QR decomposition

Any real square matrix A may be decomposed as

${\displaystyle A=QR,\,}$

where Q is an orthogonal matrix (its columns are orthogonal unit vectors meaning ${\displaystyle Q^{\textsf {T}}Q=QQ^{\textsf {T}}=I}$) and R is an upper triangular matrix (also called right triangular matrix, hence the name). If A is invertible, then the factorization is unique if we require the diagonal elements of R to be positive.

If instead A is a complex square matrix, then there is a decomposition A = QR where Q is a unitary matrix (so ${\displaystyle Q^{*}Q=QQ^{*}=I}$).

If A has n linearly independent columns, then the first n columns of Q form an orthonormal basis for the column space of A. More generally, the first k columns of Q form an orthonormal basis for the span of the first k columns of A for any 1 ≤ k ≤ n.[1] The fact that any column k of A only depends on the first k columns of Q is responsible for the triangular form of R.[1]

More generally, we can factor a complex m×n matrix A, with m ≥ n, as the product of an m×m unitary matrix Q and an m×n upper triangular matrix R. As the bottom (m−n) rows of an m×n upper triangular matrix consist entirely of zeroes, it is often useful to partition R, or both R and Q:

${\displaystyle A=QR=Q{\begin{bmatrix}R_{1}\\0\end{bmatrix}}={\begin{bmatrix}Q_{1},Q_{2}\end{bmatrix}}{\begin{bmatrix}R_{1}\\0\end{bmatrix}}=Q_{1}R_{1},}$

where R₁ is an n×n upper triangular matrix, 0 is an (m − n)×n zero matrix, Q₁ is m×n, Q₂ is m×(m − n), and Q₁ and Q₂ both have orthogonal columns.

Golub & Van Loan (1996, §5.2) call Q₁R₁ the thin QR factorization of A; Trefethen and Bau call this the reduced QR factorization.[1] If A is of full rank n and we require that the diagonal elements of R₁ are positive then R₁ and Q₁ are unique, but in general Q₂ is not. R₁ is then equal to the upper triangular factor of the Cholesky decomposition of A* A (= A^TA if A is real).

Analogously, we can define QL, RQ, and LQ decompositions, with L being a lower triangular matrix.

Consider the Gram–Schmidt process applied to the columns of the full column rank matrix ${\displaystyle A=\left[\mathbf {a} _{1},\ldots ,\mathbf {a} _{n}\right]}$, with inner product ${\displaystyle \langle \mathbf {v} ,\mathbf {w} \rangle =\mathbf {v} ^{\textsf {T}}\mathbf {w} }$ (or ${\displaystyle \langle \mathbf {v} ,\mathbf {w} \rangle =\mathbf {v} ^{*}\mathbf {w} }$ for the complex case).

Define the projection:

${\displaystyle \operatorname {proj} _{\mathbf {u} }\mathbf {a} ={\frac {\left\langle \mathbf {u} ,\mathbf {a} \right\rangle }{\left\langle \mathbf {u} ,\mathbf {u} \right\rangle }}{\mathbf {u} }}$

then:

${\displaystyle {\begin{aligned}\mathbf {u} _{1}&=\mathbf {a} _{1},&\mathbf {e} _{1}&={\mathbf {u} _{1} \over \|\mathbf {u} _{1}\|}\\\mathbf {u} _{2}&=\mathbf {a} _{2}-\operatorname {proj} _{\mathbf {u} _{1}}\,\mathbf {a} _{2},&\mathbf {e} _{2}&={\mathbf {u} _{2} \over \|\mathbf {u} _{2}\|}\\\mathbf {u} _{3}&=\mathbf {a} _{3}-\operatorname {proj} _{\mathbf {u} _{1}}\,\mathbf {a} _{3}-\operatorname {proj} _{\mathbf {u} _{2}}\,\mathbf {a} _{3},&\mathbf {e} _{3}&={\mathbf {u} _{3} \over \|\mathbf {u} _{3}\|}\\&\vdots &&\vdots \\\mathbf {u} _{k}&=\mathbf {a} _{k}-\sum _{j=1}^{k-1}\operatorname {proj} _{\mathbf {u} _{j}}\,\mathbf {a} _{k},&\mathbf {e} _{k}&={\mathbf {u} _{k} \over \|\mathbf {u} _{k}\|}\end{aligned}}}$

