Question

Let t be the trace of the matrix \[\Large A=\left[\begin{matrix}\frac{|x+y|}{|x|+|y|} & \alpha_1 & \beta_1 \\ \alpha_2 & \frac{|y+z|}{|y|+|z|}& \beta_2 \\ \alpha_3 & \beta_3 & \frac{|z+x|}{|z|+|x|}\end{matrix}\right]\] then
1). \(0 \le t \le 3 \)
2). \(1 \le t \le 2\)
3). \(1 \le t \le 3\)
4). \(-1 \le t \le 1\)

Answer 1

@ganeshie8

Answer 2

the trace is the sum of the diagonals

Answer 3

also, consider https://en.wikipedia.org/wiki/Triangle_inequality

Answer 4

1) obviously the trace is positive

2) (@inkyvoyd) Take \(x=y=z=1\), what is the trace?

3) is the lower bound 0 or 1?
\(\underline{\text{Hint}}\): if \(a\) and \(b\) have the same sign, then \(\frac{|a+b|}{|a|+|b|} = 1\).

Answer 5

The right answer is I believe
\[
1\leq t\le3
\]

Answer 6

If x,y,z have the some sign then t=3
if two of them are positive say x and y, then
\[\frac{\left| x+y\right| }{\left| x\right| +\left| y\right| }+\frac{\left| x+z\right|
}{\left| x\right| +\left| z\right| }+\frac{\left| y+z\right| }{\left| y\right| +\left|
z\right| }\geq \frac{\left| x+z\right| }{\left| x\right| +\left| z\right|
}+\frac{\left| y+z\right| }{\left| y\right| +\left| z\right| }+1\geq 1
\]
By symmetry this will do it

Answer 7

That's also how I would finish it @eliesaab @jiteshmeghwal9