# Determinant and symmetric positive definite matrix

1. May 26, 2015

### Incand

As a step in a solution to another question our lecture notes claim that the matrix (a,b,c,d are real scalars).

\begin{bmatrix}
2a & b(1+d) \\
b(1+d)& 2dc \\
\end{bmatrix}

Is positive definite if the determinant is positive. Why? Since the matrix is symmetric it's positive definite if the it got positive (real) eigenvalues and $det (A) = \prod \lambda_i$ but two negative eigenvalues would give a positive determinant too.

Earlier in the text its given that the matrix
$S = \begin{bmatrix} a & b \\ b& c \\ \end{bmatrix}$
is positive definitive while $d$ is without any restrictions if that is somehow relevant.

2. May 26, 2015

### Incand

Edit: nevermind I solved it

If i calculate the eigenvalues I get
$0 = (2a-\lambda )(2dc-\lambda) - b^2(1+d)^2 = 4adc +\lambda^2 - (2a+2dc)\lambda - b^2(1+d)^2 = det(A) + \lambda^2 -(2a+2dc)\lambda$
then
$\lambda^2-(2a+2dc)\lambda < 0$ since $det(A) > 0$
equal when
$(\lambda-(a+dc))^2-(a+dc)^2 = 0 \Longleftrightarrow \lambda = (a+dc) \pm (a+dc)$
so the smallest eigenvalue is always $\lambda> 0$.

Last edited: May 26, 2015