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I have two symmetric positive definite (SPD) matrices. I would like to prove that the sum of these two matrices is still SPD. Symmetry is obvious, but what about PD-ness? Any clues, please?

Rodrigo de Azevedo
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Richard Dong
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3 Answers3

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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, $u^T(A+B)u = \ldots?$

user7530
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    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? – kaka Jan 25 '17 at 03:35
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    @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. – user7530 Jan 25 '17 at 09:48
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    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. – kjetil b halvorsen Mar 25 '18 at 16:58
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    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. – MaudPieTheRocktorate Feb 12 '19 at 02:41
  • Does it also hold the opposite? $(A+B)$ nonnegative definite $\implies$ A and B nonnegative definite? – Mr Frog Apr 26 '22 at 08:28
  • @MrFrog No. For an easy counterexample consider $A = 2I$ and $B = -I$. – user7530 Apr 26 '22 at 18:50
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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.

Christian Prince
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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!

Zixiao Xu
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