Computing $ \mathbf A^{-1/2}$, where $ \mathbf A$ is a Diagonal Matrix. I have the following matrix :
$$ \mathbf A =\begin{bmatrix}
        100 & 0 \\
        0 & 1 \\
     \end{bmatrix}$$
I have to compute $ \mathbf A^{-1/2}$.
So I need spectral decomposition, $$ \mathbf A =  \mathbf P \mathbf \Lambda\mathbf P',$$
$\mathbf P$ be a matrix with normalized eigenvectors and $\mathbf \Lambda$ is a diagonal matrix with diagonal elements be eigenvalues. 
Eigenvalues of $ \mathbf A$ is $100$ and $1$. 
But I stumbled to calculate eigenvector. 
The characteristic equation is :
$$ \begin{bmatrix}
        100 & 0 \\
        0 & 1 \\
     \end{bmatrix} \begin{bmatrix}
        x_1 \\
        x_2 \\
     \end{bmatrix}= 100\begin{bmatrix}
        x_1 \\
        x_2 \\
     \end{bmatrix}$$
$$\Rightarrow 100x_1 = 100x_1$$
$$x_2 = 100 x_2$$ 
How is $x_2 = 100 x_2$ possible ?
And is there a simpler way to calculate any power of a diagonal matrix , for example, $ \mathbf A^{-1/2}$ ?
 A: For diagonal matrices, the normalized eigenvectors are always standard basis vectors, i.e. vectors of all zeros except a single 1 in a particular coordinate, e.g., ${\bf{e}}_1=(1,0)$ and ${\bf{e}}_2=(0,1)$ in your example. For eigenvalue 100, the corresponding eigenvector is $(1,0)$ which means that $x_2=0$ (note that solves $x_2=100x_2$). 
The power $k$ of a diagonal matrix $$
D=\left\|\begin{array}{ccc}d_1 &\cdots &0\\
\vdots & &\vdots\\
0 &\cdots &d_n\end{array}\right\|
$$ is found as
$$
D=\left\|\begin{array}{ccc}d^k_1 &\cdots &0\\
\vdots & &\vdots\\
0 &\cdots &d^k_n\end{array}\right\|
$$
A: $A=SDS^{-1}$, and more generally $A^k=SD^kS^{-1}$ where D is the diagonal matrix composed from eigenvalues, S is the matrix of eigenvectors, and $S^{-1}$ is the inverse of S.
D is $
\begin{bmatrix}
        100&0\\0&1
        \end{bmatrix}$as you have already found.
To find eigenvectors, you need to follow this format  $\left[
\begin{array}{cc|c}
        100-\lambda&0&0\\0&1-\lambda&0
        \end{array}\right]$
When $\lambda=100$, $\left[\begin{array}{cc|c}
        0&0&0\\0&-99&0
        \end{array}\right]$, $v_1=\begin{bmatrix}1\\0\end{bmatrix}$
When $\lambda=1$, $\left[\begin{array}{cc|c}
        100&0&0\\0&0&0
        \end{array}\right]$, $v_2=\begin{bmatrix}0\\1\end{bmatrix}$
Thus $S=\begin{bmatrix}
        1&0\\0&1
        \end{bmatrix}=S^{-1}$
$A^{1/2}=SD^{1/2}S^{-1}=\begin{bmatrix}
        1&0\\0&1
        \end{bmatrix}\begin{bmatrix}
        100^{1/2}&0\\0&1^{1/2}
        \end{bmatrix}\begin{bmatrix}
        1&0\\0&1
        \end{bmatrix}=\begin{bmatrix}
        10&0\\0&1
        \end{bmatrix}$
And $A^{-1/2}=(A^{1/2})^{-1}$, 
$A^{-1/2}$ is the inverse of $A^{1/2}$, which is $\begin{bmatrix}
        \frac{1}{10}&0\\0&1
        \end{bmatrix}$
A: Your matrix is in Jordan Normal Form (diag) thus:
$$f(A) = \begin{bmatrix}f(a_{11})& \\&f(a_{22})\end{bmatrix}$$
So yes you can use:
$$A^{-1/2} = \begin{bmatrix}a_{11}^{-1/2}& \\&a_{22}^{-1/2}\end{bmatrix}$$
