integral of positive definite matrix is positive definite? Say $A_{t}$ is a positive definite matrix for all t, do we have that $\int_{0}^{1}A_{t}dt$ is a positive definite matrix. Is there an obvious counterexample?
The arguement in the book was that the weighted average of positive definite matrices is also positive definite.
http://books.google.ca/books?id=k4ODAwAAQBAJ&pg=PA38&lpg=PA38&dq=maximum+principle+minimal+surfaces+interior&source=bl&ots=D8AGQK9QXi&sig=vWX5-I_c2QyAKsT7CQKGXuTEECQ&hl=en&sa=X&ei=28J_VITVLIS2yASt7YCoBw&ved=0CEQQ6AEwBg#v=onepage&q=maximum%20principle%20minimal%20surfaces%20interior&f=false
page 37
A condition for a matrix to be positive definite is $det(A)\geq0$ and  $Tr(A)\geq 0$.
So since we have a Riemann sum, we should have $Tr(\int_{0}^{1}A_{t}dt)=lim_{n\to \infty}\sum_{i}^{n}Tr(A_{t_{i}})(t_{i}-t_{i-1})>0$.
However determinants don't decompose into sums.
Thank you
 A: This result seems to apply to vector spaces over both $\Bbb R$ and $\Bbb C$, so I'll resort to the complex case in the following; the real case is then implied.  Thus I take $\langle \cdot, \cdot \rangle$ to be a hermitian inner product. Furthermore, I'll simply assume $A(t)$ has the necessary integrability properties to make things fly; continuity should be sufficient.  
Let
$A(t) = [A_{ij}(t)]; \tag{1}$
then by definition
$\int_0^1 A(t) dt = [\int_0^1 A_{ij}(t) dt]. \tag{2}$
Let $\vec x \ne 0$ be any constant vector:
$\vec x = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}; \tag{3}$
for any fixed $t \in [0, 1]$ we have
$0 < \langle \vec x, A(t) \vec x \rangle = \sum_{i, j} A_{ij}(t) \bar x_i x_j, \tag{4}$
since the $A(t)$ are positive definite.  Integrating (4) over $[0, 1]$ we see that
$0 < \int_0^1 \langle \vec x, A(t) \vec x \rangle dt = \int_0^1 (\sum_{i, j} A_{ij}(t) \bar x_i x_j) dt; \tag{5}$
by linearity of the integral we have
$\int_0^1 (\sum_{i, j} A_{ij}(t) \bar x_i x_j) dt = \sum_{i, j} \bar x_i x_j \int_0^1 A_{ij}(t) dt = \langle x, (\int_0^1 A(t) dt) x \rangle. \tag{6}$
Combining (5) and (6) yields
$\langle x, (\int_0^1 A(t) dt) x \rangle > 0 \tag{7}$
for every non-zero vector $x$; $\int_0^1 A(t) dt$ is thus positive definite.
Note:  I don't see exactly how looking at $\text{Tr}(A)$ and $\det(A)$ can help much with this one, since $\int \det A(t) dt$ and $\det \int A(t) dt$ are not obviously related.  In any event, it appears that $\text{Tr}(A) > 0$ and $\det (A) > 0$ are determinative for positive definiteness in only the $2 \times 2$ case; the above argument binds for matrices of any size $n$.  End of Note.
Hope this helps.  Cheers,
and as ever,
Fiat Lux!!!
