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Givens rotation matrix

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The Givens rotation matrix (or plane rotation matrix) is an orthogonal matrix that is often used to transform a real matrix into an equivalent one, typically by annihilating the entries below its main diagonal.

Table of Contents


Here is a definition.

Definition Let $c$ and $s$ be two real numbers such that $c^{2}+s^{2}=1$. Let [eq1] be the $K	imes K$ identity matrix. Let $m$ and n be two integers such that [eq2]. The Givens rotation matrix [eq3]is the $K	imes K$ matrix whose entries are all equal to the corresponding entries of [eq4], except for[eq5]

Let us immediately see some examples.

Example The following is a $5	imes 5$ Givens matrix:[eq6]In this case, $K=5$, $m=2$ and $n=4$. Thus, the matrix is obtained by modifying the second and fourth rows of a $5	imes 5$ identity matrix.

Example Another example is [eq7]which has been obtained by modifying the first and fourth rows of a $4	imes 4$ identity matrix.

Example A smaller rotation matrix follows:[eq8]


Givens matrices are orthogonal (i.e., their columns are orthonormal).

Proposition A Givens matrix $G$ is orthogonal, that is,[eq9]


Let [eq3]We need to prove that [eq11]By using the definition of matrix product, we can see that the latter equation holds if and only if[eq12]where $G_{kullet }$ denotes the k-th row of $G$. We are going to prove that this is true. When $k
eq m$ and $l
eq n$, then[eq13]because the rows of the identity matrix are orthonormal. When $k=l=m$, then[eq14]When $k=l=n$, then[eq15]When $k=m$ and $l=n$ (or $k=n$ and $l=m$), so that $k
eq l$, then we have[eq16]Finally, when either k or $l$ is equal to one of $m$ or n (and the other one is not equal to $m$ or n), we have either[eq17]or[eq18]which completes the proof.

Equivalent transformations

Let [eq3]be a $K	imes K$ Givens rotation matrix.

Let A be a $K	imes L$ matrix.

What happens when we compute the product[eq20]that is, when we use $G$ to perform an equivalent transformation on A?

By the usual interpretation of matrix products as linear combinations, we can see that the product is a new matrix whose rows are all equal to the corresponding rows of A, except for the $m$-th and n-th. In particular, if $l
eq m$ and $l
eq n$, then[eq21]

But we have[eq22]

Example Let [eq23]and[eq24]Then, the product is [eq25]

Annihilation of entries

Suppose that, as in the previous section, we perform an equivalent transformation of a $K	imes L$ matrix A by using a Givens rotation [eq26].

Further suppose that $A_{mm}
eq 0$.

How can we set $c$ in such a way that the transformation annihilates the entry $A_{nm}$?

We already know that, after the transformation, the n-th row is[eq27]

Therefore, [eq28]

In order to have [eq29], we must have[eq30]

But we must also satisfy the constraint[eq31]

These two equations are solved by[eq32]

Example Define [eq33]Let us find a rotation matrix that allows us to annihilate the entry $A_{31}=4$. First of all, we need a non-zero entry to use as a pivot. We choose $A_{11}=3$. Thus, the rows involved in the rotation are the first and the third one. As a consequence, our Givens matrix has the form[eq34]The numbers $c$ and $s$ are chosen as follows:[eq35]Thus, the transformation is[eq36]which achieves the desired annihilation.

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Define the matrix[eq37]Find a Givens rotation matrix that transforms A into an upper triangular matrix.


We need to annihilate the entry $A_{32}=-3$. We can do so by pivoting on $A_{22}=4$. Thus, $m=2$ and $n=3$. Moreover,[eq38]Therefore, the rotation matrix is[eq39]The equivalent transformation is[eq40]

How to cite

Please cite as:

Taboga, Marco (2017). "Givens rotation matrix", Lectures on matrix algebra.

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