We can now express the ${\displaystyle \mathbf {a} _{i}}$s over our newly computed orthonormal basis:

${\displaystyle {\begin{aligned}\mathbf {a} _{1}&=\langle \mathbf {e} _{1},\mathbf {a} _{1}\rangle \mathbf {e} _{1}\\\mathbf {a} _{2}&=\langle \mathbf {e} _{1},\mathbf {a} _{2}\rangle \mathbf {e} _{1}+\langle \mathbf {e} _{2},\mathbf {a} _{2}\rangle \mathbf {e} _{2}\\\mathbf {a} _{3}&=\langle \mathbf {e} _{1},\mathbf {a} _{3}\rangle \mathbf {e} _{1}+\langle \mathbf {e} _{2},\mathbf {a} _{3}\rangle \mathbf {e} _{2}+\langle \mathbf {e} _{3},\mathbf {a} _{3}\rangle \mathbf {e} _{3}\\&\vdots \\\mathbf {a} _{k}&=\sum _{j=1}^{k}\langle \mathbf {e} _{j},\mathbf {a} _{k}\rangle \mathbf {e} _{j}\end{aligned}}}$

where ${\displaystyle \left\langle \mathbf {e} _{i},\mathbf {a} _{i}\right\rangle =\left\|\mathbf {u} _{i}\right\|}$. This can be written in matrix form:

${\displaystyle A=QR}$

where:

${\displaystyle Q=\left[\mathbf {e} _{1},\ldots ,\mathbf {e} _{n}\right]}$

and

${\displaystyle R={\begin{pmatrix}\langle \mathbf {e} _{1},\mathbf {a} _{1}\rangle &\langle \mathbf {e} _{1},\mathbf {a} _{2}\rangle &\langle \mathbf {e} _{1},\mathbf {a} _{3}\rangle &\ldots \\0&\langle \mathbf {e} _{2},\mathbf {a} _{2}\rangle &\langle \mathbf {e} _{2},\mathbf {a} _{3}\rangle &\ldots \\0&0&\langle \mathbf {e} _{3},\mathbf {a} _{3}\rangle &\ldots \\\vdots &\vdots &\vdots &\ddots \end{pmatrix}}.}$

Consider the decomposition of

${\displaystyle A={\begin{pmatrix}12&-51&4\\6&167&-68\\-4&24&-41\end{pmatrix}}.}$

Recall that an orthonormal matrix ${\displaystyle Q}$ has the property

${\displaystyle Q^{\textsf {T}}\,Q=I.}$

Then, we can calculate ${\displaystyle Q}$ by means of Gram–Schmidt as follows:

${\displaystyle {\begin{aligned}U={\begin{pmatrix}\mathbf {u} _{1}&\mathbf {u} _{2}&\mathbf {u} _{3}\end{pmatrix}}&={\begin{pmatrix}12&-69&-58/5\\6&158&6/5\\-4&30&-33\end{pmatrix}};\\Q={\begin{pmatrix}{\frac {\mathbf {u} _{1}}{\|\mathbf {u} _{1}\|}}&{\frac {\mathbf {u} _{2}}{\|\mathbf {u} _{2}\|}}&{\frac {\mathbf {u} _{3}}{\|\mathbf {u} _{3}\|}}\end{pmatrix}}&={\begin{pmatrix}6/7&-69/175&-58/175\\3/7&158/175&6/175\\-2/7&6/35&-33/35\end{pmatrix}}.\end{aligned}}}$

Thus, we have

${\displaystyle {\begin{aligned}Q^{\textsf {T}}A&=Q^{\textsf {T}}Q\,R=R;\\R&=Q^{\textsf {T}}A={\begin{pmatrix}14&21&-14\\0&175&-70\\0&0&35\end{pmatrix}}.\end{aligned}}}$

The RQ decomposition transforms a matrix A into the product of an upper triangular matrix R (also known as right-triangular) and an orthogonal matrix Q. The only difference from QR decomposition is the order of these matrices.

