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## Homework Statement

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

## Homework Equations

## 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

z

^{2}- z(a + d) + (ad - b

^{2}) = 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 z

_{1/2}= ((a + b) +/- sqrt[(a - d)

^{2}+4b

^{2}])/2 (sorry for the messy equation :-S)

Since (a - d)

^{2}+4b

^{2}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?