I'm on my preperation for algebra's exam and I've been searching for information about several exercises. Some of them I would like to ask if I solved it correct.
- There is given a map $F:\Bbb{R}[x] \rightarrow \Bbb{R}[x]$ and $F(a(x))=a(x+1), \quad a(x) \in \Bbb{R}[x]$. Is $F$ a linear transformation?
- $F:\Bbb{R}^3 \rightarrow \Bbb{R}^3$, $F$ - linear transformation and $F((x,y,z))=(x+y-z,x-2y,3x-2y+3z)$. Is linear subspace $U=L((1,0,0))$ F-invariant?
$\mathbf{My \space solutions:}$
- Let's take $a(x)=x \in \Bbb{R}[x]$. Then we can rewrite $F(a(x))=a(x+1)$ as $F(x)=x+1$. Now it could be clearly seen that $F(\alpha x)=\alpha x+1 \neq \alpha (a+1)=\alpha F(x) \quad \forall\alpha \in \Bbb{R}$. So F is not a linear transformation.
- I found $F(1,0,0)=(1,1,3)$ . The transformation matrix is $$ A=\begin{pmatrix} 1 & 1 & 3 \\ 1 & -2 & -2 \\ -1 & 0 & 3 \\ \end{pmatrix} $$ and we can find the image of the transformation $F$ : $Im(F)=\left<(1,1,3),(1,-2,-2),(-1,0,3) \right>$. Now we can see that $F(1,0,0)=(1,1,3) \subset Im(F).$ Also we know that $Im(F)$ and $Ker(F)$ are linear transformations $F$-invariant subspaces.
Can anyone check if I was right? I would be truely grateful.
Your approach is correct, but what you claim is not true for $\alpha=1$. You only need to show that it doesn't hold for some $\alpha\in\mathbb{R}$, so just choose $\alpha=0$, then you obtain $1=0$, which is false of course.
Using left-multiplication for vectors, your matrix is correct. I don't understand why you are considering $Ker(F)$ and what you meant with the last line, but the answer should be no: $U$ is an invariant subspace under $F$ if $F(U)\subseteq U$, which is not true, as for example $F(1,0,0)=(1,1,3)\not\in L((1,0,0))$.