In this post, All matrices are assumed to be in the ring of matrices with complex entries. Also, is the matrix with -entry and all other entries

Recall that a matrix is said to be *upper triangular* if for all Similarly, is *lower triangular* if for all If is either upper triangular or lower triangular, then is said to be *triangular*. If is triangular and also for all then we say that is *strictly* triangular.

**Problem 1**. Let be an matrix with the adjugate Show that

i) if is strictly upper triangular, then

ii) if is strictly lower triangular, then

**Solution**. I only prove i) and leave the proof of ii), which is similar to the proof of i), as an easy exercise.

Let be the -minor of i.e. the matrix that results from deleting the -th row and the -th column of Let

If then, since both the first column and the last row of are zero, either the first column or the last row or both the first column and the last row of is zero and hence in this case. But if then is just an upper triangular matrix with the diagonal entries and hence Thus

**Problem 2**. Let be a strictly triangular matrix with the adjugate Show that if and for some scalar then Also, show that the result does not always hold if

**Solution**. I’ll assume that is strictly upper triangular; the argument for strictly lower triangular is similar. So we are given that by Problem 1, and hence for all and Thus if then, since we have and hence implying So for all and hence

The result does not always hold if because, for example, but

**Remark**. In the solution of the next problem, we are going to use this important fact that if and then are *simultaneously triangularizable*, i.e. there exists an invertible matrix and upper triangular matrices such that and

**Problem 3**. Let be an matrix and suppose that there exists a scalar such that

Show that

**Solution**. It follows from the relations given in the problem that and hence, by the above Remark, there exists an invertible matrix and upper triangular matrices such that and Since the eigenvalues of are all zero and hence is strictly upper triangular. Also

and thus which gives

Hence and therefore, by Problem 2, giving