# Sum of a diagonal matrix and a symmetric positive semidefinite matrix

Let the following symmetric matrix $$A = \left[ \matrix{a_{12}+a_{13} & -a_{12} & -a_{13} \\ -a_{12} & a_{12}+a_{23} & -a_{23} \\ -a_{13} & -a_{23} & a_{13}+a_{23} } \right]$$,

where each $$a_{ij}$$ is a nonnegative real number, and hence, $$A$$ is a symmetric positive semidefinite matrix, because:

$$x^T A x = (a_{12}+a_{13}) x_{1}^2 + (a_{12}+a_{23}) x_{2}^2 + (a_{13}+a_{23}) x_{3}^2 - 2( a_{12} x_1 x_2 + a_{13} x_1 x_3 + a_{23} x_2 x_3 ) \\ = a_{12}(x_1 - x_2)^2 + a_{13}(x_1 - x_3)^2 + a_{23}(x_2 - x_3)^2 \geq 0$$

On the other hand:

Let $$B = A + D = \left[ \matrix{a_{12}+a_{13} & -a_{12} & -a_{13} \\ -a_{12} & a_{12}+a_{23} & -a_{23} \\ -a_{13} & -a_{23} & a_{13}+a_{23} } \right] + \left[ \matrix{d_{11} & & \\ & d_{22} & \\ & & d_{33} } \right]$$,

where $$d_{ii} \in \mathbb{R}$$, that is, $$D$$ is a diagonal matrix with positive and negative real numbers.

What are the limits of $$d_{ii}$$ so that B is still a positive semidefinite matrix?

• What is $\mathbb R^+$? Does it contain nonnegative real numbers or only positive real numbers? Jul 28, 2020 at 18:17
• I am sorry. Nonnegative real numbers. I will modify it. Jul 28, 2020 at 18:21
• The problem arises from an electrical engineering question. Where the matrix $A$ is the reactance matrix of the system, and $B$ represents values ​​of capacitors and reactances of lines connected to a substation. Jul 28, 2020 at 18:35

Note that $$A$$ is positive semidefinite is not positive definite. In other words, $$A$$ is positive semidefinite and singular.

First, note that if all three $$d_{ii}$$ are positive, then $$A + D$$ will necessarily fail to be positive semidefinite.

So, we must suppose that at least on of these is zero. Without loss of generality, suppose that $$d_{33} = 0$$. If both $$d_{11},d_{22}$$ are non-zero, then we find that $$A + D$$ can only be positive semidefinite if $$A$$ is diagonal with $$a_{33} = 0$$. In this case, this only occurs when $$A = 0$$.

For the remaining case, we must suppose that one of $$d_{11},d_{22}$$ are zero. Without loss of generality, suppose that $$d_{22} = 0$$. In other words, we are looking for the maximal value of $$d_{11}$$ for which $$A - \pmatrix{d_{11} & 0 & 0\\0 &0 &0\\0 &0&0}$$ is positive semidefinite. As is shown here, for instance, this matrix will be positive definite for $$d_{11} \leq 1/A^+_{11}$$, where $$A^+_{11}$$ denotes the $$1,1$$ entry of the Moore-Penrose pseudoinverse $$A^+$$ of $$A$$.

As we established above, it is sufficient to consider the case in which at most one of the $$d_{ii}$$ are non-zero, since $$A + D$$ will fail to be positive semidefinite otherwise. We compute (via matlab) $$A^+ = \left(\begin{array}{ccc} \frac{4\,b_{1}+b_{2}+b_{3}}{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)} & -\frac{2\,b_{1}+2\,b_{2}-b_{3}}{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)} & -\frac{2\,b_{1}-b_{2}+2\,b_{3}}{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)}\\ -\frac{2\,b_{1}+2\,b_{2}-b_{3}}{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)} & \frac{b_{1}+4\,b_{2}+b_{3}}{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)} & -\frac{2\,b_{2}-b_{1}+2\,b_{3}}{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)}\\ -\frac{2\,b_{1}-b_{2}+2\,b_{3}}{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)} & -\frac{2\,b_{2}-b_{1}+2\,b_{3}}{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)} & \frac{b_{1}+b_{2}+4\,b_{3}}{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)} \end{array}\right) \\ = \frac{1}{9(b_1b_2 + b_1 b_3 + b_2 b_3)}\sum_{i=1}^3b_i\,(-2e_i + e_{i+1} + e_{i+2})(-2e_i + e_{i+1} + e_{i+2})^T,$$ where in the above we take $$b_{1} = a_{23}, b_2 = a_{13},b_3 = a_{12}$$. In the second line, $$e_i$$ denotes the $$i$$th standard basis vector and addition in the indices is modulo $$3$$.

