When are these eigenvalues non-negative? I'm trying to find a pair of real numbers $(a,b)$ which ensure that some matrix is strictly positive semi-definite.The eigenvalues of this matrix are $$\lambda=1 + a \pm \sqrt{(c+b)^2+2(x'-ax)^2}$$ and $$\lambda=1 - a \pm \sqrt{(c-b)^2+2(x'-ax)^2}.$$ 
I therefore need one of the eigenvalues to be zero and the rest non-negative.
For a fixed $c,x,x' \in \mathbb R$, what is the best way to determine some $a,b$ exist? If not, are there conditions on $x,x',c$ so that $a,b$ exist? Here I know that $0\leq x,x' \leq 1/\sqrt2$ and $0 \leq c \leq 1$.
 A: I guess you know if we put $a=m+n$, $b=m-n$, $t=x’-ax$, $a=x$, $b=x’$ then we obtain this problem with additional conditions that $0\le a,b\le 1/\sqrt{2}$ and one of $\lambda$’s is $0$. Nevertheless, these conditions turn us to a very different way. 
Let’s start. The required $a$ and $b$ exist iff all the following conditions hold
$|a|\le 1$
$(1 + a)^2\ge (c+b)^2+2(x'-ax)^2$  
$(1 – a)^2 \ge (c-b)^2+2(x'-ax)^2$ 
and one of the inequalities is an equality. 
Let $a$ is fixed. Put $d=\sqrt{2}|x’-ax|$. We need $d\le\min\{1-a,1+a\}$.  Then 
$\sqrt{(1 + a)^2-d^2}\ge |b+c|$  
$\sqrt{(1 - a)^2-d^2}\ge |b-c|$  
and one of the inequalities is an equality. That is 
$-\sqrt{(1 + a)^2-d^2}\le b+c\le \sqrt{(1 + a)^2-d^2}$ 
$-\sqrt{(1 - a)^2-d^2}\le b-c\le \sqrt{(1 - a)^2-d^2}$
and one of the inequalities is an equality. That is
$-\sqrt{(1 + a)^2-d^2}-c\le b\le \sqrt{(1 + a)^2-d^2}-c$ 
$-\sqrt{(1 - a)^2-d^2}+c\le b\le \sqrt{(1 - a)^2-d^2}+c$.
and one of the inequalities is an equality.
Such $b$ exists iff one of endpoints of the segments $[-\sqrt{(1 + a)^2-d^2}-c,\sqrt{(1 + a)^2-d^2}-c]$  and $[-\sqrt{(1 - a)^2-d^2}+c,\sqrt{(1 - a)^2-d^2}+c]$ belongs to the other, that is when the segments intersect. This holds iff the distance between their midpoints is not greater than a half of the sum of their lengths, that is when 
$$2c\le \sqrt{(1 + a)^2-d^2}+\sqrt{(1 - a)^2-d^2}\equiv f(a).$$
For the sake of simplicity we introduce variables $y=\sqrt{2}x$ and $y’=\sqrt{2}x’$ (so $0\le y,y’\le 1$). Then $d^2=(y’-ay)^2$ and 
$$f(a)=\sqrt{1+2a+a^2-y’^2+2ayy’-a^2y^2}+\sqrt{1-2a+a^2-y’^2+2ayy’-a^2y^2}.$$
From the above we see that the required $a$ and $b$ exist iff there exists $a$ such that $f(a)\ge 2c$ and $d\le\min\{1-a,1+a\}$. We claim that it suffices to look for such not-negative $a$. Indeed, if the needed conditions are satisfied then when we replace $a$ by $|a|$ we obtain that the new expression under the first (resp., the second) radical will become the old expression under the second (resp., the first) radical plus $4|a|yy’$, so $f(|a|)\ge f(a)$. The condition $$|d|\le \min\{1-|a|,1+|a|\}=\min\{1-a,1+a\}$$ will be also kept, because $$|(y’-|a|y|)|\le|y’|+|a||y|=|y’-ay|.$$
