Diagonalizability of a matrix. For a 3 X 3 upper triangular matrix,can we look at the matrix and tell whether its diagonalizable? What conditions need to be satisfied by the elements for it to be diagonalizable?
 A: The 3x3 matrix $A$ is diagonalizable if and only if it has three linearly independent eigenvectors.  When $A$ is triangular, its eigenvalues are in plain view right there on the diagonal.  Can you find a linearly independent set of eigenvectors that corresponds to these eigenvalues?
If all three eigenvalues of $A$ are the same and $A$ is not already diagonal, you will not be able to do this.  For in that case the eigenvalue equation is
$$
(rI + N) v = rv
$$
where $N$ is obtained from $A$ by setting each diagonal element $r$ to zero. Any $v$ that satisfies this equation must be in the nullspace of $N$, but this nullspace has dimension 3 if and only if $N$ is the zero matrix.
If all three eigenvalues of $A$ are different, then it's a standard result that $A$ is diagonalizable, as has been observed already in the comments.
This leaves the case in which two eigenvalues are the same and the third is different. Suppose first that $A$ has the form
$$
A = \left[\begin{array}{ccc} r & b & c \\ 0 & r & d \\ 0 & 0 & s\end{array}\right]
$$
Can we find two linearly independent solutions to $Av = rv$?  The answer is yes if and only if the nullity of the matrix
$$
\left[\begin{array}{ccc} 0 & b & c \\ 0 & 0 & d \\ 0 & 0 & s-r\end{array}\right]
$$
is 2, which holds only if $b=0$ assuming that $r$ and $s$ are different.
Other cases in which two eigenvalues are the same can be handled similarly.
