If $\mathbf{B}>0$, does it follow that $\mathbf{A} = 2\mathbf{B} - \mathbf{I} > 0$?

Question

If $\mathbf{B}>0$ (i.e. a positive-definite matrix), then is it true that $\mathbf{A} = 2\mathbf{B} - \mathbf{I}>0$ (where $\mathbf{I}$ is the identity matrix)?

This seems to stem from the convexity property of positive semi-definite matrices, which states that if $\mathbf{M}$ and $\mathbf{N}$ are positive semi-definite, then for any $\alpha \in (0,1)$ the matrix $\mathbf{L}=\alpha\mathbf{M} + (1-\alpha) \mathbf{N}$ is also positive semi-definite.

If re-arrange it to $\mathbf{M}=\frac{1}{\alpha}\mathbf{L} - \frac{1-\alpha}{\alpha}\mathbf{N}$ and assume $\mathbf{L}>0$ and $\mathbf{N}>0$, does it imply that $\mathbf{M}>0$?

Add 1

There is also a property that if $\mathbf{M} > 0$ is real, then there is a $δ > 0$ such that $\mathbf{M} > δ\mathbf{I}$.

Is there a way to find this $δ$?

Answer 1

A very simple example for why it is wrong is if $\mathbf{A}=\mathbf 0$. Then $\mathbf{B}=\frac{1}{2}\mathbf{I}>0$, and $\mathbf{B}=\frac{1}{2}(\mathbf A+\mathbf I)$, but it is not true that $\mathbf{A}=\mathbf{0}>0.$

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It's true that if $\mathbf{L}=\alpha \mathbf{M}+(1-\alpha)\mathbf{N}$ for $\alpha\in (0,1)$, then we have that

$$\mathbf{M}=\frac{1}{\alpha}\mathbf{L} - \frac{1-\alpha}{\alpha}\mathbf{N}=\beta \mathbf{L} +(1-\beta)\mathbf{N}\quad\quad\text{ where } \beta=\frac{1}{\alpha}$$

Note, however, that $\beta\not\in(0,1)$ if $\alpha\in(0,1).$

So if it is true that if $\mathbf{L},\mathbf{N}>0$ implies that $\mathbf{M}>0,$ then it would imply something much stronger than convexity, namely, for any $\mathbf{M},\mathbf{N}>0,$ and any real $\gamma$:

$$\gamma \mathbf{M}+(1-\gamma)\mathbf{N}>0$$

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If $\mathbf{M}$ has eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_k>0$, then $\mathbf{M}-\delta \mathbf{I}$ has eigenvalues $\lambda_1-\delta,\dots,\lambda_k-\delta.$ So you just need $0<\delta<\min(\lambda_i)$ to get $\mathbf{M}-\delta \mathbf{I}>0$, or $\mathbf{M}>\delta I$.

Answer 2

Rearrange $$B=\frac{1}{2}A+\frac{1}{2}I$$ then $$0<\langle Bx,x\rangle=\frac{1}{2}\langle Ax,x\rangle+\frac{1}{2}\langle x,x\rangle$$ Eventhough for all $x$ we have $\langle x,x\rangle=||x||^2\geqslant 0$ this does not guarantee us that $\langle Ax,x\rangle\geqslant 0$. In fact $\langle Ax,x\rangle$ could be strictly negative but just enough for the total sum to be strictly positive.

Answer 3

Try with scalars first. $2b-1 >0$ iff $b > {1 \over 2}$. So the answer is no.