So now we have $0\le a\le 1$ and $|y’-ay|\le 1-a$. The last condition determines our maximal possible value $a_\max$ of $a$, namely, $a_\max=\frac{1+y’}{2}$ if $y=1$ and $a_\max=\min\left\{\frac {1+y’}{1+y},\frac {1-y’}{1-y}\right\}$, if $y<1$.    
If $a_\max=0$ (which holds iff $y<1$ and $y’=1$) then the required $a$ and $b$ exist iff $f(0)=0\ge 2c$, that is iff $c=0$. Further we assume that $a_\max>0$. The function $f(a)$ is continuous at the closed segment $[0,a_\max]$, so it attains its maximum at some point $a_0\in [0,a_\max]$. Moreover, since the function $f(a)$ is differentiable on the interval $(0, a_\max)$, we have $a_0=0$ (remark that $f(0)=2\sqrt{1-y’^2}$), or $a_0=a_\max$, or $f’(a_0)=0$. Routine calculations simplify the last condition to 
$-1+a+yy’-ay^2\le 0$ and $(ay-y’)(ay^2y’-ay’+y-yy’^2)=0$. 
If $ay-y’=0$ then $d=0$, so $f(a)$ is defined and equals $2$, so the required $a$ and $b$ exist for any $0\le c\le 1$. The equality $ay-y’=0$ can be satisfied by some $0\le a\le 1$ iff $y\ge y’$.  So further we assume that $y<y’<1$. Then $a_\max=\frac {1-y’}{1-y}$ and $f(a_\max)=2\sqrt{\frac{1-y’}{1-y}}.$
Let $ay^2y’-ay’+y-yy’^2=0$. 
If $y’-y’y^2=0$ then $y-yy’^2=0$ and we have no conditions for $a$, but these equalities hold iff $y=y’=0$ (then $f(a)=2$) or $y=y’=1$ (then $f(a)=2\sqrt{a}$,  $a_\max=1$ and $f(a_\max)=2$). Anyway, the required $a$ and $b$ exist for any $0\le c\le 1$. By the way, these cases were already excluded.   
If  $y’-y’y^2>0$ then $a=\frac{y-yy’^2}{y’-y’y^2}$. Since $y<y’$, $a<a_\max$ and $-1+a+yy’-ay^2<0$, so this $a$ is an admissible value. We have $f(a)=2\sqrt{\frac{1-y’^2}{1-y^2}}$. It is easy to check that $f(a)>f(0)$ and $f(a)>f(a_\max)$. 
The summary. The required $a$ and $b$ exist iff $(1-2x^2)c^2\le (1-2x’^2)$.
A: In fact, many equations and unknowns are few.  The system can be written as.
$$\left\{\begin{aligned}&(\lambda-a-1)^2=(c+b)^2+2(x'-ax)^2\\&(\lambda+a-1)^2=(c-b)^2+2(x'-ax)^2\end{aligned}\right.$$
This can be represented as the solution of the linear equation.
$$\left\{\begin{aligned}&x'-ax=2ps=2kt\\&c+b=p^2-2s^2\\&c-b=k^2-2t^2\\&\lambda-a-1=p^2+2s^2\\&\lambda+a-1=k^2+2t^2\end{aligned}\right.$$
A: I will use the notations $X=\sqrt{2}x$ and $X^\prime=\sqrt{2}x^\prime$ as it makes every expression simpler. 
I will prove the following:

