I have a pretty simple question that confused me:
V is a vector space of a finite dimension.
$T: V \to V$ is a linear transformation.
The information that's been given in question:
$\operatorname{Im} T = \ker T$
I want to know if what I'm doing is right:
I took only for exmaple: $\dim \ker T = \dim \operatorname{Im} T = 2$
I take $D = \{v_1,v_2\}$ as some basis for $\ker T$ and $\operatorname{Im} T$
then I add vectors and let's say $B = \{v_1,v_2,v_3,v_4\}$ is basis of $V$,
and I demand that $v_3$ and $v_4$ follow these requirements: $T(v_3) = v_1$, $T(v_4) = v_2$.
(I did this for a reason, I didn't write here the whole question)
But I have to prove that this is basis of $V$, linearly independent and spanning $V$.
So I tried to prove that it is linearly independent:
I assume : $a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 = 0$
Then I applied $T$ on it :
$T(a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4) = 0$
$a1 * T(v1) + a2* T(v2) + a3*T(v3) + a4* T (v4) = 0$.
$v_1$ and $v_2$ are from $\ker T$ so $T(v_1) = 0$ and $T(v_2) = 0$,
so I get that there is a linear dependence, but I don't know where is the dependence: is the action that I did means that $\{v_1,v_2,v_3,v_4\}$ are linearly dependent or it means ONLY that $T(v_1)$, $T(v_2)$, $T(v_3)$, $T(v_4)$ are dependent?
Thank you
Best Answer
Since $\dim\mathop{\rm Ker} T > 0$, the operator $T$ is singular, hence it reduces the dimension of the space (because $\dim\mathop{\rm Im} T = n - \dim\mathop{\rm Ker} T < n$), meaning it maps any base to a linearly dependent set (of vectors that span its image).
In other words, $T(B)$ is a set of $4$ linearly dependent vectors, because $T(B) \subseteq \mathop{\rm Im} T$, which is a 2-dimensional space (so it cannot have more than 2 linearly independant vectors).