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

1. Feb 12, 2010

Kate2010

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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.

2. Feb 12, 2010

vela

Staff Emeritus
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 $(a-\lambda)^2=0$.

3. Feb 12, 2010

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.

4. Feb 13, 2010

HallsofIvy

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

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 $(a-d)^2+ 4bc\ge 0$?

5. Feb 13, 2010

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?

6. Feb 13, 2010

vela

Staff Emeritus
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.)