# Decomposition of matrix

1. Nov 30, 2013

### 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 $\mathbf R$ ?

2. Nov 30, 2013

### ShayanJ

For two matrices A and B we have:

$(AB)_{ij}=\sum_k A_{ik}B_{kj}$

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

$(AA^T)_{ij}=\sum_k A_{ik}A_{jk}$

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.

3. Nov 30, 2013

### Jano L.

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

4. Nov 30, 2013

### D H

Staff Emeritus
Since your matrix C is a symmetric real-valued square matrix, it can be decomposed as C=UƩUT, where U is a real unitary matrix and Ʃ is a diagonal matrix. If C is positive semidefinite, then R=UƩ1/2UT is a symmetric real-valued matrix such that RR=C. Since it's symmetric, one also has RRT=C since RT=R.