Given that $A \in R^{n,n}$ is symmetric and positive-definite and $B \in R^{n,n}$,show that $B^{T}AB$ is positive-definite if and only if $B$ is invertible. That's what I have done so far: We suppose that $C=B^{T}AB$ is positive-definite,that means that $\forall x\in R^{n}-\{0\}$ : $0<x^{T}Cx=x^{T}B^{T}ABx=y^{T}Ay , \text{ where } y=Bx$. But how can I conclude that $B$ is invertible?
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Since $B$ is positive then all eigen values are positive, so its determinant is positive as well – Jlamprong Jan 19 '14 at 17:07
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We have for $X\in\mathbb R^{n,1}$ and since $A$ is definite positive $$X^TB^TABX=(BX)^TA(BX)\ge0$$ hence $B^TAB$ is positive, moreover $$X^TB^TABX=(BX)^TA(BX)=0\iff BX=0$$ so
- if $B$ is invertible then $X=0$ so $B^TAB$ is definite
- and if $B^TAB$ is definite so $BX=0\iff X=0$ and then $B$ is invertible.
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1If a matrix $A$ is positive-definite,doesn't it mean that: $x^{T}Ax>0$ ? Or can it also be equal to $0$ ? – evinda Jan 19 '14 at 21:46
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Note that $C$ positive definite gives us that $Bx \neq 0$ for $x \neq 0$. Hence kernel $B={0}$.
So B is invertible.
I think you have got the converse (since you didn't ask about it). Essentially the converse follows similar line of argument.
voldemort
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Since $B^TAB$ is positive definite, it is invertible, since the inequality $x^TB^TABx > 0$, which our OP has astutely derived and graciously provided, implies $\ker B^TAB = 0$. $B^TAB$ invertible implies $\det (B^TAB) \ne 0$. But $\det (B^TAB) = \det(B^T) \det (A) \det(B)$, so $\det(B) \ne 0$; hence $B$ itself must be invertible.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
Robert Lewis
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