Show that the matrix D is invertible

[mathmari]

Hi!

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?

 

[I like Serena]

Hi!

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 $Dx = 0$...

 

[mathmari]

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 $0$?

 

[I like Serena]

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!

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 $Dx=y$...

 

[mathmari]

Right!
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!