Consider the (2x2) symmetric matrix A = [a b; b d]
Show that there are always real scalars z such that A-zI is singular [Hint: Use quadratic formula for the roots from the previous exercise ] t^2 - (a + d)t + (ad - bc) =0
The Attempt at a Solution
I just want to see if I did the problem correctly..
First, i did the whole cross multiplication, and simplified to the required quadratic formula, given in the hint
z2 - z(a + d) + (ad - b2) = 0
Then I found roots to the above equation, but thatss where I want for you guys to double check my logic..
The roots for z1/2 = ((a + b) +/- sqrt[(a - d)2+4b2])/2 (sorry for the messy equation :-S)
Since (a - d)2+4b2 is always positive, then the root for the above equation does exist, therefore there are real scalars z such that A-zI is singular...
Is that good, or should I add something else?