QR decomposition is Gram–Schmidt orthogonalization of columns of A, started from the first column.

RQ decomposition is Gram–Schmidt orthogonalization of rows of A, started from the last row.

The Gram-Schmidt process is inherently numerically unstable. While the application of the projections has an appealing geometric analogy to orthogonalization, the orthogonalization itself is prone to numerical error. A significant advantage however is the ease of implementation, which makes this a useful algorithm to use for prototyping if a pre-built linear algebra library is unavailable.

Householder reflection for QR-decomposition: The goal is to find a linear transformation that changes the vector ${\displaystyle x}$ into a vector of the same length which is collinear to ${\displaystyle e_{1}}$. We could use an orthogonal projection (Gram-Schmidt) but this will be numerically unstable if the vectors ${\displaystyle x}$ and ${\displaystyle e_{1}}$ are close to orthogonal. Instead, the Householder reflection reflects through the dotted line (chosen to bisect the angle between ${\displaystyle x}$ and ${\displaystyle e_{1}}$). The maximum angle with this transform is 45 degrees.

A Householder reflection (or Householder transformation) is a transformation that takes a vector and reflects it about some plane or hyperplane. We can use this operation to calculate the QR factorization of an m-by-n matrix ${\displaystyle A}$ with m ≥ n.

Q can be used to reflect a vector in such a way that all coordinates but one disappear.

Let ${\displaystyle \mathbf {x} }$ be an arbitrary real m-dimensional column vector of ${\displaystyle A}$ such that ${\displaystyle \|\mathbf {x} \|=|\alpha |}$ for a scalar α. If the algorithm is implemented using floating-point arithmetic, then α should get the opposite sign as the k-th coordinate of ${\displaystyle \mathbf {x} }$, where ${\displaystyle x_{k}}$ is to be the pivot coordinate after which all entries are 0 in matrix A's final upper triangular form, to avoid loss of significance. In the complex case, set

${\displaystyle \alpha =-e^{i\arg x_{k}}\|\mathbf {x} \|}$

(Stoer & Bulirsch 2002, p. 225) and substitute transposition by conjugate transposition in the construction of Q below.

Then, where ${\displaystyle \mathbf {e} _{1}}$ is the vector (1 0 … 0)^T, ||·|| is the Euclidean norm and ${\displaystyle I}$ is an m-by-m identity matrix, set

${\displaystyle {\begin{aligned}\mathbf {u} &=\mathbf {x} -\alpha \mathbf {e} _{1},\\\mathbf {v} &={\mathbf {u} \over \|\mathbf {u} \|},\\Q&=I-2\mathbf {v} \mathbf {v} ^{\textsf {T}}.\end{aligned}}}$

Or, if ${\displaystyle A}$ is complex

${\displaystyle Q=I-2\mathbf {v} \mathbf {v} ^{*}.}$

${\displaystyle Q}$ is an m-by-m Householder matrix and

${\displaystyle Q\mathbf {x} ={\begin{pmatrix}\alpha &0&\cdots &0\end{pmatrix}}^{\textsf {T}}.\,}$

This can be used to gradually transform an m-by-n matrix A to upper triangular form. First, we multiply A with the Householder matrix Q₁ we obtain when we choose the first matrix column for x. This results in a matrix Q₁A with zeros in the left column (except for the first row).

${\displaystyle Q_{1}A={\begin{bmatrix}\alpha _{1}&\star &\dots &\star \\0&&&\\\vdots &&A'&\\0&&&\end{bmatrix}}}$

This can be repeated for A′ (obtained from Q₁A by deleting the first row and first column), resulting in a Householder matrix Q′₂. Note that Q′₂ is smaller than Q₁. Since we want it really to operate on Q₁A instead of A′ we need to expand it to the upper left, filling in a 1, or in general:

${\displaystyle Q_{k}={\begin{pmatrix}I_{k-1}&0\\0&Q_{k}'\end{pmatrix}}.}$

After ${\displaystyle t}$ iterations of this process, ${\displaystyle t=\min(m-1,n)}$,

${\displaystyle R=Q_{t}\cdots Q_{2}Q_{1}A}$

is an upper triangular matrix. So, with

${\displaystyle Q=Q_{1}^{\textsf {T}}Q_{2}^{\textsf {T}}\cdots Q_{t}^{\textsf {T}},}$

${\displaystyle A=QR}$ is a QR decomposition of ${\displaystyle A}$.