We see then, for instance, that the maximal $$d_{11}$$ for which $$A + D$$ is positive semidefinite is given by $$d_{11} = \frac{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)}{4\,b_{1}+b_{2}+b_{3}}.$$ Similar conditions hold for $$d_{22},d_{33}$$.

• Thank you very much. I have added the following information, which may be useful... $A$ is a symmetric positive semidefinite matrix, because: $x^T A x = (a_{12}+a_{13}) x_{1}^2 + (a_{12}+a_{23}) x_{2}^2 + (a_{13}+a_{23}) x_{3}^2 - 2( a_{12} x_1 x_2 + a_{13} x_1 x_3 + a_{23} x_2 x_3 ) \\ = a_{12}(x_1 - x_2)^2 + a_{13}(x_1 - x_3)^2 + a_{23}(x_2 - x_3)^2 \geq 0$ Jul 27, 2020 at 17:11
• @Jon See my latest edit Jul 27, 2020 at 17:39
• Thanks! I have modified the question. Please can you see it? Jul 28, 2020 at 18:10
• If the entries in $D$ are all positive, then $A+D$ will still be positive semidefinite, since $x^{T}(A+D)x \geq 0$ for all $x$. Jul 28, 2020 at 18:13
• You're right. But, if $D$ can have positive and negative values, what is the condition for $B$ to be positive semi-defined? Jul 28, 2020 at 18:19

By Sylvester's criterion, $$A+D$$ is positive semidefinite if and only if all principal minors of $$A+D$$ are nonnegative. In other words, $$A+D$$ is PSD iff \begin{aligned} &d_1+a_{12}+a_{13}\ge0,\\ &d_2+a_{12}+a_{23}\ge0,\\ &d_3+a_{13}+a_{23}\ge0,\\ &(d_1+a_{12}+a_{13})(d_2+a_{12}+a_{23})\ge a_{12}^2,\\ &(d_1+a_{12}+a_{13})(d_3+a_{13}+a_{23})\ge a_{13}^2,\\ &(d_2+a_{12}+a_{23})(d_3+a_{13}+a_{23})\ge a_{23}^2\\ \end{aligned} and \begin{aligned} &(d_1+a_{12}+a_{13})(d_2+a_{12}+a_{23})(d_3+a_{13}+a_{23})\\ -&a_{23}^2(d_1+a_{12}+a_{13}) -a_{13}^2(d_2+a_{12}+a_{23}) -a_{12}^2(d_3+a_{13}+a_{23}) -2a_{12}a_{13}a_{23} \ge0. \end{aligned} The region bounded by these surfaces is typically not a translated octant of $$\mathbb R^3_+$$. Therefore, we cannot find a vector $$(d_1',d_2',d_3')$$ such that $$A+D$$ is positive semidefinite if and only if $$(d_1,d_2,d_3)\ge(d_1',d_2',d_3')$$.

To illustrate, consider the analogous case where $$A$$ is $$2\times2$$. If $$A$$ is nonzero, by scaling $$A$$, we may assume that $$A=\pmatrix{1&-1\\ -1&1}.$$ In this case, if $$D=\operatorname{diag}(d_1,d_2)$$, then $$A+D$$ is positive semidefinite if and only if $$d_1+1\ge0,\ d_2+1\ge0$$ and $$(d_1+1)(d_2+1)\ge1$$. That is, the feasible region of $$(d_1,d_2)$$ is the region above the upper right branch of the hyperbola $$(d_1+1)(d_2+1)\ge1$$. The hyperbola simply hasn't any "bottom left corner" $$(d_1',d_2')$$ that is entrywise less than or equal to all $$(d_1,d_2)$$ in the feasible region.

• you are right. $A$ is a positive semidefinite matrix. I will modify the question Jul 27, 2020 at 14:02
• I have modified the question. Please can you see it? Jul 28, 2020 at 18:09
• if $d_{1}, d_{2}, d_{3} > 0$, then $e^{T}De>0$. You seem to be assuming that the diagonal entries in $D$ are all negative, but why do you make that assumption? Jul 28, 2020 at 18:12
• @BrianBorchers The OP has changed the question. $d$ was $\le0$ in the original question. Jul 28, 2020 at 18:13
• Thank you very much for your good explanation. Is it possible to extend this for dimensions greater than 3? That is, is it possible to express the determinant in an easy way? Or define conservative limits in a simpler way? Jul 30, 2020 at 10:49