Assume we have 2 positive definite matrices A and B . Show that there exists a non-singular matrix S such that –
SAS’ = I
SBS’ = L
Here I is the Identity matrix and L is a diagonal matrix. S’ is the transpose of S.
Since $B$ is SPD, there is a $C$ such that $B=CC^T$. Consider the matrix $G=C^{-1}AC^{-T}$. Note that $G$ is SPD too. Let $G=UDU^T$ be the spectral decomposition of $G$ with $U$ orthogonal and $D$ diagonal. Then you can verify, that with $S=D^{1/2}U^TC^{-1}$, you get what you need.
This is Theorem 7.6.4 on Pg 465 in Horn & Johnson’s Matrix Analysis 1985. The proof is essentially what Algebraic Pavel suggested. Here is another dig at the same:
Since $B$ is positive-definite, $B = KK’$ for some invertible $K$ ($K’$ is the transpose of $K$).
Define $G = K^{-1}A(K^{-1})’$. We can see that $G$ is symmetric since
$$G’ = (K^{-1}A(K^{-1})’)’$$
$${}= ((K^{-1})’)'(K^{-1}A)’$$
$${}= K^{-1}A'(K^{-1})’$$
$${}= K^{-1}A(K^{-1})’$$
$${} = G$$
Thus, diagonalize $G$ into $UDU’$. Now define $S = D^{-1/2} U’ K^{-1}$. Then, $S’ = (K^{-1})’U D^{-1/2}$
Claim that $S$ satisfies the required conditions:
$SAS’ = D^{-1/2} U’ K^{-1} A (K^{-1})’U D^{-1/2} = D^{-1/2} U’ G U D^{-1/2} = D^{-1/2} D D^{-1/2} = I$
$SBS’ = D^{-1/2} U’ K^{-1} B (K^{-1})’U D^{-1/2} = D^{-1/2} U’ K^{-1} K K’ (K^{-1})’U D^{-1/2} = D^{-1/2} U’ U D^{-1/2} = D^{-1/2} D^{-1/2} = D^{-1}$. Define $L$ to be equal to $D^{-1}$.
Thus, we have constructed a non-singular $S$ such that $SAS’ = I$ and $SBS’ = L$ as needed.
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