Problem regarding Hamel basis

Question

I'm doing a problem for a homework assignment and the problem is as follows:

Let $\{e_{\alpha}\}_{\alpha\in I}$ denote a Hamel basis for a vector space $X$.

(i) Let $T:X\rightarrow X$ be a bijective linear operator. Show that $\{Te_{\alpha}\}_{\alpha\in I}$ is also a Hamel basis for $X$.

(ii) Can $\{Te_{\alpha}\}_{\alpha\in I}$ be a Hamel basis for $X$ if $T$ is not surjective?

(iii) Can $\{Te_{\alpha}\}_{\alpha\in I}$ be a Hamel basis for $X$ if $T$ is not injective?

I'm having trouble with this and I think it has mainly to due with my understanding of what a Hamel basis really is. Being this is a homework assignment I'm not looking for solutions but more an idea of how to grasp the problem. If anyone can give any hints for what I need to show that would be greatly appreciated!

Answer 1

Asserting that $\{T(e_\alpha)\mid\alpha\in I\}$ is a Hamel basis is the same thing as assertint that every element of $X$ can be writen as a (finite) linear combination of elements of that set in one and only one way. So, take $w\in X$. Since $T$ is surjective, $w=T(v)$, for some $v\in X$. But $v=a_1e_{\alpha_1}+\cdots+a_ne_{\alpha_n}$ for some $\alpha_1,\ldots,\alpha_n\in I$ and therefore$$w=T(v)=a_1T(e_{\alpha_1})+\cdots+a_nT(e_{\alpha_n}).$$

Now, suppose that you also have $w=b_1T(e_{\alpha_1})+\cdots+b_nT(e_{\alpha_n})$. Then $w=T\bigl(b_1e_{\alpha_1}+\cdots+b_ne_{\alpha_n}\bigr)$. But $w=T(v)$ too and $T$ is injective. Therefore$$b_1e_{\alpha_1}+\cdots+b_ne_{\alpha_n}=v=a_1e_{\alpha_1}+\cdots+a_ne_{\alpha_n}$$and so $b_i=a_i$ for each $i\in\{1,2,\ldots,n\}$.

So, yes, $\{T(e_\alpha)\mid\alpha\in I\}$ is a Hamel basis of $X$.

Answer 2

Prove the following facts:

Fact 1. If $\mathscr{B}$ is a spanning set of the vector space $X$ and $T\colon X\to Y$ is a linear map to another vector space, then the image of $T$ is the span of $T[\mathscr{B}]=\{T(v):v\in\mathscr{B}\}$.

Fact 2. A linear map $T\colon X\to Y$ is injective if and only if, for every linearly independent set $\mathscr{B}$ in $X$, the set $T[\mathscr{B}]$ is linearly independent in $Y$.

Now you have the answer to your three questions.

Side note: the implication “if $T[\mathscr{B}]$ is linearly independent in $Y$ then $\mathscr{B}$ is linearly independent in $X$” is true with no assumption on $\mathscr{B}$.