Note that we have a commutative diagram
Here, $I_\beta$ is the linear map defined by
$$
I_\beta(v)=(\lambda_1,\dotsc,\lambda_n)
$$
where $\beta=\{v_1,\dotsc,v_n\}$ and $\lambda_1,\dotsc,\lambda_n$ are the unique scalars satisfying
$$
v=\lambda_1 v_1+\dotsb+\lambda_n v_n
$$
The map $I_\gamma$ is defined similarly.
Fact 1. $I_\beta$ and $I_\gamma$ are invertible.
Fact 2. $I_\gamma\circ T=L_A\circ I_\beta$
See if you can prove these facts yourself! Once these two facts have been established, your claim is easy to prove.
Suppose that $\{k_1\dotsc,k_\ell\}$ is a basis for $\ker T$, so $\DeclareMathOperator{nullity}{nullity}\nullity(T)=\ell$.
We claim that $\{I_\beta(k_1),\dotsc,I_\beta(k_\ell)\}$ is a basis for $\ker L_A$. The proof of this involves two steps.
Step 1. $\{I_\beta(k_1),\dotsc,I_\beta(k_\ell)\}$ spans $\ker L_A$.
Proof. Suppose that $x\in\ker L_A$ so $L_A(x)=0$. It follows that $$ 0=(I_\gamma^{-1}\circ L_A)(x)=(T\circ I_\beta^{-1})(x) $$ so
$I_\beta^{-1}(x)\in\ker T$. Since $\{k_1,\dotsc,k_\ell\}$ is a basis
for $\ker T$, it follows that $$ I_\beta^{-1}(x)=\lambda_1
> k_1+\dotsb+\lambda_\ell k_\ell $$ Hence
$$ x=\lambda_1 I_\beta(k_1)+\dotsb+\lambda_\ell I_\beta(k_\ell)
> $$ This proves that
$\{I_\beta(k_1),\dotsc,I_\beta(k_\ell)\}$ spans $\ker L_A$. $\Box$
Step 2. $\{I_\beta(k_1),\dotsc,I_\beta(k_\ell)\}$ is linearly independent.
Proof. Suppose $$ \lambda_1 I_\beta(k_1)+\dotsb+\lambda_\ell I_\beta(k_\ell)=0 $$ Then $$ I_\beta(\lambda_1 k_1+\dotsb+\lambda_\ell
> k_\ell)=0\tag{1} $$ Applying $I_\beta^{-1}$ on the left of (1) gives
$$ \lambda_1 k_1+\dotsb+\lambda_\ell k_\ell=0 $$ Since
$\{k_1,\dotsc,k_\ell\}$ is linearly independent, it follows that
$$\lambda_1=\dotsb=\lambda_\ell=0 $$ Hence
$\{I_\beta(k_1),\dotsc,I_\beta(k_\ell)\}$ is linearly independent.
$\Box$
These two steps combine to prove that $\nullity(L_A)=\ell=\nullity(T)$. That $\DeclareMathOperator{rank}{rank}\rank(L_A)=\rank(T)$ then follows from the rank-nullity theorem.
