Scalars z such that A-zI is singular

[dlevanchuk]

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

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 z_1/2 = ((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?

 

[vela]

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

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 z_1/2 = ((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?

That's fine, though you should have (a+d) instead of (a+b) in your expression for the roots.