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Show that the matrix D is invertible

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,109
Hi!!! :eek:

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?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,854
Hi!!! :eek:

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

Well-known member
MHB Site Helper
Apr 14, 2013
4,109
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 $0$?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,854
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 $Dx=y$...
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,109
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!!!! :eek: