I have the following 2-part question:

  1. Find all $n \times n$ matrices that are both orthogonal and upper triangular, with positive diagonal entries.

  2. Show that the $QR$ factorization of an invertible $n \times n$ matrix is unique. Hint: if $A=Q_1R_1=Q_2R_2$, then the matrix $Q_2^{-1}Q_1=R_2R_1^{-1}$ is both orthogonal and upper triangular, with positive diagonal entries.

I realize that the general form of the $R$ matrix is upper triangular, with diagonal entries as vector lengths, which by definition must be positive. I'm not sure about the big picture though. Anyone kind enough to nudge me in the right direction? Thanks!

  • 1
    $\begingroup$ identity matrix? $\endgroup$ – DDaren Mar 12 '15 at 20:41
  • $\begingroup$ I agree. Any idea where to find a formal proof? $\endgroup$ – Benjamin Loya Mar 12 '15 at 20:44
  • $\begingroup$ Use the definition of being an orthogonal matrix: the columns (say) form an orthonormal basis. The first column looks like so $$\begin{pmatrix}1\\0\\\vdots\\0\end{pmatrix}$$ and this forces all the other coefficients in the first row to be zero. Hence the second column must be $$\begin{pmatrix}0\\1\\0\\\vdots\\0\end{pmatrix}$$ This pattern continues, and shows that your matrix was the identity matrix. $\endgroup$ – Olivier Bégassat Mar 12 '15 at 20:46
  • $\begingroup$ You might need to use $A^TA=R^TR$ and Cholesky to say about the uniqueness of Q. $\endgroup$ – DDaren Mar 12 '15 at 20:51
  • $\begingroup$ I was overthinking this. The hint was clear. $R_2R_1^{-1}$ is a diagonal matrix. $\endgroup$ – DDaren Mar 12 '15 at 20:58

Hint for $1$:

Let $T$ be upper triangular. Then $T$ is normal (satisfies $T^*T = TT^*$) if and only if it is diagonal.

Or, for ease of proof, use the following:

Let $A$ be partitioned as $$ A = \pmatrix{A_{11} & A_{12}\\0&A_{22}} $$ then $A$ is normal iff $A_{11},A_{22}$ are normal and $A_{12} = 0$.

Once you've proven this, the statement above on $T$ follows by clever application of the theorem on block matrices. (This trick is from Horn and Johnson).

  • $\begingroup$ Nice. We haven't gone over eigenvalues in class yet though. Any way to explicitly prove without them? $\endgroup$ – Benjamin Loya Mar 12 '15 at 20:56
  • $\begingroup$ There you go; I was being a bit more complicated than necessary. $\endgroup$ – Omnomnomnom Mar 12 '15 at 20:57
  • $\begingroup$ Almost there. I'm not familiar with this concept of "normal" matrices and components. Any good references? $\endgroup$ – Benjamin Loya Mar 12 '15 at 21:03
  • $\begingroup$ Horn and Johnson works for example. Again, normal just means that $A^*A = AA^*$; certainly this is true for orthogonal matrices ($^*$ means "conjugate-transpose" here, which for real matrices is just the transpose). $\endgroup$ – Omnomnomnom Mar 12 '15 at 21:05
  • $\begingroup$ Actually, here's another idea: if $M$ is upper-triangular, write $$ M^T M = I $$ Try to solve for the entries of $M$. You'll find that the fact that $M$ is upper triangular makes this surprisingly easy. $\endgroup$ – Omnomnomnom Mar 12 '15 at 21:06

(1) There's nothing simpler than to realize that $Q^T=Q^{-1}$ implies that $Q$ is both upper and lower triangular. Hence diagonal.

(2) Note that $Q_2^TQ_1=R_2R_1^{-1}$ implies that $R_2R_1^{-1}$ is

  1. orthogonal -- because it is a product of two orthogonal matrices,
  2. diagonal -- because it is orthogonal and triangular.

The only real orthogonal diagonal matrix is a matrix with $\pm 1$ on the diagonal. Consequently, for any two QR factorizations of $A$, the R-factors are related by such a simple diagonal matrix and in particular, their diagonals differ just by the sign. Once you require the positive entries on the diagonal of $R$, it hence follows that $R_1=R_2$ since the diagonal of $R_2R_1^{-1}$ is nothing but the "componentwise ratio" of the diagonals of $R_1$ and $R_2$.


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