Find all real values of k for which A is diagonalizable

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In summary, diagonalizability refers to the ability to find a diagonal matrix that is similar to the original matrix, and a matrix is diagonalizable if it has n linearly independent eigenvectors. Finding all real values of k is important in determining if a matrix is diagonalizable, and this information has practical applications in various fields. To find the real values of k, we need to find the eigenvalues of A and determine if there are enough linearly independent eigenvectors. Diagonalizable matrices have significant uses in fields such as physics, engineering, and economics, allowing for efficient computation and analysis of systems involving linear transformations.
  • #1
war485
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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?
 
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  • #2
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.
 
  • #3
war485 said:

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?
 
  • #4
war485 said:
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?
 
  • #5
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.
 

1. What does it mean for a matrix to be diagonalizable?

Diagonalizability refers to the ability to find a diagonal matrix that is similar to the original matrix. This means that the original matrix and the diagonal matrix have the same eigenvalues, but the diagonal matrix has 0s in all non-diagonal entries.

2. How do you determine if a matrix is diagonalizable?

A matrix is diagonalizable if and only if it has n linearly independent eigenvectors, where n is the size of the matrix. This means that the matrix can be expressed as PDP^-1, where P is the matrix of eigenvectors and D is the diagonal matrix of eigenvalues.

3. What is the importance of finding all real values of k for which A is diagonalizable?

Finding all real values of k allows us to determine when the matrix is diagonalizable and when it is not. This information is important in various fields of mathematics, such as linear algebra and differential equations.

4. How do you find the real values of k for which A is diagonalizable?

To find the real values of k, we need to find the eigenvalues of A and then determine if there are enough linearly independent eigenvectors to form a diagonal matrix. This can be done by solving the characteristic equation and finding the null space of A-kI.

5. What is the significance of a diagonalizable matrix in practical applications?

Diagonalizable matrices have many important applications in fields like physics, engineering, and economics. They allow for efficient computation and analysis of systems that involve repeated linear transformations, making them useful in modeling various real-world scenarios.

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