M(\vec v) = -(L+M)(\vec v)$

4: $L+M (Z(\vec v)) = L(Z(\vec v)) + M(Z(\vec v)) = Z*L(\vec v) + Z*M(\vec v) = Z((L+M)(\vec v))$ and then since $Z=\vec 0$ we have $\vec 0*((L+M)(\vec v))= \vec 0$? I'm not sure this gives us the zero vector though...

  • 0
    Do you mind if I edit your statements so that they're a bit more readable? – 2017-02-23
  • 0
    Yes please do! I've been very slowly learning to format! I apologize that it's so messy! – 2017-02-23
  • 0
    No worries it takes time to understand LaTeX and you can use it here. Just remember to surround equations with $ or (double dollar signs) if you want them entered. – 2017-02-23
  • 0
    If you haven't seen it, there's a [tutorial on the formatting stuff here](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) – 2017-02-23
  • 0
    Sorry was eating dinner. What you need to prove so that you can prove something is a linear transformation? hint: there are two conditions. – 2017-02-23
  • 0
    I know it must hold under addition and scalar multiplication. For the theorem I'm trying to prove that it'll hold under negatives which it should since it holds under scalar multiplication. I think my biggest question is will multiplying by Z in 4 give me the zero vector or just zero? – 2017-02-23

1 Answers 1

0

Well if you are trying to prove that $Z$ is a linear transformation. That should be easy.

Step 1: Show $Z$ is closed under addition. That is to say that if we have $\vec u,\; \vec v \in \mathbb R^n$ then

$$Z( \vec u+\vec v)= Z(\vec u)+ Z(\vec v)$$

Step 2: Show $Z$ has the homogeneity property. That for $\vec v \in \mathbb R^n$ and $a \in \mathbb R$ we have, $$Z(a\vec v)= a Z(\vec v) $$

However, I have a feeling this isn't what you're trying to do. By what it sounds you are trying to prove that linear transformations have a unique property:

that the $\vec 0$ of one space maps to $\vec 0$ of the other.

To do this we're going to denote two different $\vec 0$'s $\vec 0_1$ for the first $\mathbb R^n$, and $\vec 0_2$ for the second $\mathbb R^n$. Even though they are the same we need to do this so it won't be so confusing.

We're already given that $L$ is a linear transformation so it has the homogeneity property. We need to make one further note here. That is $a*\vec 0= \vec 0,\; \forall a \in \mathbb R$, and namely if $a=0$.

Let look at $L(\vec 0_1)=\vec u$ for $\vec u \in \mathbb R^n$, $$\begin{array}{ccc} L(\vec 0_1)&=& L(0*\vec 0_1)\\ &=& 0*L(\vec 0_1)\\ &=& 0*\vec u \end{array}$$

Now from here lets look at what $0*\vec u$ means. As $$\vec u = \begin {bmatrix} u_1\\ u_2\\ \vdots\\ u_n \end{bmatrix},$$ this implies that, $$L(\vec 0)=0*\vec u = \begin {bmatrix} 0*u_1\\ 0*u_2\\ \vdots\\ 0*u_n \end{bmatrix}=\begin {bmatrix} 0\\ 0\\ \vdots\\ 0 \end{bmatrix}=\vec 0_2$$ as desired.