Show that the matrix D is invertible

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Discussion Overview

The discussion revolves around the conditions under which the matrix expression $D^TCD$ is positive definite, specifically exploring the equivalence between $D^TCD$ being positive definite and the invertibility of the matrix $D$. The context includes theoretical reasoning and mathematical justification.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose that if $D^TCD$ is positive definite, then for all non-zero vectors $x$, the expression $x^T D^TCD x > 0$ must hold.
  • Others argue that if $D$ is not invertible, there exists a vector $x$ such that $Dx = 0$, leading to $x^T D^TCD x = 0$, which contradicts the positive definiteness of $D^TCD$.
  • A later reply suggests that if $D$ is invertible, then $Dx = 0$ implies $x = 0$, and thus $x^T D^TCD x$ must be greater than zero for all non-zero $x$.
  • Participants discuss the necessity of showing that $x^T D^TCD x > 0$ for all non-zero $x$, but express uncertainty about how to conclude this definitively.
  • It is noted that since $C$ is positive definite, for any non-zero vector $y$, the expression $y^T C y > 0$ holds, which may be relevant to the argument.

Areas of Agreement / Disagreement

Participants generally agree on the implications of $D$ being invertible for the positive definiteness of $D^TCD$, but the discussion remains unresolved regarding the definitive conclusion that $x^T D^TCD x > 0$ for all non-zero $x$.

Contextual Notes

There are unresolved steps in the mathematical reasoning, particularly in concluding the positivity of $x^T D^TCD x$ when $D$ is invertible. The discussion also depends on the properties of the symmetric positive definite matrix $C$.

mathmari
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Hi! :o

Given that $C \in \mathbb{R}^{n,n}$ is symmetric and positive definite and $D \in \mathbb{R}^{n,n}$.
I have to show that $D^TCD$ is positive definite $\Leftrightarrow $ $D$ is invertible.

For the direction $\Rightarrow $:
$D^TCD$ is positive definite, that means that $\forall x \in \mathbb{R}^n\setminus \{0\} :$ $ x^T D^TCD x >0$.
How can I continue?
 
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mathmari said:
Hi! :o

Given that $C \in \mathbb{R}^{n,n}$ is symmetric and positive definite and $D \in \mathbb{R}^{n,n}$.
I have to show that $D^TCD$ is positive definite $\Leftrightarrow $ $D$ is invertible.

For the direction $\Rightarrow $:
$D^TCD$ is positive definite, that means that $\forall x \in \mathbb{R}^n\setminus \{0\} :$ $ x^T D^TCD x >0$.
How can I continue?

Heya! ;)

Suppose $D$ is not invertible. Then there must be some $x$ for which $Dx = 0$...
 
I like Serena said:
Heya! ;)

Suppose $D$ is not invertible. Then there must be some $x$ for which $Dx = 0$...

So for some $x$ for which $Dx = 0$: $x^TD^TCDx=0$, but it should be $x^TD^TCDx>0$.
So $D$ must be invertible. Right?

For the direction $\Leftarrow $:
$D$ is invertible, so $Dx=0 \Rightarrow x=0$
To show that $D^TCD$ is positive definite, we have to show that $x^TD^TCDx>0$ $\forall x \in \mathbb{R}\setminus \{0\}$.
$\forall x \in \mathbb{R}\setminus \{0\}$ we have that $Dx \neq 0 \Rightarrow x^TD^TCDx \neq 0$. But how can we conclude that this is greater than $0$?
 
mathmari said:
So for some $x$ for which $Dx = 0$: $x^TD^TCDx=0$, but it should be $x^TD^TCDx>0$.
So $D$ must be invertible. Right?

Right! :cool:

For the direction $\Leftarrow $:
$D$ is invertible, so $Dx=0 \Rightarrow x=0$
To show that $D^TCD$ is positive definite, we have to show that $x^TD^TCDx>0$ $\forall x \in \mathbb{R}\setminus \{0\}$.
$\forall x \in \mathbb{R}\setminus \{0\}$ we have that $Dx \neq 0 \Rightarrow x^TD^TCDx \neq 0$. But how can we conclude that this is greater than $0$?

Well, it is given that $C$ is positive definite.
So for each $y \ne 0$ we have that $y^T C y > 0$.

Now suppose we set $Dx=y$...
 
I like Serena said:
Right! :cool:
Well, it is given that $C$ is positive definite.
So for each $y \ne 0$ we have that $y^T C y > 0$.

Now suppose we set $Dx=y$...

Great! Thank you very much! :o
 

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