Find the kernel and range, and state their dimensions, of the following linear transformation $L : R^3 \rightarrow R^3, L(x)= (x_1,x_1,x_1). $
Def: Let $L: V \rightarrow W$ be a linear transformation the kernel of L denoted ker(L) is defined by $\text{ker}(L)= \{v \in V | L(v) = 0 \;_W \}
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Def: Let $L:V \rightarrow W$ be a linear transformation and let S be a subspace of V. The image of $S$, denoted $L(S)$ is defined by $L(S)= \{ w \in W| w= L(v) \text{ for some} v \in S\}$
The image of the entire space $L(v)$ is called the range of $L$.
How does one use these theories in order to solve this question? Any hints at an approach would be welcomed.
Best Answer
We have $L:\Bbb R^3\to\Bbb R^3$ mapping $(x_1,x_2,x_3)\mapsto (x_1,x_1,x_1)$.
Its kernel, by definition, consists of those real triples $(x_1,x_2,x_3)$ which are mapped to $(0,0,0)$ by $L$. For this specific $L$ above, these are exactly the ones where $x_1=0$ and $x_2,x_3$ are arbitrary.
These can be parametrized without further restrictions with $2$ real numbers (namely, by $x_2$ and $x_3$), so it will be a $2$ dimensional space.
Turning this into the language of bases, just get the vectors where one of the parameters is $1$, and the rest are $0$. In our case, $(0,1,0)$, $\ (0,0,1)$ will give a basis for $\ker L$.
Its range, by definition, consists of the possible values the function can take. These are now $(a,a,a)$ for all possible real $a$. $\ $ [Indeed, e.g. $L(a,0,0)=(a,a,a)$ for any $a\in\Bbb R$.]
It can be given with $1$ parameter, it will have dimension $1$, and a basis is given by $(1,1,1)$.