# Prove the matrix is positive definite

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 $\alpha>-1$

• Is $\alpha$ supposed to be a real number, or a complex number? – Calvin Lin Mar 30 '14 at 20:11
• I interpret that $\alpha$ must be real because of the statement $\alpha \gt -1$. – Brad Mar 30 '14 at 20:12

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.

• I think there needs to be an additional restraint on the trace, which is why I think this only holds in $\mathbb{R^{2\times 2}}$ – Ellya Mar 30 '14 at 20:17
• @ellya do you mean that $\alpha \in \mathbb{R}$? Otherwise, no idea what you're talking about here. – Omnomnomnom Mar 30 '14 at 20:42
• I mean this methodology holds only if the matrix has real entries. – Ellya Mar 30 '14 at 20:43
• I believe it holds whenever the matrix is Hermitian. – Brad Mar 30 '14 at 20:44
• @ellya this matrix does only have real entries. That fact notwithstanding, where would this methodology fail for a (Hermitian) matrix with complex entries? – Omnomnomnom Mar 30 '14 at 20:44

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

• I have not heard the term Nord-east minors. I suppose that you are referring to what I call the leading principal minors – Omnomnomnom Mar 30 '14 at 20:57
• @Omnomnomnom yes, thank you! I translated in english I term that is used in my country, sorry :-) – Ant Mar 30 '14 at 20:59
• for the record, nord = north and ovest = west. East = est, which made this a bit more confusing :) – Omnomnomnom Mar 30 '14 at 21:02

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.

• Does this suffice? – Ellya Mar 30 '14 at 20:55
• It is sufficient, but too much work for my taste. As for "a link or something": for a Hermitian matrix, it is indeed enough to verify that the eigenvalues are positive – Omnomnomnom Mar 30 '14 at 21:07
• Also, for future LaTeXing: rather than \bar{(x-y)}, try \overline{(x-y)}, which gives you $\overline{(x-y)}$. – Omnomnomnom Mar 30 '14 at 21:08
• True it is more work, and thank you! that bar really annoyed me! – Ellya Mar 30 '14 at 21:08

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!!!