28.3 How to Identify if 2x2 Matrix is Positive Definite

Types: N/A
Examples: N/A
Constructions: N/A
Generalizations: N/A

Properties: 13.2 Saddle Points, 6.7 Intro to Quadratic Surfaces, Quadric Surfaces
Sufficiencies: N/A
Questions: N/A

Exm

When is a $2 \times 2$ symmetric matrix positive definite?

Suppose we have a symmetric $2 \times 2$ matrix

K = [\begin{matrix} k_{11} & k_{12} \\ k_{21} & k_{22} \end{matrix}]

where $k_{12} = k_{21}$ since $K = K^{T}$ .

\begin{proof}[Solution.]
In order to answer this question for any general $2 \times 2$ symmetric matrix, let's introduce some simpler notation by setting

K = [\begin{matrix} a & b \\ b & c \end{matrix}]

for any real-valued scalars $a, b, c \in R$ .

& Note: $det (K) = a c - b^{2}$

Recall the definition of a 28.2 Positive Definite Matrix.

Positive Definite Matrix

We say that an $n \times n$ matrix $K$ is positive definite iff it satisfies the two conditions:

Symmetry condition: $K = K^{T}$
Positivity condition: If $\vec{x} \neq \vec{0}$ , then ${\vec{x}}^{T} K \vec{x} > 0$

For short hand, we sometimes write $K > 0$ to imply that $K$ is a symmetric, positive definite matrix.

With that definition in mind, we ask ourselves how the values of $a, b, c$ relate to the property that $2 \times 2$ symmetric

K = [\begin{matrix} a & b \\ b & c \end{matrix}]

is positive definite with ${\vec{x}}^{T} K \vec{x} > 0$ for any $\vec{x} \neq \vec{0}$ . To explore this, suppose

\vec{x} = {[\begin{matrix} x_{1} & x_{2} \end{matrix}]}^{T} \neq \vec{0}

and consider

\begin{aligned} [{\vec{x}}^{T}]_{1 \times 2} [K]_{2 \times 2} [\vec{x}]_{2 \times 1} & = {[\begin{array}{c} x_{1} & x_{2} \end{array}]}_{1 \times 2} {[\begin{array}{c} a & b \\ b & c \end{array}]}_{2 \times 2} {[\begin{array}{c} x_{1} \\ x_{2} \end{array}]}_{2 \times 1} \\ = {[\begin{array}{c} x_{1} & x_{2} \end{array}]}_{1 \times 2} {[\begin{array}{c} a x_{1} + b x_{2} \\ b x_{1} + c x_{2} \end{array}]}_{2 \times 1} \\ = {[\begin{array}{c} a x_{1} + b x_{2} & b x_{1} + c x_{2} \end{array}]}_{1 \times 2} {[\begin{array}{c} x_{1} \\ x_{2} \end{array}]}_{2 \times 1} \\ ⟹ {\vec{x}}^{T} K \vec{x} & = a x_{1}^{2} + b x_{1} x_{2} + b x_{1} x_{2} + c x_{2}^{2} \\ = a x_{1}^{2} + 2 b x_{1} x_{2} + c x_{2}^{2} \\ = a \underset{How can we transform this to perfect square?}{\underset{⏟}{(x_{1}^{2} + 2 \frac{b}{a} x_{1} x_{2})}} + c x_{2}^{2} \end{aligned}

Let's focus on the expression inside the parenthesis and remember how to complete the square:

x_{1}^{2} + 2 \frac{b}{a} x_{1} x_{2} + ? = {(x_{1} + \frac{b}{a} x_{2})}^{2}

? What do we need to add to the LHS to create a perfect square trinomial so that we can factor this as a perfect square?

Here we note that

\begin{aligned} {(x_{1} + \frac{b}{a} x_{2})}^{2} & = (x_{1} + \frac{b}{a} x_{2}) \cdot (x_{1} + \frac{b}{a} x_{2}) \\ = x_{1} \cdot (x_{1} + \frac{b}{a} x_{2}) + \frac{b}{a} x_{2} \cdot (x_{1} + \frac{b}{a} x_{2}) \\ = x_{1}^{2} + \frac{b}{a} x_{1} x_{2} + \frac{b}{a} x_{2} x_{1} + \frac{b^{2}}{a^{2}} x_{2}^{2} \\ = \underset{starting term}{\underset{⏟}{x_{1} 2 + 2 \frac{b}{a} x_{1} x_{2}}} + \underset{add for perfect square trinomial}{\underset{⏟}{\frac{b^{2}}{a^{2}} x_{2}^{2}}} \end{aligned}

Therefore,

\begin{aligned} ⟹ x_{1}^{2} + 2 \frac{b}{a} x_{1} x_{2} & = x_{1}^{2} + 2 \frac{b}{a} x_{1} x_{2} + 0 \\ = x_{1}^{2} + 2 \frac{b}{a} x_{1} x_{2} + \frac{b^{2}}{a^{2}} x^{2} - \frac{b^{2}}{a^{2}} x_{2}^{2} \\ = {(x_{1} + \frac{b}{a} x^{2})}^{2} - \frac{b^{2}}{a^{2}} x_{2}^{2} \end{aligned}

Looking back at our original problem, we had

\begin{aligned} {\vec{x}}^{T} K \vec{x} & = [\begin{array}{c} x_{1} & x_{2} \end{array}] [\begin{array}{c} a & b \\ b & a \end{array}] [\begin{array}{c} x_{1} \\ x_{2} \end{array}] \\ = a x_{1}^{2} + 2 b x_{1} x_{2} + c x_{2}^{2} \\ = a (\underset{transform this to perfect square}{\underset{⏟}{x_{1}^{2} + 2 \frac{b}{a} x_{1} x_{2}}}) + c x_{2}^{2} \\ = a (x_{1}^{2} + 2 \frac{b}{a} x_{1} x_{2} + \frac{b^{2}}{a^{2}} x_{2}^{2} - \frac{b^{2}}{a^{2}} x_{2}^{2}) + c x_{2}^{2} \\ = a {(x_{1} + \frac{b}{a} x_{2})}^{2} - \frac{b^{2}}{a^{2}} x_{2}^{2} + c x_{2}^{2} \\ = a (x_{1} + \frac{b}{a} x^{2}) + (\underset{find common denom}{\underset{⏟}{c - \frac{b^{2}}{a^{2}}}}) x_{2}^{2} \\ = a {(x_{1} + \frac{b}{a} x_{2})}^{2} + (\frac{a c - b^{2}}{a}) x_{2}^{2} \end{aligned}

& Note that $det (K) = a c - b^{2}$ .

Notice in this new form, we can say something intelligent about when the matrix $K$ is positive definite based on the entries of

K = [\begin{matrix} a & b \\ b & c \end{matrix}]

Specifically, we've just shown that for

K = [\begin{matrix} a & b \\ b & c \end{matrix}]

we can write

{\vec{x}}^{T} K \vec{x} = a {(x_{1} + \frac{b}{a} x^{2})}^{2} + (\frac{a c - b^{2}}{a}) x_{2}^{2}

where both terms are perfect squares and will therefore never be negative. Let's introduce some new notation to simplify our problem. Let

\begin{aligned} y_{1} & = x_{1} + \frac{b}{a} x_{2} \\ β & = \frac{a c - b^{2}}{a} \end{aligned}

Then, we want to know when is the two variable function

a y_{1}^{2} + β x_{2}^{2} > 0

where at least one variable $y_{1}$ or $x_{2}$ is nonzero. Note, though, that this depends on coefficients $a$ and $β$ since we know

y_{1}^{2} = {(x_{1} + \frac{b}{a} x_{2})}^{2} \geq 0 and x_{2}^{2} \geq 0

Specifically, only when both

a > 0 and \frac{a c - b^{2}}{a} = β > 0

can we guarantee positivity of our quadratic expression above. However, recall that we had

β = \frac{a c - b^{2}}{a} = \frac{det (K)}{a} = \frac{d}{a}

This scalar has a very special property:

Claim

If we know the sign of $a$ , then the sign of $(a c - b^{2}) = det (K) = d$ determines the sign of $β$ .

We can check the sign of $β$ using the table below.

	$d$ positive (with $d > 0$ and $s g n (d) = 1$ )	$d$ negative (with $d < 0$ and $s g n (d) = - 1$ )
$a$ positive (with $a > 0$ and $s g n (a) = 1$ )	$β$ positive	$β$ negative
$a$ negative (with $a < 0$ ) and $s g n (a) = - 1$	$β$ negative	$β$ positive

But we said that for

K = [\begin{matrix} a & b \\ b & c \end{matrix}],

\begin{aligned} {\vec{x}}^{T} K \vec{x} & = a {(x_{1} + \frac{b}{a} x_{2})}^{2} + (\frac{a c - b^{2}}{a}) x_{2}^{2} \\ = a y_{1}^{2} + β x_{2}^{2} \end{aligned}

was for sure positive if $\vec{x} \neq \vec{0}$ iff

\begin{aligned} ⟺ a > 0 and β > 0 \\ ⟺ a > 0 and \frac{a c - b^{2}}{a} > 0 \\ ⟺ a > 0 and a c - b^{2} > 0 \\ ⟺ a > 0 and det (K) > 0 \end{aligned}

Then, we have developed a condition to check whether any $2 \times 2$ symmetric matrix in the form

K = [\begin{matrix} a & b \\ b & c \end{matrix}]

is positive definite based on the values of each entry. Specifically, for $[\vec{x}]_{2 \times 1} \neq \vec{0}$ .

Quick check for positive definiteness of any symmetric

2 \times 2

matrix

${\vec{x}}^{T} K \vec{x} > 0 ⟺ {\begin{cases} i . & a > 0 AND \\ i i . & det (K) > 0 \end{cases}$

\end{proof}

Remark. We can geometrically determine positive and negative definiteness by analyzing the shape of our Quadric Surfaces.

Positive definite when surface is an elliptical paraboloid pointing up
Negative definite when function has a saddle point (hyperbolic paraboloid)

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