Parisa D. answered • 08/13/20

PhD student in Computer Science and Master in Mathematics

Roots of the characteristic polynomial are eigenvalues of the matrix and the characteristic polynomial of matrix D is

Char(D)= det(D - xI), where I is the identity matrix of size 3.

Write this determinant using the Laplace expansion by the first column, so you get

det( D-xI) = (4-x) [ (e-x)(6-x) + 2f ].

Note that this the polynomial is of degree 3. As all three eigenvalues are identical (the assumption), this polynomial has only one root with multiplicity 3.

Since det( D-xI) = (4-x) [ (e-x)(6-x) + 2f ] = 0, so 4 must be the only solution. So, 4 is also the solution of the part

[ (e-x)(6-x) + 2f ] . Hence

(e-4)(6-4)+2f = 0 ⇒ f = 4 - e.

As 4 is the only solution, the discriminate of the polynomial (e-x)(6-x) + 2f = x^{2} - (e+6) x + 6e+2f must be 0.

Δ = (e+6)^{2} - 24e - 8f = 0.

Replacing f = 4 - e in the equation above we get

Δ = (e+6)^{2} - 24e - 8(4-e) = (e+6)^{2} - 16e - 32 = e^{2 }- 4e + 4 = (e-2)^{2 }= 0.

Hence e=2 and so f=2.

Now you have the matrix D, and you know the eigenvalues is x=4, find the associated eigenvector by looking at the null space of the matrix D-4I, where I is the identity matrix.

You may look at the following calculator and it shows step by step solution

https://matrixcalc.org/en/vectors.html#eigenvectors%28%7B%7B2,-1,2%7D,%7B0,4,0%7D,%7B-2,7,6%7D%7D%29

Parisa D.

08/13/20

Richard P.

08/13/20