MHB Proving disjoint of Kernel and Image of a linear mapping

[rputra]

I am working on a problem that goes like this:

Show that $Ker (F) \cap I am (F) = \{0\}$ if $F: W \rightarrow W$ is linear and if $F^4 = F.$

I have the solution but there is one step which I need help: (the delineation is mine)

(1) Suppose that there exists $x$, such that $x \in Ker(F) \cap Im(F)$
(2) From $x \in Ker(F)$, we have $F(x) = 0$, by definition of kernel of a mapping
(3) From $x \in Im(F)$, there exists $w \in W$, such that $F(w) = x$
(4) Then we have
$$\begin{align}
F^4(w) &= F^3(F(w))\\
&= F^3(x)\\
&= F^2(F(x))\\
&= F^2(0)\\
&= 0.
\end{align}$$
(5) Since we have $F^4(w) = 0, F^4(w) = F(w)$, and $F(w) = x$, we conclude that $x = 0$. Hence $Ker(F) \cap Im(F) = \{0\}$, as desired.

Here is the one step I did not understand: In the last line of step (4), why is that $F^2(0) = 0$? Any help would be very much appreciated. Thank you for your time in advance.

 

[I like Serena]

I am working on a problem that goes like this:

Show that $Ker (F) \cap I am (F) = \{0\}$ if $F: W \rightarrow W$ is linear and if $F^4 = F.$

I have the solution but there is one step which I need help: (the delineation is mine)

(1) Suppose that there exists $x$, such that $x \in Ker(F) \cap Im(F)$
(2) From $x \in Ker(F)$, we have $F(x) = 0$, by definition of kernel of a mapping
(3) From $x \in Im(F)$, there exists $w \in W$, such that $F(w) = x$
(4) Then we have
$$\begin{align}
F^4(w) &= F^3(F(w))\\
&= F^3(x)\\
&= F^2(F(x))\\
&= F^2(0)\\
&= 0.
\end{align}$$
(5) Since we have $F^4(w) = 0, F^4(w) = F(w)$, and $F(w) = x$, we conclude that $x = 0$. Hence $Ker(F) \cap Im(F) = \{0\}$, as desired.

Here is the one step I did not understand: In the last line of step (4), why is that $F^2(0) = 0$? Any help would be very much appreciated. Thank you for your time in advance.

Hey Tarrant!

For a linear transformation $F$, we have that $F(\mathbf 0)=\mathbf 0$.
Proof
Suppose $F(\mathbf 0)\ne \mathbf 0$, then for any $\mathbf u$ we have that $F(\mathbf 0) = F(0\cdot \mathbf u)=0\cdot F(\mathbf u)=\mathbf 0$. Contradiction. Therefore $F(\mathbf 0)=\mathbf 0$.

 

[rputra]

Hey Tarrant!

For a linear transformation $F$, we have that $F(\mathbf 0)=\mathbf 0$.
Proof
Suppose $F(\mathbf 0)\ne \mathbf 0$, then for any $\mathbf u$ we have that $F(\mathbf 0) = F(0\cdot \mathbf u)=0\cdot F(\mathbf u)=\mathbf 0$. Contradiction. Therefore $F(\mathbf 0)=\mathbf 0$.

Ok, thanks. It follows the definition of linear mapping: For $c$ an arbitrary real number, $F(cx) = cF(x)$ which leads to $F(0) = 0$ for $c = 0.$ Thank you again for your time.