If $c^2 +{X^\prime}^2\leq 1$, then: 
  
  
*
  
*Pick any $b$ in the interval $\left[0, \sqrt{1-{X^\prime}^2}-c\right]$
  
*If $X\neq 1$ put $a=\dfrac{-\left(1+XX^\prime\right) +\sqrt{\left(X+X^\prime\right)^2+\left(1-X^2\right)\left(c+b\right)^2}}{1-X^2}$
  
*If $X=1$ put $a=\dfrac{(c+b)^2+{X^\prime}^2-1}{2\left(1+X^\prime\right)}$
  
  
  (a, b) is a solution and the vanishing eigenvalue is $1 + a - \sqrt{(c+b)^2+(X'-aX)^2}$


Note that the smallest of the four eigenvalues is either $$\lambda_1=1 + a - \sqrt{(c+b)^2+(X'-aX)^2}\qquad  or\qquad \lambda_2=1 - a - \sqrt{(c-b)^2+(X'-aX)^2}$$ 
Your problem is equivalent to finding $a,b$ such that one of these situations occurs:


*

*$0=\lambda_1\leq\lambda_2$ 

*$0=\lambda_2\leq\lambda_1$.


I'll focus on the first situation. 
First, $0=\lambda_1 $ is equivalent to 
$$a+1\geq 0 \qquad and \qquad a^2 \left(1-X^2\right)+2a\left(1+XX^\prime\right)+\left(1-{X^\prime}^2-(c+b)^2\right)=0$$
When $X=1$ (the largest allowed value) this is a degree $1$ equation whose unique solution $$a=\frac{(c+b)^2+{X^\prime}^2-1}{2\left(1+X^\prime\right)}$$
is easily seen to satisfy $a+1\geq 0$ as required.
When $X\neq 1$, you need $\left(1+XX^\prime\right)^2-\left(1-X^2\right)\left(1-{X^\prime}^2-(c+b)^2\right)$ to be non-negative (we assume this for now and will look at it closer when we will be looking for $b$), there are then two solutions for $a$:
$$a=\dfrac{-\left(1+XX^\prime\right) \pm\sqrt{\left(1+XX^\prime\right)^2-\left(1-X^2\right)\left(1-{X^\prime}^2-(c+b)^2\right)}}{1-X^2}$$
but since $a+1$ needs to be nonnegative, there is hope only for the solution with "$+$" and the condition $a+1\geq 0$ becomes
$$\left(1+XX^\prime\right)^2-\left(1-X^2\right)\left(1-{X^\prime}^2-(c+b)^2\right)\geq \left(X^2+XX^\prime\right)^2 $$
since $u\leq v$ and $u^2\leq v^2$ are equivalent if $u$ and $v$ are known to be non-negative.
After crossing out the nonnegative common factor $\left(1-X^2\right)$ this simplifies into $$(X+X^\prime)^2+(c+b)^2\geq 0$$ which is always true.

So far, we have shown that $0=\lambda_1$ is equivalent to:
  
  
*
  
*When $X=1$,  $$a=\frac{(c+b)^2+{X^\prime}^2-1}{2\left(1+X^\prime\right)}$$
  
*When $X\neq 1$,  $$a=\dfrac{-\left(1+XX^\prime\right) +\sqrt{\left(X+X^\prime\right)^2+\left(1-X^2\right)\left(c+b\right)^2}}{1-X^2}$$
  

Now let's have a look at $\lambda_1\leq\lambda_2$. Because of all the square roots, it would be extremely painful to solve exactly. However there are some ranges of parameters that allow us to make this easy.
Note that $\lambda_1\leq\lambda_2$ has the form $$2a\leq \text{(some difference of square roots)}$$ 
I will solve $$2a\leq 0\leq \text{(that difference of square roots)}$$ 
First, $a\leq 0$ is equivalent to $$b\leq \sqrt{1-{X^\prime}^2}-c$$
It is easy to check in the case $X=0$, and not too difficult in the case $X\neq 0$.
Second, the difference of square roots on the right is $$\sqrt{(c+b)^2+(X'-aX)^2} - \sqrt{(c-b)^2+(X'-aX)^2}$$ and since $c$ is nonnegative, this difference is nonnegative if and only if $b\geq 0$. Finding $b$ such that  $0\leq b\leq \sqrt{1-{X^\prime}^2}-c$ is possible if and only if $c\leq \sqrt{1-{X^\prime}^2}$.
The announced result follows.
The result was double-checked by computer.
