Given transformation $T$ in a finite unitary space $V$ and $TT^*=4T-3I$ Prove $T$ is positive definite and find all eigenvalues. (and 1 more question) 
1.Given transformation $T$ in a finite unitary space $V$. and $TT^*=4T-3I$ prove $T$ is positive definite and find all eigenvalues of T
2.Let the Transformation $T: \Bbb R^2 \to \Bbb R^2$ be a projection to $x$ axis with the line $y=\sqrt{3}x$ find $T^*$


For the first part I noticed it was asked before Here but I solved in a different way , I would like to know if it is correct
if all the eigenvalues of $T$ are positive then $T$ is a positive definite transformation (according to the textbook. I do now know what this is actually called)
so given $TT^*=4T-3I \iff 4T=TT^*+3I$ if we do "Hermitian adjoint" on both sides (Sorry if the translation is not correct) we get $(4T)^*=(TT^*+3I)^* \iff 4T^*=T^*T+3I$ from here we get that $T=T^*$
let $\lambda \in \Bbb R$ so the eigenvalue will be $T(v)= \lambda v$ if $0 \not=v \in V$
$TT^*=TT$ according to what we just got , and from $T(v)= \lambda v$ we get $\lambda^2-4\lambda v +3 =0$ which gives us $\lambda_1=3$ and $\lambda_2=1$ which are our eigenvalues and they are different and positive which means our transformation is positive definite.
for the second part -
I don't understand it , I have the short solution in the textbook which says:
for all $(x,y) \in \Bbb R^2$ we get $(x,y)=(x,0)+(0,y)=(x,0)+(\frac{1}{\sqrt{3}}y,y)-(\frac{1}{\sqrt{3}}y,0)=(x-\frac{1}{\sqrt{3}}y,0)+(\frac{1}{\sqrt{3}}y,y)$
therefore $A=[T]_E= \left(
\begin{array}{ccc|c}
  1&-\frac{1}{\sqrt3}\\
  0&0\\
\end{array}
\right)$
from here we get
$A=[T]^*_E= \left(
\begin{array}{ccc|c}
  1&0\\
  -\frac{1}{\sqrt3}&0\\
\end{array}
\right)$
and we get $T^*(x,y)=(x,-\frac{1}{\sqrt3}x)$
appreciate any explanation or different approach to the second part and if anyone can approve my first part
thank you
 A: In the first part, one can prove that $T$ is positive definite before finding the eigenvalues of $T$. (To me, this feels cleaner.) Since $T=\frac14(TT^\ast+3I)$, it is self-adjoint and
$$
\langle Tx,x\rangle
=\frac14\langle TT^\ast x,x\rangle+\frac34\langle x,x\rangle
=\frac14\underbrace{\langle T^\ast x,T^\ast x\rangle}_{\ge0}+\frac34\underbrace{\langle x,x\rangle}_{>0}
>0
$$
whenever $x\ne0$. Therefore $T$ is positive definite.
In the second part, the author means that $Tu=u$ for every vector $u$ on the $x$-axis and $Tv=0$ for every vector $v$ on the line $y=\sqrt{3}x$. In particular, $T(1,0)=(1,0)$ and $T(1,\sqrt{3})=0$. Therefore
$$
T(0,1)=T\left(\frac{(1,\sqrt{3})-(1,0)}{\sqrt{3}}\right)=\frac{1}{\sqrt{3}}\left(T(1,\sqrt{3})-T(1,0)\right)=(-\tfrac{1}{\sqrt{3}},0)
$$
and the matrix representation of $T$ with respect to the standard basis is $\pmatrix{1&-\frac{1}{\sqrt{3}}\\ 0&0}$. Taking transpose, we obtain the matrix of $T^\ast$ (with respect to the dot product on $\mathbb R^2$) and also $T^\ast$ itself.
