Let \(A\) be a real symmetric positive definite \(n\times n\) matrix.
Show there is an upper triangular matrix \(R\) with positive diagonal entries such that \(R^TR=A\).
Show that \(RR^T\) has the same eigenvalues as \(A\).
Solution
1. There is a real symmetric positive definite matrix \(B\) such that\[ B^2=A. \]
By the spectral theorem, there is an orthogonal matrix \(U\) and positive numbers \[ \lambda_1,\ldots,\lambda_n \] such that \[ A = U \operatorname{diag}(\lambda_1,\ldots,\lambda_n) U^T. \] Define \[ B \coloneqq U \operatorname{diag}(\sqrt{\lambda_1},\ldots,\sqrt{\lambda_n}) U^T. \] Then \(B\) is real and symmetric, all its eigenvalues are positive, and \[ B^2 = U \operatorname{diag}(\lambda_1,\ldots,\lambda_n) U^T = A. \]
2. The matrix \(B\) has a QR factorization\[ B=QR \]in which \(Q\) is orthogonal and \(R\) is upper triangular with positivediagonal entries.
The matrix \(B\) is invertible because its eigenvalues in step 1 are positive. Applying Gram–Schmidt to its linearly independent columns gives a QR factorization with nonzero diagonal entries in \(R\). Multiplying a column of \(Q\) and the corresponding row of \(R\) by \(-1\) when necessary makes every diagonal entry of \(R\) positive without changing the product \(QR\).
3. The matrix \(R\) from step 2 satisfies\[ \boxed{R^TR=A}. \]
Since \(B=QR\) and \(Q^TQ=I\), \[ \begin{aligned} R^TR &= R^TQ^TQR\\ &= B^TB. \end{aligned} \] By step 1, \(B^T=B\) and \(B^2=A\). Hence \[ R^TR=B^2=A. \] This proves part (a).
4. The matrix \(R\) is invertible and\[ R^T=AR^{-1}. \]
By step 2, \(R\) is triangular with positive, hence nonzero, diagonal entries, so it is invertible. From step 3, \[ R^TR=A. \] Multiplying on the right by \(R^{-1}\) gives \[ R^T=AR^{-1}. \]
5. One has\[ RR^T=RAR^{-1}. \]
Multiply the identity in step 4 on the left by \(R\): \[ RR^T = R(AR^{-1}) = RAR^{-1}. \]
6. The matrices \(RR^T\) and \(A\) have the same eigenvalues, with thesame algebraic multiplicities.
Step 5 says exactly that \(RR^T\) is similar to \(A\). Similar matrices have the same characteristic polynomial and therefore the same eigenvalues with the same algebraic multiplicities. This proves part (b).
7. Q.E.D.
Step 3 proves part (a), and step 6 proves part (b).