Eigenvalues and proof

Question

Prove that a 2x2 matrix with only one eigenvalue is diagonalizable if and only if it is already diagonal.

Answer 1

The 2x2 matrix has a quadratic characteristic equation.

Since only one eigenvalue is given, the characteristic equation has double roots.

EX (X- k)(X-k) = 0

     X=k 

 

  --------->

  It is given that A is diagonalizable for proving the if and only if statement from left to right.

 

The eigenvalue is calculated as:

 

   det(  A - lambda*I) = 0

 

    (A11- lamda)(A22-lambda) - A21*A12 = 0

 

     A11*A22 - (A11+A22)*lambda + lambda^2 - A21*A12 = 0

 

 

    A11*A22 - A21*A12 - (A11+A22)*lamda + lambda^2 = 0

 

     det(A) - (A11+A22)*lambda + lambda^2 = 0

 

     

     lambda^2 - (A11+A22)*lambda + lambda^2 = 0  <---- For any diagonizable matrix, the determinant is the

                                                                                       product of the eigenvalues, which is lambda*lambda = lambda^2

 

     lambda^2 cancels

 

      -(A11+A22)* lambda = 0

 

      A11 +A22 = 0

 

    A11 = -A22

 

   So the matrix Must be 

 

  (X   A12)

  (A21  -X)

 

  which is diagonal

 

 

<--------  diagonal implies diagonalizable