5

I am trying to show that if $B_1$ and $B_2$ are orthonormal bases for $\mathbb{R}^n$, then the change of basis matrix $P$ from $B_1$ to $B_2$ is an orthogonal matrix.

I'm a bit stuck. I started with this: Let $x,y \in \mathbb{R}^n$. Then $[x]_{B_2}^t[y]_{B_2} = (P[x]_{B_1})^tP[y]_{B_1} = [x]_{B_1}^tP^tP[y]_{B_1}$. I want to show that $P^tP = I_n$. I know I need to apply the fact that $B_1$ and $B_2$ are orthonormal, but I don't see how to apply that to this expression. Am I approaching this properly or is there a better way to think about it?

Alex137
  • 51
  • 1
  • 3

2 Answers2

4

Because $B_1$ and $B_2$ are orthonormal bases for $\mathbb{R}^n$, corresponding matrices (when expressed in a standard basis) are both orthogonal matrices. Since $B_2=PB_1$, $P=B_2 B_1^{-1}$ holds, and it is quite clear that $P$ is also an orthogonal matrix.

Henry
  • 3,125
2

Though your idea is not the only possible approach, it works. Just let $x$ and $y$ independently run thought all the standard basis vectors, and you equation says that all $n^2$ entries of $P^t P$ are equal to the corresponding entry of the identity matrix.

Marc van Leeuwen
  • 115,048
  • Hmm... I hope that works but don't quite see it, since different bases are used on each side of the equation. Where has the orthonormality of the bases been applied? – Alex137 Nov 14 '13 at 05:44
  • I may have misread your approach. I think you would need that the scalar product of two coordinate vectors on an orthonormal basis remains unchanged when transforming to coordinates on another orthonormal basis; this is so similar to what you are proving that it seems like a circular argument. Anyway it seems much easier to do this using matrix equations, see the other answer. – Marc van Leeuwen Nov 14 '13 at 07:20