Why is it that in a transformation, if the eigenvectors are chosen as the basis vectors, the transformation matrix $A$ would turn out to be that diagonal eigenvalues matrix $\Lambda$ and the transformation becomes $\Lambda \vec{x}=\vec{c}$ ?

What is the reason or cause that the transformation matrix could easily just be the eigenvalues matrix $\Lambda$?

I tried to prove it to show myself but somehow, it doesn't look very right.
Let $\vec{b\_i}$ be the eigenvectors that form the basis for the vector $\vec{v}$ and the $c\_n$ are combination to the eigenvectors...

$\vec{v} = c\_1 \vec{b\_1} + c\_2 \vec{b\_2} + ... + c\_n \vec{b\_n}
$

$
T(\vec{v})= c\_1 T(\vec{b\_1}) + c\_2 T(\vec{b\_2}) + ... + c\_n T(\vec{b\_n})
$

$
T(\vec{v})= c\_1 A\vec{b\_1} + c\_2 A\vec{b\_2} + ... + c\_n A\vec{b\_n}
$

$
T(\vec{v})= c\_1 \lambda\_1 \vec{b\_1} + c\_2 \lambda\_2\vec{b\_2} + ... + c\_n \lambda\_n\vec{b\_n}
$

$
T(\vec{v})= \begin{bmatrix}
 & & \\ 
\vec{b\_1} & \vec{b\_2} &...& \vec{b\_n}\\ 
 & & 
\end{bmatrix}
\begin{bmatrix}
\lambda\_1 & & & \\ 
 & \lambda\_2 & & \\ 
 & & \ddots & \\ 
 & & & \lambda\_n
\end{bmatrix}
\begin{bmatrix}
c\_1 & & & \\ 
 & c\_2 & & \\ 
 & & \ddots & \\ 
 & & & c\_n
\end{bmatrix}
$

The last line is wrong but I'm trying to force out that $\Lambda$ which I couldn't.