Let $T:V\rightarrow V$ be a linear transformation such that $T^2=2T$ 
Let $T:V\rightarrow V$ be a linear transformation such that $T^2=2T$.
Prove the following statements:
  
  
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*If $B=(b_1,...,b_k)$ is a basis of KerT and $C=(c_1,...c_{n-k})$ is a basis of ImT, then $(b_1,...,b_k,c_1,...,c_{n-k})$ is a basis of V. (I assumed dimV=n)
  
*Show that there exists a basis of V, A, such that the representing matrix of T- $[T]_A$ is diagonal and the diagonal entries are all in {0,2}
  

Hi all. I've only just started my Linear Algebra course and we are yet to learn diagonalizing matrices (and shouldn't use this method for this problem).
for 1, I assumed $(b_1,...,b_k,c_1,...,c_{n-k})$ is a linear dependent set and therefore there exists $\beta_j\neq 0$ such that $-\sum_{i=1}^{k}\frac{\alpha_i}{\beta_j}\vec b_i-\sum_{i\neq j}^{n-k}\frac {\beta_i}{\beta_j}\vec c_i = \vec c_j$. I've reached the point where I show that $T(\vec c_j) = \sum_{i\neq j}^{n-k}\frac {\beta_i}{\beta_j}T(\vec c_i)$
How do I continue from here? Would like some help. (and also on question 2)
thanks in advance :)
 A: Let $v\in V$ and you want to write down $$v=a_1b_1+\cdots+a_kb_k+d_1c_1+\cdots+d_{n-k}c_{n-k}$$
for some scalars $a_i,d_i$ in base field of vector space.
As $v\in V$, we have $T^2(v)=2T(v)$ i.e., $T(T(v)-2v)=0$ i.e., $T(v)-2v$ is in kernel of $T$.
So you have 
$$T(v)-2v=a_1b_1+\cdots+a_kb_k$$
for some scalars $a_i$ in base field.
Can you continue from here?
Suggestion : If you are not into Physics completely, it is good to write $v,w$ for elements of vector space than $b_i,c_i$ and not to write those arrows above elements of vector space. 
A: I like to do things as geometrically as possible as long as possible. That means writing things in terms of vectors and linear maps, not in terms of bases, coordinates and matrices.
Take an $x\in V$ such that $x$ is both in the kernel and in the image of $T$. Then there is a $y\in V$ such that $x = Ty$, and we have $$0 = Tx = T^2y = 2Ty = 2x$$so $x = 0$. Since $x$ was arbitrary, we must have $\ker T\cap \operatorname{im}T = \{0\}$. Because the $b_i$ are linearly independent, and the $c_i$ are linearly independent, the entire collection is linearly independent. Since we have a collection of $n$ linearly independent vectors, we have a basis of $V$.
For the second point, note that $Tb_i = 0$ and $Tc_i = 2c_i$. Also, recall that the matrix representation of a linear map is given by the columns being the images of the basis vectors, expressed in that basis. Thus we can simply use the basis consisting of $b_i$'s and $c_i$'s.
A: Indeed, we need to show that the set $B \cup C$ is linearly independent.  It seems that you wanted to prove this by contradiction (i.e. starting by assuming that $B \cup C$ is linearly dependent), but I don't like this approach.
A nicer approach here is to simply show that if $x \in \ker T$ and $y \in \operatorname{Im}(T)$ are such that $x + y = 0$, then it must be that $x = y = 0$ (to put it another way: we'd like to show that $\ker T \cap \operatorname{Im}(T) = \{0\}$). With this shown, we can conclude that if
$$
\sum \beta_i b_i + \sum \gamma_i c_i = 0
$$
then it must be the case that $\sum \beta_i b_i = 0$ and $\sum \gamma_i c_i = 0$, which (by the linear independence of the individual bases) implies that all $\beta_i$ and $\gamma_i$ are $0$.  So,

Claim: $x \in \ker T$ and $y \in \operatorname{Im}(T)$ are such that $x + y = 0$, then it must be that $x = y = 0$

Proof of claim: Let $v$ be such that $y = Tv$. Note that $x + y = 0$.  Thus, $T(x + y) = T(0) = 0$.  So, we have
$$
0 = T(x + y) = Tx + Ty = 0 + T(Tv) = T^2 v = 2Tv = 2y
$$
So, $2y = 0$, which means $y = 0$.  Since $x + y = 0$, conclude that $x = 0$ as well. $\square$
Next, we need to show that every vector in $v \in V$ can be written in the form $v = x + y$ where $x \in \ker(T)$ and $y \in \operatorname{Im}(T)$.  There are several approaches that work here, but there are two I like: once is to simply argue that $\ker(T) \cap \operatorname{Im}(T) = \{0\}$, which means that $\ker(T) + \operatorname{Im}(T)$ is necessarily an $n$-dimensional subspace of $V$.  The second approach is to note that every $v \in V$ can be decomposed as
$$
v = \underbrace{\left(v - \frac 12 Tv\right)}_x + \underbrace{\frac 12 Tv}_{y}
$$

Part 2 is easy if you understand what it means to write the matrix of a transformation with respect to a particular basis.  In particular, you should argue that the matrix of $T$ with respect to $A = (b_1,...,b_k,c_1,...,c_{n-k})$ is of the desired diagonal form.
A: For the first item you need to prove that Ker$T$ $\cap$ Im$T=\{0\}$. Let $\{b_1,\dots, b_k\}$ be a basis for Ker$T$, and let $v\in$ Im$T$. Suppose $v\in$Ker$T$ $\cap$ Im$T$, then $v=a_1b_1+\cdots+a_kb_k$, Hence 
$$T(v)=\sum_{i=1}^k a_i T(b_i)=0$$
On the other hand, since $T^2=2T$ and $v\in$ Im$T$, there exists $u\in V$ such that $T(u)=v$, hence $T^2(u)=T(v)=0$ and $T^2(u)=2T(u)$, that is $T(T(u)-2u)=0$. This implies that $T(v-2u)=T(v)-2T(u)=0$, hence $2v=0$, so $v=0$.
For the second point, the matrix associate to the basis $\{b_1,...,b_k,c_1,..,c_{n-k}\}$ is diagonal, because $T(b_i)=0$ and $T(c_j)=2c_j$.