This method has greater numerical stability than the Gram–Schmidt method above.

The following table gives the number of operations in the k-th step of the QR-decomposition by the Householder transformation, assuming a square matrix with size n.

Operation	Number of operations in the k-th step
Multiplications	${\displaystyle 2(n-k+1)^{2}}$
Additions	${\displaystyle (n-k+1)^{2}+(n-k+1)(n-k)+2}$
Division	${\displaystyle 1}$
Square root	${\displaystyle 1}$

Summing these numbers over the n − 1 steps (for a square matrix of size n), the complexity of the algorithm (in terms of floating point multiplications) is given by

${\displaystyle {\frac {2}{3}}n^{3}+n^{2}+{\frac {1}{3}}n-2=O\left(n^{3}\right).}$

Let us calculate the decomposition of

${\displaystyle A={\begin{pmatrix}12&-51&4\\6&167&-68\\-4&24&-41\end{pmatrix}}.}$

First, we need to find a reflection that transforms the first column of matrix A, vector ${\displaystyle \mathbf {a} _{1}={\begin{pmatrix}12&6&-4\end{pmatrix}}^{\textsf {T}}}$, into ${\displaystyle \left\|\mathbf {a} _{1}\right\|\;\mathbf {e} _{1}={\begin{pmatrix}1&0&0\end{pmatrix}}^{\textsf {T}}.}$

Now,

${\displaystyle \mathbf {u} =\mathbf {x} -\alpha \mathbf {e} _{1},}$

and

${\displaystyle \mathbf {v} ={\mathbf {u} \over \|\mathbf {u} \|}.}$

Here,

${\displaystyle \alpha =14}$ and ${\displaystyle \mathbf {x} =\mathbf {a} _{1}={\begin{pmatrix}12&6&-4\end{pmatrix}}^{\textsf {T}}}$

Therefore

${\displaystyle \mathbf {u} ={\begin{pmatrix}-2&6&-4\end{pmatrix}}^{\textsf {T}}={\begin{pmatrix}2\end{pmatrix}}{\begin{pmatrix}-1&3&-2\end{pmatrix}}^{\textsf {T}}}$ and ${\displaystyle \mathbf {v} ={1 \over {\sqrt {14}}}{\begin{pmatrix}-1&3&-2\end{pmatrix}}^{\textsf {T}}}$, and then

${\displaystyle {\begin{aligned}Q_{1}={}&I-{2 \over {\sqrt {14}}{\sqrt {14}}}{\begin{pmatrix}-1\\3\\-2\end{pmatrix}}{\begin{pmatrix}-1&3&-2\end{pmatrix}}\\={}&I-{1 \over 7}{\begin{pmatrix}1&-3&2\\-3&9&-6\\2&-6&4\end{pmatrix}}\\={}&{\begin{pmatrix}6/7&3/7&-2/7\\3/7&-2/7&6/7\\-2/7&6/7&3/7\\\end{pmatrix}}.\end{aligned}}}$

Now observe:

${\displaystyle Q_{1}A={\begin{pmatrix}14&21&-14\\0&-49&-14\\0&168&-77\end{pmatrix}},}$

so we already have almost a triangular matrix. We only need to zero the (3, 2) entry.

Take the (1, 1) minor, and then apply the process again to

${\displaystyle A'=M_{11}={\begin{pmatrix}-49&-14\\168&-77\end{pmatrix}}.}$

By the same method as above, we obtain the matrix of the Householder transformation

${\displaystyle Q_{2}={\begin{pmatrix}1&0&0\\0&-7/25&24/25\\0&24/25&7/25\end{pmatrix}}}$

after performing a direct sum with 1 to make sure the next step in the process works properly.

