<s>[INST] 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 $0 \leq c \leq 1$.
 [/INST] 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 $(1-2x^2)c^2\le (1-2x’^2)$.
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