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I am reading an introduction on linear maps in my text book on linear algebra. The following statements are made:

Suppose $G_1 (\vec{u}) = (x_1 + 2x_2 + 3x_3 + 1, 4x_1, 9x_3)$

Then we can use the following property of linear maps. Let $\lambda = 0$ and $\vec{u} = \vec{0}$

$$G(\lambda\vec{u}) = \lambda G(\vec{u})$$

And specifically:

$$G(\vec{0}) = 0 \cdot G(\vec{0}) = \vec{0}$$

This means that a linear map maps the zero vector to the zero vector. It also means that $G_1$ cannot be a linear map, this is because $G_1(0,0,0) = (1,0,0) \neq (0,0,0)$. The constant term $1$ is breaking the linearity.

My analysis

I don't understands the above statements completely. For example this statement: $G(\vec{0}) = 0 \cdot G(\vec{0}) = 0$ should be true for any function $G(\vec{u})$, since whatever result of the map $G(\vec{u})$ will be it will be multiplied by $0$ and result in $\vec{0}$. In the case above it would be $0 \cdot (1,0,0) = \vec{0}$. This would map the zero vector to the zero vector and hence be a correct linear map?

Can anyone please explain this to me?

Lukas Arvidsson
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  • For "most" functions $G$, it simply isn't true that $G(\vec{0}) = 0 \cdot G(\vec{0})$. Linear maps are special. –  Jun 07 '14 at 17:39
  • If $G$ is a linear map, it satisfies $G(\lambda\vec u)=\lambda G(\vec u)$, and so $G(\vec 0)=\vec 0$. But $G_1(\vec 0)=(1,0,0)\ne\vec 0$. So $G_1$ cannot be a linear map. –  Jun 07 '14 at 17:39
  • @Rahul But if $\lambda = 0$ then $G_1$ above would satisfy $G(\lambda\vec u)=\lambda G(\vec u)$ and be a linear map? – Lukas Arvidsson Jun 07 '14 at 17:44
  • Why don't you expand both sides of that equality and see what happens? Go on, I'll wait. –  Jun 07 '14 at 17:46
  • @Rahul Ok I'll try :). So if $\vec{u} = \vec{0}$ then $G_1(\vec{u}) = (1,0,0)$. This gives us $0 \cdot (1,0,0) = (0 \cdot 1, 0, 0) = (0,0,0) = \vec{0}$? – Lukas Arvidsson Jun 07 '14 at 17:52
  • You're focusing on the last equality, $0\cdot G(\vec 0)=\vec 0$. But that's always true. You should be looking at the whole thing, $G(\vec0)=\vec 0$, which is not true for $G_1$. –  Jun 07 '14 at 18:12

3 Answers3

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You are being confused. There are two statements going on here.

The first statement is a property of linear transformations. Specifically, it is true that $G(\lambda \mathbf{u}) = \lambda G(\mathbf{u})$ for any $\lambda$ and any $\mathbf{u}$.

A consequence of this is that every linear map must map $\mathbf{0}$ to $\mathbf{0}$, because $\mathbf{0} = 0\mathbf{v}$ for any vector $v$, so $$G(\mathbf{0}) = G(0\mathbf{v}) = 0G(\mathbf{v}) = \mathbf{0}.$$

This property holds true for any vector $\mathbf{v}$ we choose, and certainly it holds if we wish to choose $\mathbf{v} = \mathbf{0}$.

So, this is a necessary condition for a mapping $G$ to be linear.

Now, let's see what happens under your mapping $G_1$ when applied to $\mathbf{0}$:

$$G_1(\mathbf{0}) = (1,0,0)^T.$$

Since $G_1$ doesn't map $\mathbf{0}$ to $\mathbf{0}$, then it cannot be linear!

Therefore, it doesn't make sense to try to pull zero out of the function, because the function is not linear. Hence, $G_1$ does not satisfy the property that $G_1(\lambda \mathbf{u}) = \lambda G_1(\mathbf{u})$.

Emily
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The problem the OP had in this question (it appears to me), which led to the misunderstanding, was caused by not working the algebraic function out in the first place. This is a common misunderstanding stemming from the idea that the function is simply being multiplied by zero ( 0*f(v)=0 ) when what is being stated by the theorem is $$0*f(0) = 0 = f(0)$$

performing the algebra indicated when learning this idea will make the problem instantly clear $$G_1 (\vec{u}) = (x_1 + 2x_2 + 3x_3 + 1, 4x_1, 9x_3)$$ $$0*G_1 (\vec{u}) = 0*(x_1 + 2x_2 + 3x_3 + 1, 4x_1, 9x_3)=(0,0,0)$$ $$G_1 (\vec{0}) = (0_1 + 2*0_2 + 3*0_3 + 1, 4*0_1, 9*0_3)=(1,0,0)\ne (0,0,0) $$ Many people grasp the idea that the zero, being used to prove the theorem, in the function is being multiplied by every term, and forget that there is a non-x component which is being added. Terms of that nature, make the equation not linear. It is such a simple idea that it is easily overlooked when learning.

habitabitha
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The vector $\vec{0}$ can be written as $0$ . $\vec{u}$ for any $\vec{u}$, therefore

$G(\vec{0})= G(0\vec{u}) =0 \cdot G(\vec{u}) = \vec{0}$

You can also write $\vec{0}$ as $\vec{0}+\vec{0}$ and argue that

$G(\vec{0}) = G(\vec{0}+\vec{0}) = G(\vec{0}) + G(\vec{0})$ $\Rightarrow$ $G(\vec{0})=\vec{0}$

DGRasines
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  • Thank you for your answer. But i still don't understand this: My book claims that $G(\vec{0}) = 0 \cdot G(\vec{0}) = \vec{0}$ is a property of linear maps. Why does not $G_1$ above satisfy this? – Lukas Arvidsson Jun 07 '14 at 17:46
  • @LukasArvidsson because G1 is not linear – DGRasines Jun 07 '14 at 17:51