Is the Matrix R Necessarily Lower Triangular in Matrix Decomposition?

[Jano L.]

I would like to learn bit more about matrices and their decomposition. Let $\mathbf C$ be symmetric real-valued square matrix. Let $\mathbf R$ be such that

$$
\mathbf R\mathbf R^T = \mathbf C.
$$

Is the matrix $\mathbf R$ necessarily lower triangular (I suspect not)?

Cholesky decomposition leads to $\mathbf R$ that is lower triangular. Is there some other method of calculationg $\mathbf R$ ?

 

[ShayanJ]

For two matrices A and B we have:

<br /> (AB)_{ij}=\sum_k A_{ik}B_{kj}<br />

Now we have B=A^T \Rightarrow B_{kj}=A_{jk} so the above equation becomes:

<br /> (AA^T)_{ij}=\sum_k A_{ik}A_{jk}<br />

Now it is obvious that (AA^T)_{ij}=(AA^T)_{ji}

So the product of any square matrix with its transpose is a symmetric matrix.

 

[Jano L.]

OK, from your example now I see $\mathbf R$ is not necessarily lower triangular. Thanks.

 

[D H]

Cholesky decomposition leads to $\mathbf R$ that is lower triangular. Is there some other method of calculationg $\mathbf R$ ?

Since your matrix C is a symmetric real-valued square matrix, it can be decomposed as C=UƩU^T, where U is a real unitary matrix and Ʃ is a diagonal matrix. If C is positive semidefinite, then R=UƩ^1/2U^T is a symmetric real-valued matrix such that RR=C. Since it's symmetric, one also has RR^T=C since R^T=R.