Find all real values of k for which A is diagonalizable

[war485]

Homework Statement

find all real values of k for which A is diagonalizable.

A = [ 1 1 ]
[ 0 k ]

The Attempt at a Solution

let L = lamba = eigenvalue

I did this:
det(A - LI) = L² - Lk - L + k

so then it sort of looks like a quadratic so I did this:
L² - Lk - L + k = 0
L² - L(k+1) + k = 0
and used the quadratic equation
x = (-b +- sqrt(b²-4ac)) / 2a
so then I got stuck at this point:

((k+1) +- (k-1)) / 2

and this is where I got really really stuck. I'm not sure how or what to conclude about matrix A being diagonalizable for all real values of k.


or can I skip the quadratic equation method and just do this instead?

(1-L)(k-L) = 0
then
L = 1 and L = k
so then matrix A is diagonalizable for any real number k except 0?

 

[VeeEight]

What is the criteria for a matrix to be diagonalizable? If I remember correctly, one is that it's characteristic polynomial must split. If you are working in the reals then check for which real numbers k give an irreducible polynomial and exclude them from the final answer. Then check if the other criteria are satisfied for the remaining reals.

 

[HallsofIvy]

Homework Statement

find all real values of k for which A is diagonalizable.

A = [ 1 1 ]
[ 0 k ]

The Attempt at a Solution

let L = lamba = eigenvalue

I did this:
det(A - LI) = L² - Lk - L + kultiply it out? det(A- LI)= (1-L)(k- L)= 0 so the eigenvalues are L= 1 and L= k.

so then it sort of looks like a quadratic so I did this:
L² - Lk - L + k = 0
L² - L(k+1) + k = 0
an+d used the quadratic equation
x = (-b +- sqrt(b²-4ac)) / 2a
so then I got stuck at this point:

((k+1) +- (k-1)) / 2

k+1+ (k-1)= 2k. k+1- (k-1)= 2.

and this is where I got really really stuck. I'm not sure how or what to conclude about matrix A being diagonalizable for all real values of k.


or can I skip the quadratic equation method and just do this instead?

(1-L)(k-L) = 0
then
L = 1 and L = k
so then matrix A is diagonalizable for any real number k except 0?

Yes, you are allowed to do it the obvious way!

No, under what conditions is a matrix diagonalizable?

 

[D H]

A = [ 1 1 ]
[ 0 k ]

The Attempt at a Solution

let L = lamba = eigenvalue

I did this:
det(A - LI) = L² - Lk - L + k

Why did you do that? To get to that expression you had to expand (1-L)*(k-L)=0. There is no need to use the quadratic form when you already have a factorization. You can instead directly read the solutions L from (1-L)*(k-L)=0.

How do the eigenvalues indicate whether a matrix is diagonalizable?

 

[Hurkyl]

I'm not sure how or what to conclude about matrix A being diagonalizable for all real values of k.

The most straightforward way is to actually perform the diagonalization procedure A. Make sure you divide into cases where appropriate: for example, if you would divide by k, then you need to split the problem into two cases: one where k is nonzero, and then handle the case where k is zero separately.

In one (or more) cases, the diagonalization procedure will fail. Then you'll have your answer.


This is standard fare -- the simplest response to "when can't you do X" is to actually do X, and see what obstacles impede your progress.