"practice 7 problem 14"

No. There are 2 eigenvalues which have multiplicity 2 and 1 respectively.

Eigenvalue = 3 has multiplicity 2 and therefore must appear in the matrix twice (as it does).

Eigenvalue = 5 has multiplicity 1 and appears only once.

Then, for eigenvalue = 3 we can find 2 linearly independent eigenvectors which means that there will be 2 Jordan blocks for this eigenvalue.

We also know that for eigenvalue = 5 there is only 1 eigenvector so there is 1 Jordan block.

So we have 3 Jordan blocks for a 3 x 3 matrix which can only mean that each block is a 1 x 1 matrix.

If this were not the case then our Jordan matrix would be too big.

So if each block is 1 x 1, then there are no 1's on the diagonal above (since it does not exist).