A=( 3 2 4 2 0 2 4 2 3) the eigenvectors of A when λ=-1,it is (-1/2,1,0) and (-1,0,1) re the eigenvectors of A corresponding to the eigenvalue λ = −1 orthogonal with respect to the usual dot product? If not, apply the Gram-Schmidt process to turn them into an orthonormal set of eigenvectors

The usual dot product of two vectors is given by the sum of the products of their corresponding components. Two vectors are orthogonal if their dot product is zero.

Let's check if the given eigenvectors are orthogonal:

v1 = (-1/2, 1, 0) v2 = (-1, 0, 1)

The dot product of v1 and v2 is:

v1 . v2 = (-1/2)(-1) + 10 + 0*1 = 1/2

Since the dot product is not zero, the vectors are not orthogonal.

To make them orthogonal, we can apply the Gram-Schmidt process. The Gram-Schmidt process takes a set of vectors and generates an orthogonal (or orthonormal, if we also normalize the vectors) set of vectors.

Let's apply the Gram-Schmidt process to v1 and v2:

1. Start with the first vector, v1. We'll use this as our first vector in the orthogonal set:

   u1 = v1 = (-1/2, 1, 0)

2. Next, subtract the projection of v2 onto u1 from v2 to get the second vector in the orthogonal set:

   u2 = v2 - ((v2 . u1) / (u1 . u1)) * u1

   u2 = (-1, 0, 1) - ((-1, 0, 1) . (-1/2, 1, 0)) / ((-1/2, 1, 0) . (-1/2, 1, 0)) * (-1/2, 1, 0) = (-1, 0, 1) - ((1/2) / (1/4 + 1)) * (-1/2, 1, 0) = (-1, 0, 1) - (1/3) * (-1/2, 1, 0) = (-1, 0, 1) - (-1/6, 1/3, 0) = (-5/6, -1/3, 1)

So, the orthogonal set of vectors is {(-1/2, 1, 0), (-5/6, -1/3, 1)}.

To make them orthonormal (unit vectors), we can normalize each vector by dividing by its magnitude:

u1_normalized = u1 / ||u1|| = (-1/2, 1, 0) / sqrt((-1/2)^2 + 1^2 + 0^2) = (-1/3, 2/3, 0)

u2_normalized = u2 / ||u2|| = (-5/6, -1/3, 1) / sqrt((-5/6)^2 + (-1/3)^2 + 1^2) = (-5/7, -1/7, 3/7)

So, the orthonormal set of vectors is {(-1/3, 2/3, 0), (-5/7, -1/7, 3/7)}.