How to find eigenvalues of this 3x3 matrix?

Question

If I have this 3x3 matrix:
1 a b
0 1 c
0 0 1

How exactly do I find the eigenvalues of this 3x3 matrix? I can do it easily for a 2x2 using the characteristic equation of x²-(trA*x)+detA, but don't know if there is one for a 3x3. Please explain simply if possible :) thanks.

Answer 1

In a triangular matrix, the numbers on the diagonal will be the eigenvalues.

Your matrix has an eigenvalue of 1 (with multiplicity 3).



An upper triangular matrix has nothing but zeros below the diagonal going from the top left to the bottom right. This is what you have.



A lower triangular matrix has nothing but zeros above the same diagonal. The transpose of your matrix would be a lower triangular matrix.

1 0 0

a 1 0

b c 1



The theorem is valid on matrices of any size not just 3x3 matrices.

Answer 2

Eigenvalues For A 3x3 Matrix

Answer 3

The characteristic equation of any matrix A is Det(A - λI) = 0



For a 3x3 matrix {{a,b,c},{d,e,f},{g,h,i}}, the characteristic equation is Det({{a-λ,b,c},{d,e-λ,f},{g,h,i-λ}}) = 0.



For your matrix, you get -λ^3 + 3λ^2 - 3λ + 1 = 0, which factorises to (1-λ)^3 = 0.



So your eigenvalues are a triple repeated root of λ = 1. So whatever a,b,c are, the eigenvalue is just 1.

Answer 4

"A triangular matrix is a special kind of square matrix where either all the entries below or all the entries above the main diagonal are zero."



Please check the following sources for easy understanding.