Linear Algebra and Its Applications, Exercise 3.4.11

Exercise 3.4.11. If the matrix Q is both upper triangular and orthogonal, show that Q must be a diagonal matrix.

Answer: Let Q be an n by n matrix. Since Q is upper triangular we have

Q = \begin{bmatrix} q_{11}&q_{12}&\cdots&q_{1n} \\ &q_{22}&\cdots&q_{2n} \\ &&\ddots&\vdots \\ &&&q_{nn} \end{bmatrix}

where q_{ij} = 0 for i > j. Our goal is to prove that Q is also diagonal, with q_{ij} = 0 for i \ne j.

Since Q 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 \sum_i q_{i1}^2 = 1 and for row 1 we have \sum_j q_{1j}^2 = 1.

But since Q is upper triangular we also have q_{i1} = 0 for i > 1, so that for column 1 we have

1 = \sum_i q_{i1}^2 = q_{11}^2 + \sum_{i > 1} q_{i1}^2 = q_{11}^2 + \sum_{i > 1} 0^2 = q_{11}^2

Since q_{11}^2 = 1 we have q_{11} = \pm 1.

Turning to row 1 we have

1 = \sum_j q_{1j}^2 = q_{11}^2 + \sum_{j > 1} q_{1j}^2 = 1 + \sum_{j > 1} q_{1j}^2

or \sum_{j > 1} q_{1j}^2 = 0. But since the square of any nonzero number is positive this implies that q_{1j} = 0 for j > 1.

For row 1 we thus have q_{1j} = 0 for j \ne 1. In other words, for row 1 all off-diagonal elements are zero. Given the previous result that q_{11} = \pm 1 the matrix Q must therefore look as follows:

Q = \begin{bmatrix} \pm 1&&& \\ &q_{22}&\cdots&q_{2n} \\ &&\ddots&\vdots \\ &&&q_{nn} \end{bmatrix}

The argument then proceeds by induction for the other rows: Suppose that for some k \ge 1 we have all off-diagonal elements equal to zero for rows 1 through k. More formally, we assume that for 1 \le i \le k we have q_{ij} = 0 for i \ne j.

Consider the situation for row k+1. Since column k + 1 is orthonormal we have

1 = \sum_i q_{i,k+1}^2 = \sum_{i < k+1} q_{i,k+1}^2 + q_{k+1,k+1}^2 + \sum_{i > k+1} q_{i,k+1}^2

Since Q is upper triangular we have q_{i,k+1} = 0 for i > k+1. All elements in the second sum above are therefore zero:

1 = \sum_i q_{i,k+1}^2 = \sum_{i < k+1} q_{i,k+1}^2 + q_{k+1,k+1}^2 + \sum_{i > k+1} 0^2

= \sum_{i < k+1} q_{i,k+1}^2 + q_{k+1,k+1}^2

But by our assumption all off-diagonal elements in rows 1 through k are also zero. Therefore q_{i,k+1} = 0 for 1 \le i \le k (or i < k+1). All elements in the remaining sum are therefore zero, and we have

1 = \sum_{i < k+1} q_{i,k+1}^2 + q_{k+1,k+1}^2 = \sum_{i < k+1} 0^2 + q_{k+1,k+1}^2 = q_{k+1,k+1}^2

so that q_{k+1,k+1}^2 = 1 or q_{k+1,k+1} = \pm 1.

Since row k + 1 is orthonormal we have

1 = \sum_j q_{k+1,j}^2 = \sum_{j < k+1} q_{k+1,j}^2 + q_{k+1,k+1}^2 + \sum_{j > k+1} q_{k+1,j}^2

But since Q is upper triangular we have q_{k+1,j} = 0 for k + 1 > j (or j < k+1), and from above we have q_{k+1,k+1}^2 = 1. We thus have

1 = \sum_{j < k+1} q_{k+1,j}^2 + q_{k+1,k+1}^2 + \sum_{j > k+1} q_{k+1,j}^2

= \sum_j 0^2 + 1 + \sum_{j > k+1} q_{k+1,j}^2 = 1 + \sum_{j > k+1} q_{k+1,j}^2

so that \sum_{j > k+1} q_{k+1,j}^2 = 0 and thus q_{k+1,j} = 0 for j > k+1.

Since we already had q_{k+1,j} = 0 for j < k+1 we therefore have q_{k+1,j} = 0 for j \ne k+1.

In other words, all off-diagonal entries in row k+1 are zero given our assumption that off-diagonal entries in rows 1 through k were zero. Since this assumption is true for row 1, we have all off-diagonal entries equal to zero for all rows 1 through n. More formally, q_{ij} = 0 for j \ne i and 1 \le i \le n.

Since q_{ij} = 0 for i \ne j any orthogonal upper triangular matrix Q is therefore also a diagonal matrix. In addition, each element q_{ii} 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.

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