What does it mean for the characteristic polynomial to equal 0?

[Kate2010]

Homework Statement

I'm working on a problem which is show that the matrix A = ($$^{a}_{c}$$ $$^{b}_{d}$$) is diagonalisable iff (a-d)² + 4bc > 0 or a-d = b = c = 0

Homework Equations

The Attempt at a Solution

I've shown if (a-d)² + 4bc > 0 then the characteristic polynomial has 2 distinct roots so 2 distinct eigenvalues so is diagonalisable.

However, if (a-d)² + 4bc $$\geq$$ 0 then it has a repeated root, so I have split it into cases:

If a-d = b = c then A is already diagonal

However, if a-d = b = 0 (c non-zero), the characteristic polynomial of A is already equal to 0. Does this mean A has no eigenvalues? Infinitely many? I will encounter a similar problem if a-d = c = 0 (b non-zero).

I am also unsure how to tackle (a-d)² = 4bc.

 

[vela]

Does the problem statement have a greater-than sign or a greater-than-or-equal-to sign in it? As you have written it, you don't need to worry about the cases you're talking about since you've already shown in A is already diagonal or it has two distinct roots. I'm guessing it should have been the greater-than-or-equal-to sign in the original statement.

If a-d=0 and b=0 and c not zero, you have a matrix of the form

$$\begin{pmatrix}a & 0 \\ c & a\end{pmatrix}$$

which will have a characteristic polynomial of the form $(a-\lambda)^2=0$.

 

[Kate2010]

The question goes show the matrix $$\begin{pmatrix}a & b \\ c & d\end{pmatrix}$$ is diagonal if and only if a-d = b = c = 0 or (a-d)²+4bc > 0.

So I have covered the case for (a-d)²+4bc > 0

I have also shown if a-d=b=c=0 then it is diagonal.

However I need to show that if a-d=b = 0 but c non-zero, a-d=c=0 but b non-zero, and (a-d)² = -4bc (so b or c would have to be negative) do not lead to diagonal matrices.

 

[HallsofIvy]

Homework Statement

I'm working on a problem which is show that the matrix A = ($$^{a}_{c}$$ $$^{b}_{d}$$) is diagonalisable iff (a-d)² + 4bc > 0 or a-d = b = c = 0

Homework Equations

The Attempt at a Solution

I've shown if (a-d)² + 4bc > 0 then the characteristic polynomial has 2 distinct roots so 2 distinct eigenvalues so is diagonalisable.

However, if (a-d)² + 4bc $$\geq$$ 0 then it has a repeated root, so I have split it into cases:

You mean the case (a-d)²+ 4bc= 0 since you have already handled (a-d)²+ 4bc> 0.

If a-d = b = c then A is already diagonal

However, if a-d = b = 0 (c non-zero), the characteristic polynomial of A is already equal to 0. Does this mean A has no eigenvalues? Infinitely many? I will encounter a similar problem if a-d = c = 0 (b non-zero).

You are only talking about 2 by 2 matrices so there cannot be more than 2 eigenvalues. In this particular case, 0 is a double eigenvalue.
The real problem is that you are trying to prove something that is not true.
The matrix
$$\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$$
satisifies the condition that $(a-d)^2+ 4bc= 0^2+ 0= 0$ but is NOT diagonalizable.

Your original statement said $(a-d)^2+ 4bc> 0$ and you have already proved that. Why did you change to $(a-d)^2+ 4bc\ge 0$?

I am also unsure how to tackle (a-d)² = 4bc.

 

[Kate2010]

The way I've tried to answer this question is to find the characteristic polynomial of the matrix which is x² - (a+d)x + ad-bc

This has 2 distinct real roots if (a-d)² +4bc > 0 (so is definitely diagonalisable)

It has a repeated real root if (a-d)² + 4bc = 0, in which case don't know if it is diagonalosable. This is the part I'm trying to split into cases. I think I have managed the first 3, but I don't know how to show if (a-d)² = -4bc (could happen if b or c was negative) then the matrix is not diagonalisable?

 

[vela]

You've shown the RHS of the iff implies the LHS. What you want to show now is the LHS implies the RHS. It was a little confusing why you were still messing around with cases on the RHS, but your approach seems to be to prove the contrapositive, ~RHS -> ~LHS, which is logically fine. It might be easier, however, to prove LHS->RHS directly, depending on what you know about diagonalizable matrices. (I don't know how the proof goes; I'm just throwing the idea out there for you to ponder.)