3
$\begingroup$

Let $U = $ ${\vec{u_1}...\vec{u}_n} $ and $V = $ ${\vec{v_1}...\vec{v}_n} $ be two orthonormal bases fo $R^n$. Prove that the change-of-basis matrix $S$ from U to V is orthogonal.

I know that something is orthogonal is its dot-product is zero. Therefore, by definition, if we have U, and V which are orthonormal bases, then all of their vectors are orthogonal to each other, by definition. But I'm not sure if this would be a sufficient proof, or even be a proof, at all.

$\endgroup$
1
  • 1
    $\begingroup$ You need to show that $S$ is invertible and $S^{-1}=S^T$ (transpose of $S$) $\endgroup$
    – sinbadh
    Mar 15, 2016 at 19:11

3 Answers 3

2
$\begingroup$

Let $P$ the change matrix from $V$ to $U$ i.e. we have

$$Pu_i=v_i,\;\forall i$$ We have

$$\delta_{i,j}=\langle Pu_i,Pu_j\rangle=(Pu_i)^T(Pu_j)=u_i^T(P^TP)u_j,\;\forall i,j$$ so we get

$$P^TP=I$$

$\endgroup$
1
$\begingroup$

An orthogonal matrix is a matrix $M$ such that $M^{-1} = M^T$. In English, the transpose of $M$ is the inverse of $M$. Write down the change of basis matrix $S$ from $U$ to $V$ and show that $S^T = S^{-1}.$

$\endgroup$
1
$\begingroup$

An orthogonal matrix should be thought of as a matrix whose transpose is its inverse. The change of basis matrix $S$ from $U$ to $V$ is

$$S_{ij} = \vec{v_i}\cdot\vec{u_j}$$

The reason this is so is because the vectors are orthogonal; to get components of vector $\vec{r}$ in any basis we simply take a dot product:

$$r_i = (\vec{r}\cdot\vec{u_i})$$

So $$S_{ij}r_j = \vec{v_i}\cdot(r_j\vec{u_j}) = \vec{v_i}\cdot\vec{r}$$ where repeated indices are summed over. We get the components in the $v$ basis, just like the change of basis matrix should.

Now to show the change of basis matrix is orthogonal. You need to show $$SS^{T} = S^{T}S=I$$ the identity matrix. Hint: use the first equation and the completeness property of an orthonormal basis.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .