Show that the matrix $B$ is positive definite without using the positive pivot and determinant test.

Question

I'm a little confused about definite matrices. We learned about $4$ different types of definite matrices: positive semidefinite, positive definite, negative semidefinite and negative semidefinite. In a positive semidefinite matrix we have that the scalar $x^TBx \geq 0$. In a positive definite matrix the scalar $x^TBx > 0$. The opposite is true for the negative matrices.

Let's take the example of a $3 \times 3$ matrix $B = \left(\begin{array}{ccc} 25 & 15 & -5 \\ 15 & 18 & 0 \\ -5 & 0 & 11 \\ \end{array}\right)$. To check if $B$ is a positive definite matrix we can see if $x^TBx > 0$.

$$x^TBx = \left(\begin{array}{ccc} a & b & c \\ \end{array}\right) \cdot \left(\begin{array}{c} 25 & 15 & -5 \\ 15 & 18 & 0 \\ -5 & 0 & 11 \\ \end{array}\right) \cdot \left(\begin{array}{ccc} a \\ b \\ c \\ \end{array}\right)= 25a^2+30ab-10ac+18b^2+11c^2$$

How can I continue on from here? I know $B$ is positive definite because after using the Gauß Elimination technique all the pivot values are greater than $0$ and all upper-left sub-matrices are also positive. I'm just trying to see how the original version works.

Answer 1

$$ 25a^2+30ab-10ac+18b^2+11c^2 =\\ 25(1-\frac 12 -\frac 1{11}) a^2 + (\frac 5{\sqrt 2}a + 3\sqrt 2 b )^2 + (\frac 5{\sqrt {11}}a - \sqrt{11}c)^2> 0 $$

Answer 2

The easiest way (in my opinion) to see that the matrix is positive definite is to check two things: is the matrix symmetric (yes and easily checkable) and are all the eigenvalues positive (yes, but less obvious).

You can quickly see that the eigenvalues are all positive using the Gershgorin circle theorem (https://en.wikipedia.org/wiki/Gershgorin_circle_theorem). In short, all the eigenvalues (for your matrix) live in (at least) one of the balls: $B(25, 20)$ (ball centered at 25 of radius 20), $B(18, 15)$, and $B(11, 5)$. All of those balls are strictly positive, so all your eigenvalues are positive.

Symmetry combined with all positive eigenvalues implies positive definite.

Answer 3

Here is an expanded proof that the form is positive definite using Gaussian elimination. Using simultaneous row-and-column operations, we have $$ \pmatrix{25 & 15 & -5 \\ 15 & 18 & 0 \\ -5 & 0 & 11} \leadsto \pmatrix{25&0&-5\\0&9&3\\-5 & 3 & 12 } \leadsto \pmatrix{25&0&0\\0&9&3\\0 & 3 & 12 } \leadsto \pmatrix{25&0&0\\0&9&0\\0&0&11}. $$ So, the matrix is positive definite. But why does this method work? The key is to see that every row-reduction step corresponds to grouping variables to remove coefficients.

For instance, the elementary corresponding to the first step was $$ E = \pmatrix{1&0&0\\-3/5 & 1 & 0\\0 & 0 & 0}, $$ which is to say that $$ M_1 = \pmatrix{25&0&-5\\0&9&3\\-5 & 3 & 12 } = EBE^T. $$ We want to see why $B$ will be positive definite if and only if $B$ is positive definite. To that end, note that $$ M_1 = EBE^T \implies B = E^{-1}M_1E^{-T}. $$ Note that $$ E^{-T}x = \pmatrix{1&3/5&0\\0&1&0\\0&0&1}\pmatrix{a\\b\\c} = \pmatrix{a + 3b/5\\ b\\ c} $$ So, the quadratic form $$ Q(x) = x^TBx = x^TE^{-1}BE^{-T}x = (E^{-T}x)^TB (E^{-T}x) $$ can be written as $$ Q(x) = 25(a + 3b/5)^2 + 10(a + 3b/5)c + 18b^2 + 11c^2. $$ With that, we can see that if the quadratic form $$ x^TM_1x = 25a^2 + 10ac + 18b^2 + 11c^2 $$ is positive definite, then so is $Q(x)$. This connection is maintained until the final step, where we arrive that $M_3$ and the associated quadratic form $$ x^TM_3x = 25a^2 + 9b^2 + 11c. $$ Because this quadratic form was positive definite, so was the one associated with $M_2,M_1,$ and $B$.