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Suppose I have a matrix:

$\mathbf{Q} = \mathbf{Z}^T\mathbf{B}\mathbf{Z}$

where $\mathbf{Z} \in \mathbb{R}^{n\times m},\mathbf{B}\in \mathbb{R}^{n\times n}, m \leq n$

Can I say that if $\mathbf{Q}$ is positive definite, then $\mathbf{B}$ is also positive definite?

Is this conclusion possibly only valid in certain special cases such as if $\mathbf{Z}$ were a 1D vector, or if $m=n$?

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  • $\begingroup$ What do you mean by positive definite ? Are $Q$ and $B$ symmetric (as your title suggets) ? $\endgroup$ – M. Dus Aug 14 '18 at 18:37
  • $\begingroup$ I mean that for $Q$ the eigenvalues are all positive, for it to be positive definite. You can assume symmetry for $Q$ and $B$ if you require it for any conclusions you form. But if $Q$ has positive eig values, does it imply $B$ also has positive eig values for example? $\endgroup$ – pche8701 Aug 14 '18 at 19:04
  • $\begingroup$ What do you mean by the eigenvalues are all positive ? Are you assuming that $Q$ is diagonalizable? My answer below is for symmetric matrices. $\endgroup$ – M. Dus Aug 14 '18 at 19:09
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I assume that your matrices are symmetric, as suggested by your title, so that definite positiveness of a matrix $A$ is equivalent to the property that for all nonzero vector $X$, $X^TAX>0$

Simple construction of counter-example: take $B=\begin{pmatrix}1 & 0 & 0 \\ 0& 1& 0\\ 0&0&-1 \end{pmatrix}$ and $Z=\begin{pmatrix}1&0\\0& 1\\0&0\end{pmatrix}$. Then, you have $$Z^TBZ=\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 0& 1& 0\\ 0&0&-1\end{pmatrix}\begin{pmatrix}1&0\\0& 1\\0&0\end{pmatrix}=\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}\begin{pmatrix}1&0\\0& 1\\0&0\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$

Changing a bit this example, you can have any counter-example you want for any $n$ and any $m\leq n$.

Actually, your claim true if and only if $Z$ is surjective. Indeed, if $Z$ is surjective, assume that $B$ has a nonpositive eigenvalue $\lambda$ with eigenvector $Y$ and let $X\in \mathbb{R}^m$ such that $ZX=Y$. Then, $X^TQX=\lambda \|Y\|\leq 0$, a contradiction since $X\neq 0$.

Now, if $Z$ is not surjective, it's false. Choose $B$ which is definite positive in the range of $Z$ and definite negative in the orthogonal complement. Then for all $X\neq 0$, $X^TQX>0$ but $B$ is not definite positive.

By the way, note that the opposite claim (if $B$ is definite positive, then so is $Q$) is true if and only if $Z$ is one-to-one (exercise).

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  • $\begingroup$ Could you please clarify just a little bit more about how the surjectivity assumption comes into play (in the first paragraph)? $\endgroup$ – pche8701 Aug 15 '18 at 6:10
  • $\begingroup$ I'm also a little confused because in one line you state that $Z$ is surjective, and in other propose $Z$ to be one-to-one. $\endgroup$ – pche8701 Aug 15 '18 at 6:56
  • $\begingroup$ Okay I tried to clarify. I changed the order and first included my counter-example. I then proved the statement about surjectivity. Now, the end about one-to-one is not directly related to your question, it's the opposite statement, I clarified it as well. Finally, I use symmetry for the equivalence between positive eigenvalues and the fact that $X^TAX>0$ (this only holds for symmetric matrices). Anyway, if you are interrested in quadratic forms, your matrices are indeed symmetric. $\endgroup$ – M. Dus Aug 15 '18 at 7:36
  • $\begingroup$ Thanks so much. The changes have helped a lot. Just one final clarification in the second last paragraph if you can for me. Could you briefly explain why $B$ is guaranteed to be not definite positive? $\endgroup$ – pche8701 Aug 15 '18 at 17:01

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