Find the eigenvalues and eigenvectors of A = ([2, 0 , 0], [0, 1, 1], [0, 3, 3])

We can use the characteristic equation, det(A - kI) = 0 to find the eigenvalues of A. Performing this, we see that(2-k) * ( (1-k)(3-k) - 3 ) = 0.Immediately, we can see a root is k = 2, which is our first eigenvalue, so we now just need to solve (1-k)(3-k) - 3 = 0 for our other two. This gives k2 - 4k = 0, which we can factorise to obtain (k -4)(k ) = 0.So our other two roots are 4 and 0.Now, we perform Ax = k x for a general vector x = (a, b, c) to obtain our eigenvectors, x_1, x_2, x_3. k = 2 gives x_1 = (1, 0 , 0), x_2 = (0,1,3), and x_3 = (0, -1, 1). We can see this for x_1 as A(a,b,c) = (2a+ 2b + 2b, b+c, 3b + 3c) = 2*(a+b+c, (b+c)/2, 2/3*(b+c)). So b + c must be equal to zero, and we are free to choose a = 1.