Orthonormal Basis times a real Matrix

[linearishard]

Hi!

I have an orthonormal basis for vector space $V$, $\{u_1, u_2, ..., u_n\}$. If $(v_1, v_2, ..., v_n) = (u_1, u_2, ... u_n)A$ where $A$ is a real $n\times n$ matrix, how do I prove that $(v_1, v_2, ... v_n)$ is an orthonormal basis if and only if $A$ is an orthogonal matrix?

Thanks!

 

[S.G. Janssens]

Could you explain how $(u_1, u_2, \ldots, u_n)A$ is defined? I don't understand, since the $u_j$ are vectors in an arbitrary $n$-dimensional (real) vector space, while $A$ is an $n \times n$ matrix.

Do you mean to write $v_j = Au_j$ where $A$ is an endomorphism of a real inner product space $V$ and then prove that the $v_j$ form an orthonormal basis iff $A$ is an orthogonal transformation?

 

[linearishard]

Hi, yes that is what I meant! Sorry!

 

[Euge]

Hi linearishard,

First suppose $A$ is orthogonal. Then $A^TA = I$, so for all $i,j\in \{1,2,\ldots, n\}$, $$\langle Au_i, Au_j\rangle = \langle A^TAu_i, u_j\rangle =\langle Iu_i, u_j\rangle = \langle u_i, u_j\rangle = \delta_{ij}$$ where the last equality follows from orthonormality of $\{u_1,\ldots, u_n\}$. Therefore, $\{v_1,\ldots, v_n\}$ is an orthonormal set. The set $\{v_1,\ldots, v_n\}$ is also linearly independent, for if $$\sum_{i = 1}^n c_i v_i = 0$$ is a linear dependence relation, then $$A\left(\sum_{i = 1}^n c_i u_i\right) = 0$$ Pre-multiplying both sides of the equation by $A^T$ and using the fact $A^TA = I$, one receives $$\sum_{i = 1}^n c_i u_i = 0$$ whence $c_1 = \cdots = c_n = 0$ by linear independence of $\{u_1,\ldots, u_n\}$. Thus, $\{v_1,\ldots, v_n\}$ is a basis for $V$.

Conversely, suppose $\{v_1,\ldots, v_n\}$ an orthonormal basis. For all indices $i$ and $j$, $\langle Au_i, Au_j\rangle = \delta_{ij}$. So since $\{u_1,\ldots, u_n\}$ is an orthonormal basis then for every $i$ $$A^T Au_i = \sum_{j = 1}^n \langle A^T A u_i, u_j\rangle u_j = \sum_{j = 1}^n \langle Au_i, Au_j\rangle u_j = \sum_{j = 1}^n \delta_{ij}u_j = u_i$$ Consequently $A^TA = I$, i.e., $A$ is orthogonal. $\blacksquare$

 

[linearishard]

Thank you for your response, could you please explain to me the logic behind the last line? How do you go from <ATAui,uj> to <Aui,Auj>?

 

[Euge]

Since $A^T$ is the adjoint of $A$, $\langle A^Tx,y\rangle = \langle x, Ay\rangle$ for all $x,y\in V$. In particular, the equality holds when $x = Au_i$ and $y = u_j$.

 

[I like Serena]

Since we're apparently working with real numbers, we can write:
$$\langle A^Tx,y\rangle
= (A^Tx)^T \cdot y
= (x^TA)\cdot y
=x^T\cdot (Ay)
=\langle x, Ay\rangle
$$
This is generally true for the dot product of vectors with real numbers.
For complex numbers the same thing holds, just with the conjugate transpose or adjoint (usually written as $A^\dagger$, $A^*$, or $A^H$) instead of the transpose $A^T$.