Answer
a. True
b. True
c. False
d. False
Work Step by Step
a. If a matrix A is orthogonally diagonalizable, we should be able to write $A=PDP^T$ where $A^T=(PDP^T)^T=(P^T)^TD^TP=PD^PP=PDP^T=A$. So A is symmetric
b. The two equations show us that u and v are eigenvectors of A corresponding to eigenvalues 3 and 4. Because eigenvectors corresponding to different eigenvalues are orthogonal, $\vec{u}\cdot\vec{v}=0$
c. It needs to have n eigenvalues counting multiplicity such that each eigenvalue with multiplicity a has a eigenvectors as a basis for its eigenspace.
d. $\vec{v}$ needs to be a unit vector that is in the basis.
