As the title, let $(V,\langle,\rangle)$ be a complex inner product space and assume $S_1=(u_1,\ldots,u_n)$, $S_2=(v_1,\ldots,v_n)$ are orthonormal bases of $V$. Prove that the change of basis matrix $M_ IV(S_2,S_1)$ is a unitary matrix.

(There is a hint that let $S$ be the operator s.t. $S(u_i)=v_i$ and prove this is a unitary operator.)

  • $\begingroup$ It's always a good idea to show a bit of your work, then show us where you're stuck and ask a specific question (as opposed to just stating the problem straight out of the book). $\endgroup$
    – icurays1
    Dec 2, 2012 at 5:49

1 Answer 1


We need to show that $\langle Mx,My\rangle=\langle x,y\rangle$ for all $x,y$. By definition, the change of basis matrix $M$ has the property that

$$ Mu_i=v_i $$

Let $x=\alpha_1u_1+\ldots+\alpha_nu_n$ and $y=\beta_1u_1+\ldots\beta_nu_n$. Then, since $\{u_i\}$ is an orthonormal basis, $\langle u_i,u_j\rangle=\delta_{ij}$, the Kronecker delta. Thus, by expanding out the inner product $\langle x,y\rangle$, we see that $$ \langle x,y\rangle=\sum_j\alpha_j\bar{\beta}_j $$

Now compute $\langle Mx,My\rangle$:

$$ \begin{align*} \langle Mx,My\rangle&=\langle M(\alpha_1u_1+\ldots+\alpha_nu_n),M(\beta_1u_1+\ldots+\beta_nu_n)\rangle\\ &=\langle \alpha_1Mu_1+\ldots+\alpha_nMu_n,\beta_1Mu_1+\ldots+\beta_nMu_n\rangle\\ &=\langle \alpha_1v_1+\ldots+\alpha_nv_n,\beta_1v_1+\ldots+\beta_nv_n\rangle\\ &=\sum_i\sum_j\alpha_i\bar{\beta}_j\langle v_i,v_j\rangle \end{align*} $$

Since $\{v_i\}$ is orthonormal, $\langle v_i,v_j\rangle=\delta_{ij}$, and so the above is exactly $\langle x,y\rangle$

  • $\begingroup$ why is this true? " By definition, the change of basis matrix M has the property that : Mui=vi"? is $Mui=av_1+bv_2....cv_n$(a,b.. scalars) ? $\endgroup$
    – KIMKES1232
    Jun 22, 2019 at 16:00
  • 1
    $\begingroup$ Not quite, u_i is a single vector. M would convert it to one single vector in the other basis. I.e. M * u_i = v_i, for any i, but a single one. $\endgroup$ Sep 29, 2020 at 16:36

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