When is a 2x2 matrix diagonalizable based on its entries?

[simplyderp]

Homework Statement

Prove that A=[a,b;c,d]
is diagonalizable if -4bc < (a-d)^2
is not diagonalizale if -4bc > (a-d)^2

Homework Equations

For an nxn matrix, if there are n distinct eigenvalues then the matrix is diagonalizable.
For an nxn matrix, if there are n linearly independent eigenvetors then the matrix is diagonalizable.

The Attempt at a Solution

Characteristic equation for eigenvalues:
|λ-a,-b;-c,λ-d| = λ^2 + λ(a-d) + (ab - bc) = 0
λ = 0.5 * (d - a plus-or-minus sqrt((a-d)^2 - 4ad + 4bc))

For part 1, I need to show that there are two distinct solutions to this quadratic equation (b^2 - 4ac > 0: a,b,c from general quadratic eq - not from this problem)
I know that (a-d)^2 + 4bc > 0

However, I do not know how to show that (a-d)^2 - 4ad + 4bc > 0

 

[Mark44]

Homework Statement

Prove that A=[a,b;c,d]
is diagonalizable if -4bc < (a-d)^2
is not diagonalizale if -4bc > (a-d)^2

Homework Equations

For an nxn matrix, if there are n distinct eigenvalues then the matrix is diagonalizable.
For an nxn matrix, if there are n linearly independent eigenvetors then the matrix is diagonalizable.

The Attempt at a Solution

Characteristic equation for eigenvalues:
|λ-a,-b;-c,λ-d| = λ^2 + λ(a-d) + (ab - bc) = 0
λ = 0.5 * (d - a plus-or-minus sqrt((a-d)^2 - 4ad + 4bc))

For part 1, I need to show that there are two distinct solutions to this quadratic equation (b^2 - 4ac > 0: a,b,c from general quadratic eq - not from this problem)
I know that (a-d)^2 + 4bc > 0

However, I do not know how to show that (a-d)^2 - 4ad + 4bc > 0

Expand the (a - d)² term and then combine like terms. You're almost there!

 

[I like Serena]

You've got a minus sign wrong in your characteristic equation.

 

[Rockoz]

I had trouble with the same exact problem recently. Your process is correct. You just have a minus sign error.