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# orthogonal matrices

A real square $n\times n$ matrix $Q$ is orthogonal if $Q^{{\mathrm{T}}}Q=I$, i.e., if $Q^{{-1}}=Q^{{\mathrm{T}}}$. The rows and columns of an orthogonal matrix form an orthonormal basis.

Orthogonal matrices play a very important role in linear algebra. Inner products are preserved under an orthogonal transform: $(Qx)^{{\mathrm{T}}}Qy=x^{{\mathrm{T}}}Q^{{\mathrm{T}}}Qy=x^{{\mathrm{T}}}y$, and also the Euclidean norm $||Qx||_{2}=||x||_{2}$. An example of where this is useful is solving the least squares problem $Ax\approx b$ by solving the equivalent problem $Q^{{\mathrm{T}}}Ax\approx Q^{{\mathrm{T}}}b$.

Orthogonal matrices can be thought of as the real case of unitary matrices. A unitary matrix $U\in\mathbb{C}^{{n\times n}}$ has the property $U^{*}U=I$, where $U^{*}=\overline{U^{{\mathrm{T}}}}$ (the conjugate transpose). Since $\overline{Q^{{\mathrm{T}}}}=Q^{{\mathrm{T}}}$ for real $Q$, orthogonal matrices are unitary.

An orthogonal matrix $Q$ has $\det(Q)=\pm 1$.

Important orthogonal matrices are Givens rotations and Householder transformations. They help us maintain numerical stability because they do not amplify rounding errors.

Orthogonal $2\times 2$ matrices are rotations or reflections if they have the form:

$\begin{pmatrix}\cos(\alpha)&\sin(\alpha)\\ -\sin(\alpha)&\cos(\alpha)\end{pmatrix}\text{or}\begin{pmatrix}\cos(\alpha)&% \sin(\alpha)\\ \sin(\alpha)&-\cos(\alpha)\end{pmatrix}$ |

respectively.

This entry is based on content from The Data Analysis Briefbook (http://rkb.home.cern.ch/rkb/titleA.html)

# References

- 1 Friedberg, Insell, Spence. Linear Algebra. Prentice-Hall Inc., 1997.

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## Comments

## Q^T Q = Id <=> Q^-1 = Q^T ?

Hi

An inverse of A is matrix A^-1 such that

A A^-1 = A^-1 A = Id.

In this entry, does it follow from Q^T Q = Id

that Q^T = Q^-1? Or more generally, is it possible that

A B = Id for some matrices A, B, but B A \neq Id?

Matte

## Re: Q^T Q = Id <=> Q^-1 = Q^T ?

> is it possible that A B = Id for some matrices A, B,

> but B A \neq Id?

If A and B are squre matrices, then it is impossible I think. Because, consider

A B= id

take det() and you get

det(A) det(B) = 1

so det(A) and det(B) are unequal to zero, so invertable, and so

from A B = id follows that B = A^-1 due to due to the uniqueness of inverse matrix and thus B A = id.

Now, if you have non-square matrices then, first of all

> is it possible that A B = Id for some matrices A, B,

> but B A \neq Id?

there are different Id. And it is fairly possible that

A B = Id_1 but B A \neq Id_2,

example:

(1 0 0)

(0 1 0)

x

(1 0)

(0 1)

(0 0)

=

(1 0)

(0 1)

BUT

(1 0)

(0 1)

(0 0)

x

(1 0 0)

(0 1 0)

\neq

(1 0 0)

(0 1 0)

(0 0 1).

Serg.

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## Re: Q^T Q = Id <=> Q^-1 = Q^T ?

thanks!