## 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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