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Let ${\bf A} \in {\Bbb R}^{d \times d}$ be a negative definite matrix and let ${\bf B} \in {\Bbb R}^{d \times r}$, where $r<d$. Let ${\bf x} \in {\Bbb R}^{d}$ be defined by ${\bf x} := {\bf B} {\bf w} - {\bf B}^* {\bf w}^*$, where ${\bf w} \in {\Bbb R}^{r}$ and both ${\bf B}^*$ and ${\bf w}^*$ are given. I would like to know if $$ {\bf x}^\top {\bf B} {\bf B}^\top {\bf A} {\bf x}$$ is non-positive.

What I have tried was the following.

  1. to identify definiteness of a positive semidefinite (PSD) matrix times a negative definite (ND) matrix. Doesn't work.

  2. Cholesky decomposition. $-{\bf A} = {\bf L} {\bf L}^\top$ since $-{\bf A}$ is positive definite (PD). Then, ${\bf x}^\top {\bf B} {\bf B}^\top {\bf A} {\bf x} = -{\bf x}^\top {\bf B} {\bf B}^\top {\bf L} {\bf L}^\top x$. I still cannot say anything about the sign.

  3. I have run some random numerical tests. It seems it is always negative.

Leafstar
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    Not much contrxt nor effort shown so far here, I must say. [tour] – Mike Apr 01 '25 at 01:20
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    Do you mean "$\ B\in\mathbb{R}^{d\times\color{red}{k}},k<d\ $", or "$\ B\in\mathbb{R}^{d\times r},\color{red}{r}<d\ $"? As the question currently stands, the inequality $\ k<d\ $ doesn't seem to provide any relevant information because it's the only place where the variable $\ k\ $ appears. – lonza leggiera Apr 01 '25 at 07:59
  • Usually, definiteness is about quadratic forms and the associated symmetric matrices. However, neither is ${\bf x}^\top {\bf B} {\bf B}^\top {\bf A} {\bf x}$ a quadratic form nor is the associated matrix symmetric-looking. In other words, I do not understand your question at all. – Rodrigo de Azevedo Apr 01 '25 at 09:27
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    @Rodrigo de Azevedo Because $\ \mathbf{x^\top BB^\top Ax}\ $ is a $1\times1$ matrix, it's symmetric: $$ \mathbf{x^\top BB^\top Ax}=\mathbf{x^\top ABB^\top x}\ ,$$ and therefore $$\mathbf{x^\top BB^\top Ax}=\mathbf{x^\top}\left(\frac{1}{2}\big(\mathbf{BB^\top A+ABB^\top}\big)\right)\mathbf{x}$$ which looks to me like a quadratic form with symmetric matrix $\ \frac{1}{2}\big(\mathbf{BB^\top A+ABB^\top}\big)\ .$ I'm assuming here that negative definiteness of $\ \mathbf{A}\ $ implies that it's symmetric. – lonza leggiera Apr 01 '25 at 10:33
  • @lonzaleggiera Well, IMHO, it should be written in terms of lower-dimensional $\bf w$, not in terms of $\bf x$. In other words, we have affine terms, that is, we do not have a quadratic form. Sure, taking the symmetric part makes the matrix symmetric. We agree on that. – Rodrigo de Azevedo Apr 01 '25 at 10:41
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    @Roger de Alzevedo Yes, I agree that once you replace $\ \mathbf{x}\ $ with $\ \mathbf{Bw-B^*w}\ ,$ the result isn't a quadratic form in $\ \mathbf{w}\ .$ Neverthelss, I took the OP's question to be whether $\ \mathbf{x^\top BB^\top Ax}\ $ (regarded as a number) is non-positivefor all $\ \mathbf{w}\ ,$ which seems to me to be a perfectly legitimate question. – lonza leggiera Apr 01 '25 at 10:56
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    In case $B=b$ is a column vector, you are asking whether $(x^Tb)(b^TAx)$ is always non-positive. It should be easy to see that the answer is no. Consider e.g. $x=(1,1)^T$, $b=(1,-2)^T$ and $A=diag(-3,-1)$. – user1551 Apr 01 '25 at 11:28
  • @lonzaleggiera Yes, that is a typo. Thank you for pointing this out! – Leafstar Apr 01 '25 at 19:33
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    Thanks for your edits by the way.... – Mike Apr 01 '25 at 23:04

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