Why eigenvectors basis then transformation matrix is $\Lambda$? 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.
 A: The matrix of a linear transformation with respect to a basis has as columns the coordinates of the images of the basis elements under the transformation. If you have a basis of eigenvectors then the corresponding matrix will be diagonal because $A v_i = \lambda_i v_i$, that is, there is only one non-zero coordinate (more precisely, at most one, because $\lambda_i$ may be $0$).
A: Let me back up a bit. Remember what the point is of the "coordinate matrix of a linear transformation with respect to a basis."
If $T\colon\mathbf{V}\to\mathbf{V}$ is a linear transformation, and $\beta=[\mathbf{v}_1,\ldots,\mathbf{v}_n]$ is a basis for $T$, then the values of $T$ at every vector in $\mathbf{V}$ is completely determined by its values on $\mathbf{v}_1,\ldots,\mathbf{v}_n$. Why? Because given any $\mathbf{x}\in\mathbf{V}$, we know there exist (unique) scalars $c_1,\ldots,c_n$ such that $\mathbf{x}=c_1\mathbf{v}_1+\cdots+c_n\mathbf{v}_n$, so
$$T(\mathbf{x}) = T(c_1\mathbf{v}_1+\cdots+c_n\mathbf{v}_n) = c_1T(\mathbf{v}_1)+\cdots+c_nT(\mathbf{v}_n).$$
So if you know what $T(\mathbf{v}_1),\ldots,T(\mathbf{v}_n)$ are, you know that $T(\mathbf{x})$ is for all $\mathbf{x}\in\mathbf{V}$.
Now, since $\beta$ is a basis, each $T(\mathbf{v}_i)$ can be expressed as a linear combination of the $\mathbf{v}_i$ in a unique way; that is, there are scalars $a_{11}, a_{12},\ldots,a_{nn}$ such that
$$\begin{align*}
T(\mathbf{v}_1) &= a_{11}\mathbf{v}_1 + a_{21}\mathbf{v}_2 + \cdots +a_{n1}\mathbf{v}_n\\
T(\mathbf{v}_2) &= a_{12}\mathbf{v}_1 + a_{22}\mathbf{v}_2 + \cdots + a_{n2}\mathbf{v}_n\\
&\vdots\\
T(\mathbf{v}_n) &= a_{1n}\mathbf{v}_1 + a_{2n}\mathbf{v}_2 + \cdots + a_{nn}\mathbf{v}_n.
\end{align*}$$
Then
$$\begin{align*}
G(\mathbf{x}) &= c_1T(\mathbf{v}_1 + \cdots + c_nT(\mathbf{v}_n)\\
&= c_1\left(a_{11}\mathbf{v}_1 + a_{21}\mathbf{v}_2 + \cdots +a_{n1}\mathbf{v}_n\right)\\
&\qquad+c_2\left(a_{12}\mathbf{v}_1 + a_{22}\mathbf{v}_2 + \cdots + a_{n2}\mathbf{v}_n\right)\\
&\qquad+\cdots+c_n\left(a_{1n}\mathbf{v}_1 + a_{2n}\mathbf{v}_2 + \cdots + a_{nn}\mathbf{v}_n\right)\\
&= \Bigl(c_1a_{11} + c_2a_{12} + \cdots + c_na_{1n}\Bigr)\mathbf{v}_1\\
&\qquad +\Bigl(c_1a_{21} + c_2a_{22} + \cdots + c_na_{2n}\Bigr)\mathbf{v}_2\\
&\qquad+\cdots+ \Bigl(c_1a_{n1}+c_2a_{n2}+\cdots+c_na_{nn}\Bigr)\mathbf{v}_n\\
&= \left(\begin{array}{cccc}
\mathbf{v}_1 & \mathbf{v}_2 & \cdots & \mathbf{v}_n\end{array}\right) \left(\begin{array}{cccc}
a_{11} & a_{12} & \ldots & a_{1n}\\
a_{21} & a_{22} & \ldots & a_{nn}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \ldots & a_{nn}
\end{array}\right) \left(\begin{array}{c}c_1\\c_2\\\vdots\\c_n\end{array}\right).
\end{align*}$$
The column vector
$$\left(\begin{array}{c}c_1\\c_2\\\vdots\\c_n\end{array}\right)$$
represents the vector $\mathbf{x}$; it is the "coordinate vector of $\mathbf{x}$ with respect to $\beta$". The matrix 
$$\left(\begin{array}{ccc}
a_{11} & \cdots & a_{1n}\\
\vdots & \ddots & \vdots\\
a_{n1} & \cdots & a_{nn}\end{array}\right)$$
is the matrix that "codifies" the images of $\mathbf{v}_i$ under $T$ in terms of $\beta$: the $i$th column is the coordinate vector of $T(\mathbf{v}_i)$ relative to $\beta$. This is the "coordinate matrix of $T$ relative to $\beta$."
What happens if $\mathbf{v}_1$ is an eigenvector corresponding to $\lambda_1$? Then
$$T(\mathbf{v}_1) = \lambda_1\mathbf{v}_1 = \lambda_1\mathbf{v}_1 + 0\mathbf{v}_2 + 0\mathbf{v}_3 + \cdots + 0\mathbf{v}_n.$$
That is, the first column of the coordinate matrix is just
$$\begin{array}{c}
\lambda_1 \\
0 \\
0\\
\vdots \\
0\\
0\end{array}$$
What if $\mathbf{v}_2$ is an eigenvector corresponding to $\lambda_2$? Then $T(\mathbf{v}_2)=\lambda_2\mathbf{v}_2$. To write it as a linear combination of $\beta$, we just put $\lambda_2$ for the coefficient of $\mathbf{v}_2$, and $0$s for every other coefficient:
$$T(\mathbf{v}_2) = \lambda_2\mathbf{v}_2 = 0\mathbf{v}_1 + \lambda_2\mathbf{v}_2 + 0\mathbf{v}_3 + \cdots + 0\mathbf{v}_n.$$
So the second column of the coordinate matrix is just
$$\begin{array}{c}0\\\lambda_2\\0\\\vdots\\0\\0\end{array}$$
And so on: if $\mathbf{v}_i$ is an eigenvector corresponding to $\lambda_i$, then
$$T(\mathbf{v}_i) = \lambda_i\mathbf{v}_i = 0\mathbf{v}_1+0\mathbf{v}_2+\cdots+0\mathbf{v}_{i-1} + \lambda_i\mathbf{v}_i + 0\mathbf{v}_{i+1} + \cdots+0\mathbf{v}_n,$$
so the $i$th column of the coordinate matrix will have $0$s everywhere except in the $i$th row, where it will have $\lambda_i$.
And so, if the entire basis is made up of eigenvectors, then the coordinate matrix will be such that the $i$th column has zeros everywhere except (perhaps) in the $i$th row, where it has the eigenvalue of $\mathbf{v}_i$; that is, we get a diagonal matrix
$$\Lambda = \left(\begin{array}{cccc}
\lambda_1 & 0 & \cdots & 0\\
0 & \lambda_2 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & \lambda_n
\end{array}\right).$$
And that is the coordinate matrix of $T$ with respect to $\beta$.
