Prove that if A is an upper triangular matrix with distinct values on the main diagonal, then A is diagonalizable.

Question

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.

Answer 1

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.

Answer 2

To clarify, since the matrix is upper triangular, it is already in Schur form and thus its eigenvalues are on the diagonal; every matrix A is orthonormally similar to an upper triangular matrix U with the eigenvalues on the diagonal s.t. $A = QUQ^T$, which is known as the Schur form of a matrix. Then, as mentioned in another comment, you just need to prove that eigenvectors of distinct eigenvalues are linearly independent (see here: How to prove that eigenvectors from different eigenvalues are linearly independent). Diagonalizability obviously follows from this.

Answer 3

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.