the weights of the induced $\mathbb C^*$-action on the tangent space to $\mathbb P^m$ at fixed points Let $\mathbb C^*$ act on $\mathbb C^{m+1}$ with generic weights $\lambda_0,\cdots, \lambda_r$, and this action can be naturally regarded as an action on $\mathbb P^m$.
Let $p_i\in \mathbb P^m$ ($i=0,1,\cdots m$) be the points determined by the basis vectors, i.e. the fixed points of the $\mathbb C^*$-action on $\mathbb P^m$. 
Now at $p_i$ the action on $\mathbb P^m$ will also induce an action on the tangent space $T_{p_i}\mathbb P^m$.
Question 1 Show that this induced action on the tangent space at $p_i$ has weights $\lambda_i-\lambda_j$ for $j\neq i$.
I am confused about some basic points. For example, I don't know how to interpret and understand the weights of the action, and what is the precise definition of the weights in this case? Also, what is the induced action on the tangent spaces?
The following is my attempt. Assume $p_0=[1,0,\cdots, 0]\in \mathbb P^m$. Consider $U_0=\{[x_0,\cdots,x_m]\in \mathbb P^m | x_0\neq 0\}$, and the corresponding local chart $\psi_0 : U_0 \to \mathbb C^m, [x_0,\cdots,x_m]\mapsto (\frac{x_1}{x_0}, \cdots, \frac{x_m}{x_0})$. Let $\rho:\mathbb C^*\to \text{Aut}(\mathbb P^m), t \mapsto \rho_t$ be the action. For example, if we assume $\rho_t[x_0,\cdots, x_m]=[t^{a_0}x_0, \cdots, t^{a_m}x_m]$, then in the above local chart, $\rho_t(y_1,\cdots, y_m)=(t^{a_1-a_0}y_1, \cdots t^{a_m-a_0}y_m)$.
Question 2 What if we replace the $\mathbb C^*$ action by a $\mathbb T:=\mathbb C^*\times \cdots \times \mathbb C^*= (\mathbb C^{*})^{m+1}$ action such as $\rho_{(t_0, \cdots, t_m)}[x_0,\cdots, x_m]=[t_0x_0, \cdots, t_mx_m]$? In this case, how to understand the weight of this action?
PS: There is a relevant question.
 A: It seems that you essentially have all the ingredients in place except for a couple of definitions.
Weights of $\mathbb{C}^*$: An action of $\mathbb C^*$ on $\mathbb{C}^n$ is defined by a group homomorphism: $\rho:\mathbb{C}^* \to \mathrm{Aut}(\mathbb{C}^n) = GL_n$. Since $\mathbb{C}^*$ is an abelian group, its irreducible representations are all one dimensional. So let's consider the representation $\rho: \mathbb{C}^* \to GL_1 = \mathbb{C}^*$, i.e., for any $z \in \mathbb{C}^*$, $\rho(z)$ is a nonzero complex number. We can use a Laurent series around $z=0$ to write $\rho(z)$:
$$ \rho(z) = \sum_{k=-\infty}^\infty a_k z^k\,. \tag{L}$$
Since $\rho$ is a group homomorphism, it has to satisfy:
$$\rho(z) \rho(w) = \rho(wz)\,.$$
If we use the expansion (L) on both sides of the above equation, we find:
$$ \sum_{k=-\infty}^\infty \sum_{l=-\infty}^\infty a_k a_l w^k z^l = \sum_{k=-\infty}^\infty a_k (wz)^k\,. $$
Since this has to be satisfied for all $w, z \in \mathbb{C}^*$, we must impose that $a_k$ is zero for all but one integer value of $k$ and if $a_k \ne 0$ then we must have $a_k=1$, i.e., irreducible representations of $\mathbb{C}^*$ can be parametrized by integers, and they are all of the form:
$$\rho_k(z) = z^k\,, \qquad k \in \mathbb Z\,.$$
This number $k$ is called the weight of the representation.
So, when it's said that the weight of the $\mathbb{C}^*$-action on $\mathbb{C}^{m+1}$ is $(\lambda_0, \cdots, \lambda_m)$, it's implied that the action is described as follows: for any $z \in \mathbb{C}^*$ and $(v_0, \cdots, v_m) \in \mathbb C^{m+1}$:
$$\rho(z): (v_0, \cdots, v_m) \mapsto (z^{\lambda_0} v_0, \cdots, z^{\lambda_m} v_m)\,.$$
Induced action: I am not going to give a formal definition of the induced action, rather I will give a "computational" definition. But if you haven't seen it already in texts then I suggest that you read it from a proper text (I am familiar with the one by Nakahara (Geometry, Topology and Physics) but surely there are many options).
