# Rows of orthogonal matrix from an orthonormal basis of $R^n$

The question is:

Let $U$ be an $n \times n$ orthogonal matrix. Show that the rows of U form an orthonormal basis of $\mathbb R ^n$.

So far I have stated: Since $U$ is orthogonal its column vectors are linearly independent and by the Invertible Matrix theorem $U$ is invertible and $U^T$ is invertible.

I know that I need to somehow show that $U^T U=I$, but I don't know how to get there from where I left off.

• What is the definition of an orthogonal matrix? – Ameet Sharma Apr 25 '14 at 6:45
• A set is orthogonal if each pair of distinct vectors within it are orthogonal to each other. An orthogonal matrix is a matrix whose column vectors form an orthogonal set. – Victoria Potvin Apr 25 '14 at 6:48
• Hmm... I'd double check the definition. this is the definition of orthogonal matrix that I learned: mathworld.wolfram.com/OrthogonalMatrix.html. So you'd start knowing that $U^{T}U=I$. then use that to prove the rows form an orthonormal basis. – Ameet Sharma Apr 25 '14 at 6:56
• The thing is, these 3 statements can be shown to be equivalent definitions of orthogonal matrix: 1) $U^{T}U=I$. 2) The rows of matrix U form an orthonormal set(they are orthogonal, plus have magnitude one). 3) The columns of matrix U form an orthonormal set. So the question is... is the problem asking you to use 1) to prove 2)... or to use 3) to prove 2)... – Ameet Sharma Apr 25 '14 at 7:10

## 1 Answer

Hint: This is long (and can be very much shortened), but I think it will be a good learning curve for you to make it long. It became this long, because we start with the way you have understood orthogonal matrices (as seen in comments). I am reproducing it here "A set is orthogonal if each pair of distinct vectors within it are orthogonal to each other. An orthogonal matrix is a matrix whose column vectors form an orthogonal set". This is infact one of the many equivalent ways to define a orthogonal matrix. We will start with this definition without any other assumptions.

We will individually denote both columns and rows of $U$. Let columns of $U$ be given by $N\times 1$ vectors, $c_1,\dots,c_N$. Let its rows be given by $N\times 1$ vectors $r_1,\dots,r_N$. So you have $$U=[c_1|\dots|c_N]$$ and $$U=\begin{bmatrix}r_1^T \\ \vdots \\ r_N^T \end{bmatrix}$$

(notice how I used the transpose). Now note that $$A=U^TU$$ can be viewed upon as a matrix whose $(i,j)$th entry is given as $$A_{ij}=c_i^Tc_j$$ Similarly $$B=UU^T$$ is a matrix whose $(i,j)$th entry is given as $$B_{ij}=r_i^Tr_j$$

Now notice the following things

• prove $U^TU=A=I$ (why?)
• use the fact that $rank{(PQ)}\leq \min\{rank(P),rank(Q)\}$ to argue that $U$ and $U^T$ are invertible. $P$ and $Q$ are any arbitrary $N\times N$ matrices. Substitute $P$ and $Q$ with suitable matrices.
• Use above two facts to prove that $U^{-1}=U^T$. Hint: Perhaps a right multiplication on the LHS of first fact with a suitable matrix.
• Use above fact to prove $UU^T=I$. Hint: For this you also need that if $P$ are $Q$ are inverses of each other $PQ=I$ and $QP=I$
• Now using the above fact and definition of $B$, look at $r_i^Tr_j$