Respuesta :
Answer:
Step-by-step explanation:
We are given the matrix
[tex] A = \left[\begin{matrix}4&0&0 \\ 1&3&0 \\-2&3&-1 \end{matrix}\right] [/tex]
a) To find the characteristic polynomial we calculate [tex]\text{det}(A-\lambda I)=0[/tex] where I is the identity matrix of appropiate size. in this case the characteristic polynomial is
[tex]\left|\begin{matrix}4-\lambda&0&0 \\ 1&3-\lambda&0 \\-2&3&-1-\lambda \end{matrix}\right|=0[/tex]
Since this matrix is upper triangular, its determinant is the multiplication of the diagonal entries, that is
[tex](4-\lambda)(3-\lambda)(-1-\lambda)=(\lambda-4)(\lambda-3)(\lambda+1)=0[/tex]
which is the characteristic polynomial of A.
b) To find the eigenvalues of A, we find the roots of the characteristic polynomials. In this case they are [tex]\lambda=4,3,-1[/tex]
c) To find the base associated to the eigenvalue lambda, we replace the value of lambda in the expression [tex]A-\lambda I[/tex] and solve the system [tex](A-\lambda I)x =0[/tex] by finding a base for its solution space. We will show this process for one value of lambda and give the solution for the other cases.
Consider [tex]\lambda = 4[/tex]. We get the matrix
[tex]\left[\begin{matrix}0&0&0 \\ 1&-1&0 \\-2&3&-5 \end{matrix}\right] [/tex]
The second line gives us the equation x-y =0. Which implies that x=y. The third line gives us the equation -2x+3y-5z=0. Since x=y, it becomes y-5z =0. This implies that y = 5z. So, combining this equations, the solution of the homogeneus system is given by
[tex](x,y,z) = (5z,5z,z) = z(5,5,1)[/tex]
So, the base for this eigenspace is the vector (5,5,1).
If [tex]\lambda = 3[/tex] then the base is (0,4,3) and if [tex]\lambda = -1[/tex] then the base is (0,0,1)
