Exercise 3.4.11. If the matrix is both upper triangular and orthogonal, show that must be a diagonal matrix.
Answer: Let be an by matrix. Since is upper triangular we have
where for . Our goal is to prove that is also diagonal, with for .
Since is an orthogonal matrix all of its columns are orthonormal, and all of its rows are also orthonormal. (See Remark 2 on page 169.) So, in particular, for column 1 we have and for row 1 we have .
But since is upper triangular we also have for , so that for column 1 we have
Since we have .
Turning to row 1 we have
or . But since the square of any nonzero number is positive this implies that for .
For row 1 we thus have for . In other words, for row 1 all off-diagonal elements are zero. Given the previous result that the matrix must therefore look as follows:
The argument then proceeds by induction for the other rows: Suppose that for some we have all off-diagonal elements equal to zero for rows 1 through . More formally, we assume that for we have for .
Consider the situation for row . Since column is orthonormal we have
Since is upper triangular we have for . All elements in the second sum above are therefore zero:
But by our assumption all off-diagonal elements in rows 1 through are also zero. Therefore for (or ). All elements in the remaining sum are therefore zero, and we have
so that or .
Since row is orthonormal we have
But since is upper triangular we have for (or ), and from above we have . We thus have
so that and thus for .
Since we already had for we therefore have for .
In other words, all off-diagonal entries in row are zero given our assumption that off-diagonal entries in rows 1 through were zero. Since this assumption is true for row 1, we have all off-diagonal entries equal to zero for all rows 1 through . More formally, for and .
Since for any orthogonal upper triangular matrix is therefore also a diagonal matrix. In addition, each element on the diagonal must be either 1 or -1.
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fifth Edition and the accompanying free online course, and Dr Strang’s other books.