I have two matrices, which are square, symmetric, and positive definite. I would like to prove that the sum of the two matrices still have the same properties, that is square, symmetric, and positive definite. The first two properties are obvious, what about the positive definite property. Any clue to the proof? Thank you.

  • $\begingroup$ Guys, his question has nothing to do with matrix products... $\endgroup$ – user7530 Oct 29 '13 at 14:41

A real matrix $M$ is positive-definite if and only if it is symmetric and $u^TMu > 0$ for all nonzero vectors $u$.

Now if $A$ and $B$ are positive-definite, $u^T(A+B)u = \ldots?$

  • $\begingroup$ Why is symmetric condition necessary, for a real matrix, to be Pos. def. ? How about $\begin{pmatrix}1&1\\-1&1\end{pmatrix}$, which is non-symmetric but still Pos. definite? $\endgroup$ – kaka Jan 25 '17 at 3:35
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    $\begingroup$ @kaka The definition of a positive-definite matrix includes that it is symmetric (or Hermitian, if complex). You can talk about matrices which satisfy some positive-definite-like properties, such as $u^TMu > 0$ for nonzero $u$, without being symmetric, but this is not a standard definition. $\endgroup$ – user7530 Jan 25 '17 at 9:48
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    $\begingroup$ The reason behind that definition is that it comes from quadratic forms $x^T A x$. That is positive definite if it is positive for all $x \not = 0$. You can check that for such quadtatic forms, we only need to consider symmetric matrices. That is, if $A$ is not symmetric, you can replace it with $\frac{A+A^T}{2}$ without changing the form. $\endgroup$ – kjetil b halvorsen Mar 25 '18 at 16:58
  • $\begingroup$ Geometrically, it means that $Mu\in H_u$, where $H_u = \{v: u \cdot v > 0 \}$ is the positive half-space defined by $u$. Then $Au + Bu \in H_u$ since $H_u + H_u = H_u$. Similarly for semi-definite matrices. $\endgroup$ – MaudPieTheRocktorate Feb 12 at 2:41

This is an more detailed answer.

Now let $ A $ and $B$ be positive definite matrices, that is for all $ h \; \in \mathbb{R} ^{n}$ we must have $ h^{T}Ah > 0 $ and $ h^{T}Bh > 0 $.

From properties of real numbers

$0 < h^{T}Bh + h^{T}Ah $

Now from the distributive laws of matrix multiplication we must have

$0< h^{T}Bh + h^{T}Ah =h^{T} (B + A)h $

This implies that $0<h^{T} (B + A)h $ meaning $ B+A \succ 0$.

That is $ B+A $ is positive definite.


A+B, or B+A, is positive definite if both A and B are positive definite. Suppose A is a m1*n1 matrix and B is a m2*n2 matrix. Because you can sum them up, m1=m2, n1=n2. Since then, as you add up these two matrices, the properties of leading principal minors will not change from the old. Since then, A+B is positive definite. Good luck!


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