0
$\begingroup$

Prove that if $U$ is an orthogonal $n\times n$ matrix, then the rows of $U$ form an orthonormal basis for $\mathbb{R}^n$

I'm unsure how to proceed with proving this. Basically my idea is as follows.

Let $U=\begin{pmatrix} u_1 & \cdots & u_n\end{pmatrix}$. Since $U$ is an orthogonal matrix, $\{u_1,\dots,u_n\}$ is a linearly independent set and $u_i^{T}u_i=1$ for all $i$ and $u_i^{T}u_j=0$ whenever $i\neq j$.

From here, my idea is to simply show that since the same relationship: "$u_i^{T}u_i=1$ for all $i$ and $u_i^{T}u_j=0$ whenever $i\neq j$" holds for the set $\{u^{T}_1,\dots,u^{T}_n\}$, the rows of $U$ form an orthonormal basis for $\mathbb{R}^n$?

$\endgroup$
  • $\begingroup$ If you know that the columns of an orthogonal matrix is an ON-basis then just apply this to the orthogonal (why?) matrix $U^T$. P.S. It is easy to prove by definition that the orthogonal system of vectors is linear independent, but it is not obvious and may need some explanation. $\endgroup$ – A.Γ. Aug 2 '15 at 9:10
1
$\begingroup$

The coefficient at place $(i,j)$ in $U^TU$ is $u_i^Tu_j$ which is precisely the scalar product of $u_i$ by $u_j$. So the set is orthogonal and its elements have norm $1$; along with the fact that this set is a basis because $U$ is invertible this suffices to prove the assertion.

Note that this is an orthonormal basis of $\mathbb{R}^n$ under the standard scalar product.

$\endgroup$
0
$\begingroup$

In general if you have $n$ vectors linearly independent in $\mathbb{R}^n$ they form a basis of $\mathbb{R}^n$. You know that $UU^T=I$ this implies $|u_j|=1$ for every $1\leq j\leq n$ and $(u_i,u_j)=0$ if $i\neq j$. It follows that $(u_1,...,u_n)$ is an orthonormal basis.

$\endgroup$
0
$\begingroup$

define $y = U^T x$, it is the orthogonal projection of $x$ onto the vectors $\{u_1, \cdots u_n\}$ the rows of $U$. thus :

$$||y||^2 = ||U^T x ||^2 = x^T U U^T x = x^T I x = x^T x = ||x||^2$$

by definition of $||.||^2$ and that $U$ is orthogonal.

this is one of the definitions of an orthonormal basis : the projection of a vector onto the basis keeps the same norm.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy