Prove that if A is an upper triangular matrix with distinct values on the main diagonal, then A is diagonalizable. I know that for a matrix to be diagonalizable, the eigenvectors of its eigenvalues must be linearly independent. However, I am unable to prove the theorem in the title.
It is clear that the eigenvalues of this matrix are listed in the diagonal entries. To see this, consider the matrix $\lambda I-A$ which is also triangular. 

Now this matrix must have $n$ different eigenvalue and must correspond to $n$ different eigenvectors. The linear independency could be proved by using linearity of matrix multiplications.
Given the matrix $A$ is a $k × k$ upper triangular matrix with distinct diagonal entries, $a_1, ~a_2, \cdots, ~a_k$. 

The determinant of an upper triangular matrix is the product of its diagonal entries. 

So $$f(t)= \det(A-tI)=(a_1-t)(a_2-t)\cdots(a_k-t) $$
Setting that to $~0~$, your $~k~$ eigenvalues are all distinct with multiplicity of $~1~$. 

The dimension of your eigenspace must be $~1~$. 

So by the test for diagonalizability, since your characteristic polynomial splits and the dimension of your eigenspace is equal to the algebraic multiplicity for each eigenvalue, $A$ is diagonalizable.