Prove the matrix is positive definite

Question

Prove that the matrix $\left(\begin{array}{cc} 2+\alpha & -1\\-1&2+\alpha\end{array}\right)$ is positive definite in $\mathbb{C}^2$ for any $\alpha>-1$

Answer 1

For simplicity let $\beta = 2 + \alpha$.

$$\left(\begin{array}{cc} 2+\alpha & -1\\-1&2+\alpha\end{array}\right) = \left(\begin{array}{cc} \beta & -1\\-1&\beta\end{array}\right)$$

To find the eigenvalues solve:

$$\left|\begin{array}{cc} \beta - \lambda & -1\\-1&\beta - \lambda\end{array}\right| = (\beta-\lambda)^2 - 1 = 0$$

We find that $\beta - \lambda = 1$ and $\beta - \lambda = -1.$ Substituting our expression for $\beta$ and solving for $\lambda$ we find that:

$$\lambda = 1 + \alpha$$ $$\lambda = 3 + \alpha$$

In order for both of these to be positive, $\alpha$ must be greater than $-1$.

A (Hermititan) matrix is positive definite if both of the eigenvalues are positive.

Answer 2

you just need to show that the leading principal minors are all positive

$2 + \alpha > 0 \Rightarrow \alpha > -2$

Plus the determinant must be positive, so

$(2 + \alpha)^2 - 1 > 0 \Rightarrow \alpha > -1$ (or $\alpha < -3$)

So if $\alpha > -1$ both conditions are satisfied and the matrix is definite positive

Answer 3

Here the quadratic form is: $(2+\alpha)|x|^2-x\bar{y}-y\bar{x}+(2+\alpha)|y|^2=(1+\alpha)|x|^2 +(1+\alpha)|y|^2+\bar{x}(x-y)+\bar{y}(y-x)=(1+\alpha)|x|^2 +(1+\alpha)|y|^2+(\bar{x}-\bar{y})(x-y)=(1+\alpha)|x|^2 +(1+\alpha)|y|^2+\overline{(x-y)}(x-y)=(1+\alpha)|x|^2 +(1+\alpha)|y|^2+|x-y|^2\ge0$.

(if $\alpha \gt -1$)

Done.

Answer 4

I take it as implicit in the phrase "$\alpha > -1$" that $\alpha \in \Bbb R$, i.e., $\alpha$ is real.

Set

$A = \begin{bmatrix} 2+\alpha & -1 \\-1 & 2+\alpha \end{bmatrix}. \tag{0}$

Look at the characteristic polynomial; it is

$\det(\begin{bmatrix} 2+\alpha -\lambda & -1\\-1&2+\alpha\ - \lambda \end{bmatrix}) = \lambda^2 -2(2 + \alpha) + (2 + \alpha)^2 - 1$ $ = \lambda^2 - (2 \alpha + 4) \lambda + (\alpha^2 + 4\alpha + 3), \tag{1}$

and we have by the quadratic formula

$\lambda_+, \lambda_- = \dfrac{1}{2}((2(\alpha + 4) \pm \sqrt{(2 \alpha + 4)^2 - 4(\alpha^2 + 4\alpha + 3)})$ $= ((\alpha + 2) \pm 1) = \alpha + 3, \alpha + 1. \tag{2}$

We next observe that the given matrix $A$ is symmetric; thus there exists a real orthogonal matrix $O$ which brings $A$ into diagonal form, thus

$O^TAO = \begin{bmatrix} \lambda_+ & 0 \\ 0 & \lambda_- \end{bmatrix} = D. \tag{3}$

For any $x = (x_1, x_2)^T \in \Bbb C^2$, we have

$\langle x, O^TAOx \rangle = \langle x, Dx \rangle = x^\dagger Dx = \bar x_1 x_1 \lambda_+ + \bar x_2 x_2 \lambda_-$ $= \bar x_1 x_1 (\alpha + 3) + \bar x_2 x_2 (\alpha + 1); \tag{4}$

it is easy to see that this expression is positive for all $x \in \Bbb C^2$ if and only if $\alpha + 1 > 0$, or $\alpha > -1$. Now choose any $y \in \Bbb C^2$ and set $x = O^Ty$. Then, since $OO^T = O^TO = I$, $y = Ox$ and

$\langle y, Ay \rangle = \langle y, OO^TAOO^T y \rangle = \langle O^Ty, O^TAO(O^Ty) \rangle = \langle x, Dx \rangle > 0\tag{5}$

if and only if $\alpha > -1$, as we have seen. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!