Scalars z such that A-zI is singular

  • Thread starter dlevanchuk
  • Start date
  • #1
29
0

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 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?

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,995
1,572


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

Related Threads on Scalars z such that A-zI is singular

Replies
5
Views
2K
Replies
5
Views
9K
Replies
5
Views
683
Replies
7
Views
9K
  • Last Post
Replies
22
Views
6K
Replies
0
Views
2K
  • Last Post
Replies
2
Views
3K
Replies
3
Views
3K
Replies
5
Views
24K
Top