Suppose we are in a local chart on some (locally) complex manifold $M$ with a group action. So we have local coordinates $\mathbf{x}=(x^1, \cdots, x^d)$ and the group action is given as a coordinate transformation:
$$g:(x^1, \cdots, x^d) \mapsto (g^1(\mathbf x), \cdots, g^d(\mathbf x)) =: g(\mathbf x). \tag{G}$$
At any point $\mathbf p = (p^1, \cdots, p^d) $ in this local chart, the tangent space $T_\mathbf{p} M$ is spanned by the partial derivatives at that point:
$$ T_\mathbf{p}M = \mathrm{Span}_\mathbb{C} \left( \left.\frac{\partial}{\partial x^1}\right|_{\mathbf{x} = \mathbf{p}}\,, \cdots, \left.\frac{\partial}{\partial x^d}\right|_{\mathbf{x} = \mathbf{p}} \right)\,. $$
Rcall from chain rule that the transformation (G) induces the following transformation for partial derivative:
$$ \left.\frac{\partial}{\partial x^i}\right|_{\mathbf{x} = \mathbf{p}} \mapsto \sum_{j=1}^d \left.\frac{\partial g^j(\mathbf x)}{\partial x^i}\right|_{\mathbf{x} = \mathbf{p}} \left.\frac{\partial}{\partial x^j}\right|_{\mathbf{x} = \mathbf{g(p)}}\,. \tag{T}$$
Noting that $\left.\frac{\partial}{\partial x^i}\right|_{\mathbf{x} = \mathbf{g(p)}}$ are basis vectors of $T_{g(\mathbf p)}M$, we thus get a map (which I denote here by $g_*$):
$$ g_*(\mathbf p): T_\mathbf{p}M \to T_{g(\mathbf p)}M\,.$$
Let us define the following matrix:
$$J(\mathbf p)_i^j := \left.\frac{\partial g^j(\mathbf x)}{\partial x^i}\right|_{\mathbf{x} = \mathbf{p}}. \tag{J}$$
Now, if we want to write down the matrix form of the linear transformation (T) then we should write the new basis vectors $\left\{\left.\frac{\partial}{\partial x^i}\right|_{\mathbf{x} = \mathbf{g(p)}}\right\}$ in terms of the old basis vectors $\left\{\left.\frac{\partial}{\partial x^i}\right|_{\mathbf{x} = \mathbf{p}}\right\}$, inverting the chain rule we get:
$$ \left. \frac{\partial}{\partial x^i} \right|_{\mathbf x = \mathbf g(p)} = \sum_{j=1}^d \left(J(\mathbf p)^{-1}\right)_i^j \left. \frac{\partial}{\partial x^j} \right|_{\mathbf x = \mathbf p}\,. $$
Therefore, $J(\mathbf p)^{-1}$ is the matrix form of the induced action on the tangent spaces, given an action on a manifold. If $\mathbf p$ happens to be a fixed point of the group action then we have a linear representation of the group $g_*(\mathbf p):T_\mathbf{p}M \to T_\mathbf{p} M$.
Question 1: The weights of the $\mathbb C^*$ action (let's denote the action by $\rho^{\mathbb C^{m+1}}$) on $\mathbb C^{m+1}$ is $(\lambda_0, \cdots, \lambda_m)$, which means, for any $z \in \mathbb{C}^*$ and $(x_0, \cdots, x_m) \in \mathbb C^{m+1}$ we have:
$$ \rho^{\mathbb C^{m+1}}(z):(x_0, \cdots, x_m) \mapsto (z^{\lambda_0} x_0, \cdots, z^{\lambda_m} x_m)\,.$$
Then your own computation shows that, on the local chart $U_0$ of $\mathbb P^m$, the action of $\mathbb C^*$ (let's denote it by $\rho^{\mathbb{P}^m}$) is given by:
$$\rho^{\mathbb P^m}(z):(y_1, \cdots, y_m) \mapsto (z^{\lambda_1-\lambda_0} y_1, \cdots, z^{\lambda_m-\lambda_0} y_m)\,.$$
This is the analogue of the coordinate transformation (G) and therefore  we have $\left(\rho^{\mathbb P^m}(z)\right)^i (\mathbf y) = z^{\lambda_i-\lambda_0} y_i$ and the inverse of the matrix $J$ (defined in (J)) is given by:
$$J^{-1}(\mathbf y) = \mathrm{diag}(z^{\lambda_0 -\lambda_1}, \cdots, z^{\lambda_0 -\lambda_m})\,.$$
The induced action on tangent spaces is just multiplication by the above matrix, so we see that the weights are given by $(\lambda_0-\lambda_1, \cdots, \lambda_0 - \lambda_m)$. (Note that it's a representation only at the fixed point $\mathbf y = 0$, otherwise it's a linear map between two different vector spaces.)
Question 2: If you have the action of $(\mathbb C^*)^m$ on $\mathbb C^n$ then for each factor of $\mathbb C$ you will need an $m$-tuple of weights, let's say if we denote the $m$-tuple of weights corresponding to the $a$'th factor of $\mathbb C$ by $(w^a_1, \cdots, w^a_m)$, then we will write the action of $(t_1, \cdots, t_m) \in (\mathbb C^*)^m$ on $(x_1, \cdots, x_n) \in \mathbb C^n$ as:
$$ \rho(t_1, \cdots, t_m): (x_1, \cdots, x_n) \mapsto (t_1^{w^1_1} \cdots t_m^{w^1_m} x_1, \cdots, t_1^{w^n_1} \cdots t_m^{w^n_m} x_n)\,. $$
