When a matrix isn't diagonalizable

[says]

Homework Statement

Determine if the matrix is diagonalizable or not.

A=
[ 3 -1 ]
[ 1 1 ]

Homework Equations

Eigenvalues = det(A-Iλ)
determinant of a 2x2 matrix = ad-bc

The Attempt at a Solution

Eigenvalues = det(A-Iλ)

[ 3 -1 ] - [ λ 0 ] = [ 3 -λ -1 ]
[ 1 1 ] [ 0 λ ] [ 1 1-λ ]

det =

(3 -λ*1-λ) -1*(-1)
λ²-4λ+4
(λ-2)(λ-2) = 0
λ = 2

Only one eigenvalue -- for this matrix A to be diagonalizable we need two eigenvalues so that

D=
[ λ1 0 ]
[ 0 λ2 ]

where λ1 and λ2 are 2 eigenvalues.

Am I correct in my assumptions?

 

[Samy_A]

Only one eigenvalue -- for this matrix A to be diagonalizable we need two eigenvalues so that

Take the matrix $$B=\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}$$

What are the eigenvalues of B?
Is B diagonalizable?

Not saying that your final conclusion is necessarily wrong, but what about the argumentation ?

 

[says]

The eigenvalue of your question would be = 1.

 

[Mark44]

Homework Statement

Determine if the matrix is diagonalizable or not.

A=
[ 3 -1 ]
[ 1 1 ]

Homework Equations

Eigenvalues = det(A-Iλ)
determinant of a 2x2 matrix = ad-bc

The Attempt at a Solution

Eigenvalues = det(A-Iλ)

[ 3 -1 ] - [ λ 0 ] = [ 3 -λ -1 ]
[ 1 1 ] [ 0 λ ] [ 1 1-λ ]

det =

(3 -λ*1-λ) -1*(-1)
λ²-4λ+4
(λ-2)(λ-2) = 0
λ = 2

Only one eigenvalue -- for this matrix A to be diagonalizable we need two eigenvalues so that

D=
[ λ1 0 ]
[ 0 λ2 ]

where λ1 and λ2 are 2 eigenvalues.

Am I correct in my assumptions?

Not necessarily. Sometimes a single eigenvalue can be associated with more than one eigenvector.

 

[Samy_A]

The eigenvalue of your question would be = 1.

Yes, and surely you'll accept that B is diagonalizable.
This shows that not having 2 eigenvalues doesn't necessarily mean that a 2*2 matrix is not diagonalizable.

What are the eigenvectors of your matrix A?

 

[says]

In my question I get an eigenvalue = 2

If I plug that into (A-λI)*v1 = 0
[ 3 -1 ] - [ 2 0 ]
[ 1 1 ] [ 0 2 ] * v1 = 0

[ 1 -1]
[ 1 -1] = 0

[1 -1] (x)
[1 -1] (y) = 0

x-y = 0
x-y = 0

iff x = 1 y = 1

eigenvector 1 = (1,1)

Just reading some more theory now. Because the matrix is 2x2, this means if it is to be diagonalizable it needs to have 2 linearly independent eigenvectors. I only have one.

 

[Samy_A]

In my question I get an eigenvalue = 2

If I plug that into (A-λI)*v1 = 0
[ 3 -1 ] - [ 2 0 ]
[ 1 1 ] [ 0 2 ] * v1 = 0

[ 1 -1]
[ 1 -1] = 0

[1 -1] (x)
[1 -1] (y) = 0

x-y = 0
x-y = 0

iff x = 1 y = 1

eigenvector 1 = (1,1)

Just reading some more theory now. Because the matrix is 2x2, this means if it is to be diagonalizable it needs to have 2 linearly independent eigenvectors. I only have one.

I think this is essentially correct.
But it is so difficult to parse what you write.

What you proved is that the eigenspace of the eigenvalue 2 has dimension 1. And it should have been 2 for the matrix to be diagonalizable.

 

[Mark44]

Just reading some more theory now. Because the matrix is 2x2, this means if it is to be diagonalizable it needs to have 2 linearly independent eigenvectors. I only have one.

Yes, that's correct. Since you have only one eigenvector, but need two of them, your matrix is not diagonalizable.