# QR decomposition: From orthogonal vectors to orthonormal ones

I'm using Strang's Introduction to Linear Algebra, and I'm a bit confused about QR decomposition.

I understand how to turn $\begin{pmatrix} a & b & c\end{pmatrix}$ into $\begin{pmatrix} A & B & C\end{pmatrix}$, in which A, B and C are perpendicular to each other. Then the textbook goes on to talk about $QR$ $decomposition$.

$\begin{pmatrix} a & b & c\end{pmatrix}$ = $\begin{pmatrix} q_1 & q_2 & q_3 \end{pmatrix}$$\begin{pmatrix} q_1^Ta & q_1^Tb & q_1^Tc \\ 0 & q_2^Tb & q_2^Tc \\ 0 & 0 & q_3^Tc\end{pmatrix} , in which q_1, q_2 and q_3 are normalized version of A, B and C. This is definitely related to the idea that projecting b onto the whole space: b=q_1(q_1^Tb)+q_2(q_2^Tb)+...+q_n(q_n^Tb), and here is where I get stuck: Before we obtain q_2, we obtain B, which is the error vector when we project b onto a, and then we normalize B to obtain q_2. I don't get why b = q_1(q_1^Tb)+q_2(q_2^Tb). I'm only projecting b onto a, not onto the plane spanned by a and b. I'm also not sure how to normalize B to reach q_2. I guess these two questions are connected. I can only go as far as B=b(I-q_1q_1^T), but how do I calculate its length and normalize it? ## 2 Answers Let's use slightly different notation, so that our observations will extend to an arbitrary number of vectors. Start with three vectors a_0, a_1, a_2 . The Gram-Schmidt process starts by taking q_0 in the direction of a_0 , but of unit length: • \rho_{0,0} = \| a_0 \|_2 • q_0 = a_0 / \rho_{0,0} Next, you compute the component of a_1 in the direction of q_0 , q_0^T a_1 q_0 and then subtract this from a_1 to be left with the component of a_1 orthogonal to q_0 . But you do it in steps: • \rho_{0,1} = q_0^T a_1 • a_1^\perp = a_1 - \rho_{0,1} q_0 (the component perpendicular to q_0 .) And then you take that, and make it of unit length to compute q_1 : • \rho_{1,1} = \| a_1^\perp \|_2 • q_1 = a_1^\perp / \rho_{1,1} And then you move on: you compute the components of a_2 in the direction of q_0 , q_0^T a_2 q_0 , and q_1 , q_1^T a_2 q_1 and you subtract off those components to be left with the component orthogonal to q_0 and q_1 : • \rho_{0,2} = q_0^T a_2 • \rho_{1,2} = q_1^T a_2 • a_2^\perp = a_2 - \rho_{0,2} q_0 - \rho_{1,2} q_1 (the component perpendicular to q_0 .) And then you take that, and make it of unit length to compute q_2 : • \rho_{2,2} = \| a_2^\perp \|_2 • q_2 = a_2^\perp / \rho_{2,2} Now, if you look at this carefully, you will find that if you • make a_0 , a_1 , and a_2 the columns of matrix A , • make q_0 , q_1 , and q_2 the columns of matrix Q , • \rho_{i,j} the elements of upper triangular matrix R then A = Q R . Here is another way of looking at this as an algorithm. Consider A = Q R . Partition A , Q , and R so that$$ \left( \begin{array}{c | c c} A_0 & a_1 & A_2 \end{array} \right) = \left( \begin{array}{c | c c} Q_0 & q_1 & Q_2 \end{array} \right) \left( \begin{array}{c | c c} R_{00} & r_{01} & R_{02} \\ \hline 0 & \rho_{11} & r_{12}^T \\ 0 & 0 & R_{22} \end{array} \right) $$Now, assume that the orthonormal columns of Q_0 have already been computed, as has upper triangular R_{00} (the coefficients we discussed before). What we want to do is to compute the elements in r_{01} and \rho_{11} as well as the next column of Q , q_1 . How do we do this? Multiplying out part of the right-hand side of$$\left( \begin{array}{c | c c} A_0 & a_1 & A_2 \end{array} \right) = \left( \begin{array}{c | c c} Q_0 & q_1 & Q_2 \end{array} \right) \left( \begin{array}{c | c c} R_{00} & r_{01} & R_{02} \\ \hline 0 & \rho_{11} & r_{12}^T \\ 0 & 0 & R_{22} \end{array} \right) $$we find that$$ a_1 = Q_0 r_{01} + q_1 \rho_{11} $$. Here we know a_1 , Q_0 and we know that q_1 will be orthogonal to Q_0 . So, apply Q_0^T from the left to both side:$$ Q_0^T a_1 = Q_0^T Q_0 r_{01} + Q_0^T q_1 \rho_{11} = r_{01} $$so, we know how to compute the coefficients in vector r_{01} . When we go back to$$ a_1 = Q_0 r_{01} + q_1 \rho_{11} $$and compute$$ a_1^\perp = \rho_{11} q_1 = a_1 - Q_0 r_{01}$$the component of$ a_1 $orthogonal to the space spanned by the columns of$ Q_0 $. (Notice:$ a_1^\perp = a_1 - Q_0 r_{01} = a_1 - Q_0 ( Q_0^T a_1 ) $which is the formula for the component orthogonal to...) All that is left then is to compute the length of$a_1^\perp$as$ \rho_{11} = \| a_1^\perp \|_2 $and normalize:$ q_1 = a_1 / \rho_{11} $. Bingo! we have computed the next columns of$ Q $and$ R \$.

• Hi thanks for the reply! Sorry for being late. I'll read your notes carefully and get back to you if I still cannot grasp the idea. – Nicholas Humphrey Apr 4 '18 at 0:04

Hint: write out the formula for Gram-Schmidt process, and try to observe linear relations.

• Thanks for the reply! I'll review the notes carefully. – Nicholas Humphrey Apr 4 '18 at 0:04