Scaling a matrix to make its eigenvalues fall within a certain interval Suppose I have a diagonalizable matrix $M$ which has all its eigenvalues between $a$ and $b$. Is it possible to scale $M$ to $M_S$ such that all the eigenvalues of $M_s$ lie in the interval $[-1,1]$?  
One method I came across:
Scale such that
$$ 
M_s=\frac{M-(b+a)/2}{(b-a)/2}.
$$
But, this is not working. Does anyone know anything better?
 A: I found out that the following transformation works,
$$Ms=\frac{M−((b+a)/2) I}{(b−a)/2}$$ where I is a Identity matrix.
$M_s$ also has the same eigenvectors as the original matrix $M$.
A: So the added constraint to your problem is that eigenvalues $a$ and $b$ must be mapped to $-1$ and $1$.
One possible solution is what follows.
If T is the diagonalizing matrix of M, then
$M=TDT^{-1}$, where $D=\begin{bmatrix}
a&0&0&\dots\\
0&b&0&\dots\\
0&0&c&\dots\\
\vdots&\vdots&\vdots&\ddots
\end{bmatrix}
$
The rescaling transformation for a diagonal D matrix that maps eigenvalues $a$ and $b$ to $-1$ and $1$ would be $R_D$ such that the following holds (let's call $D_1$ the output of such a rescaling):
$D_1 = R_DD$, where $R_D = \begin{bmatrix}
-1/a&0&0&\dots\\
0&1/b&0&\dots\\
0&0&1&\dots\\
\vdots&\vdots&\vdots&\ddots
\end{bmatrix}$
Now let's call $M_1$ the rescaling of M such that eigenvalues $a$ and $b$ are mapped to $-1$ and $1$. It must be:
$M_1 = TD_1T^{-1} = TR_DDT^{-1} = TR_DT^{-1}TDT^{-1} = R_MM$.
So the rescaling transformation you are looking for is given by the multiplication on the left by the matrix:
$R_M = TR_DT^{-1}$
A: $M_S=sign(a+b)/max(|a|, |b|) M$