Now, we find

${\displaystyle Q=Q_{1}^{\textsf {T}}Q_{2}^{\textsf {T}}={\begin{pmatrix}6/7&-69/175&58/175\\3/7&158/175&-6/175\\-2/7&6/35&33/35\end{pmatrix}}}$

Or, to four decimal digits,

${\displaystyle {\begin{aligned}Q&=Q_{1}^{\textsf {T}}Q_{2}^{\textsf {T}}={\begin{pmatrix}0.8571&-0.3943&0.3314\\0.4286&0.9029&-0.0343\\-0.2857&0.1714&0.9429\end{pmatrix}}\\R&=Q_{2}Q_{1}A=Q^{\textsf {T}}A={\begin{pmatrix}14&21&-14\\0&175&-70\\0&0&-35\end{pmatrix}}.\end{aligned}}}$

The matrix Q is orthogonal and R is upper triangular, so A = QR is the required QR-decomposition.

The use of Householder transformations is inherently the most simple of the numerically stable QR decomposition algorithms due to the use of reflections as the mechanism for producing zeroes in the R matrix. However, the Householder reflection algorithm is bandwidth heavy and not parallelizable, as every reflection that produces a new zero element changes the entirety of both Q and R matrices.

QR decompositions can also be computed with a series of Givens rotations. Each rotation zeroes an element in the subdiagonal of the matrix, forming the R matrix. The concatenation of all the Givens rotations forms the orthogonal Q matrix.

In practice, Givens rotations are not actually performed by building a whole matrix and doing a matrix multiplication. A Givens rotation procedure is used instead which does the equivalent of the sparse Givens matrix multiplication, without the extra work of handling the sparse elements. The Givens rotation procedure is useful in situations where only a relatively few off diagonal elements need to be zeroed, and is more easily parallelized than Householder transformations.

Let us calculate the decomposition of

${\displaystyle A={\begin{pmatrix}12&-51&4\\6&167&-68\\-4&24&-41\end{pmatrix}}.}$

First, we need to form a rotation matrix that will zero the lowermost left element, ${\displaystyle \mathbf {a} _{31}=-4}$. We form this matrix using the Givens rotation method, and call the matrix ${\displaystyle G_{1}}$. We will first rotate the vector ${\displaystyle {\begin{pmatrix}12&-4\end{pmatrix}}}$, to point along the X axis. This vector has an angle ${\displaystyle \theta =\arctan \left({-(-4) \over 12}\right)}$. We create the orthogonal Givens rotation matrix, ${\displaystyle G_{1}}$:

${\displaystyle {\begin{aligned}G_{1}&={\begin{pmatrix}\cos(\theta )&0&-\sin(\theta )\\0&1&0\\\sin(\theta )&0&\cos(\theta )\end{pmatrix}}\\&\approx {\begin{pmatrix}0.94868&0&-0.31622\\0&1&0\\0.31622&0&0.94868\end{pmatrix}}\end{aligned}}}$

And the result of ${\displaystyle G_{1}A}$ now has a zero in the ${\displaystyle \mathbf {a} _{31}}$ element.

${\displaystyle G_{1}A\approx {\begin{pmatrix}12.64911&-55.97231&16.76007\\6&167&-68\\0&6.64078&-37.6311\end{pmatrix}}}$

We can similarly form Givens matrices ${\displaystyle G_{2}}$ and ${\displaystyle G_{3}}$, which will zero the sub-diagonal elements ${\displaystyle a_{21}}$ and ${\displaystyle a_{32}}$, forming a triangular matrix ${\displaystyle R}$. The orthogonal matrix ${\displaystyle Q^{\textsf {T}}}$ is formed from the product of all the Givens matrices ${\displaystyle Q^{\textsf {T}}=G_{3}G_{2}G_{1}}$. Thus, we have ${\displaystyle G_{3}G_{2}G_{1}A=Q^{\textsf {T}}A=R}$, and the QR decomposition is ${\displaystyle A=QR}$.

The QR decomposition via Givens rotations is the most involved to implement, as the ordering of the rows required to fully exploit the algorithm is not trivial to determine. However, it has a significant advantage in that each new zero element ${\displaystyle a_{ij}}$ affects only the row with the element to be zeroed (i) and a row above (j). This makes the Givens rotation algorithm more bandwidth efficient and parallelisable than the Householder reflection technique.