This problem is made harder by the fact that is not the characteristic polynomial of the given matrix. I suspect that the matrix should be (with a +2 in the bottom left corner). That matrix at least appears to have the right eigenvalues.

The key fact that you need to solve the problem is that if is a unit vector (I'll write vectors as rows rather than columns, to save space), then the projection onto the one-dimensional space spanned by has matrix (with respect to the standard basis).

For the matrix , a unit eigenvector corresponding to the eigenvalue 1 is . The projection onto the eigenspace of the eigenvalue x=1 is therefore .

The projection onto the one-dimensional eigenspace of the eigenvalue x=–1 is calculated in a similar way.

For the two-dimensional eigenspace of the repeated eigenvalue x=3, there are various possible methods to find the projection onto it.

Method 1 (the slow but safe method). Find two orthonormal eigenvectors for the eigenvalue x=3, form the projections onto the one-dimensional spaces spanned by each of them, then add those two projections to get .

Method 2 (the lazy way). Use the fact that sum of , and must be the identity matrix. Since you have already found and , you can just write down .

Method 3 (the belt and braces method). Use Method 1 to find . Then use Method 2 to check that your answer is right.

The advantage of Method 2 is that it is quick. The disadvantage is that it assumes that the eigenspace of the eigenvalue x=3 is two-dimensional. There is in principle no guarantee of this. If the eigenvalue x=3 only has one linearly independent eigenvector then the matrix A will not have a spectral decomposition at all.