- #1

mathmari

Gold Member

MHB

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I saw the below sentence in some notes:

Let $A\in \mathbb{R}^{n\times n}$ be a not necessarily symmetric, strictly positive definite matrix, $x^TAx>0$, $x\neq 0$ und $Q\in \mathbb{R}^{n\times n}$ an orthogonal matrix, then $B=Q^TAQ$ has a LU decomposition.

I want to understand this implication, so I thought that this could hold for the following reason:

Since the matrix $A$ is strictly positive definite matrix, we know that the determinant is always positiv, i.e. not zero.

The determinant of $Q$ and $Q^T$ can neither be zero.

Therefore we get that the determinant of $B$ is not zero.

In this case a LU decomposition exists if the leading principal minors are not $0$, right? :unsure: