The formulation of the problem is the following:

Let $u_1=(1,1,1)$, $u_2=(2,0,0)$, and $u_3=(0,3,0)$ be a basis of $\mathbb{R}^3$ and let $\{e_1, e_2, e_3\}$ be the canonical basis of $\mathbb{R}^3$. Calculate the coordinates of $w=2e^{*2}+3e^{*3}$ in dual basis $\{u^{*1}, u^{*2}, u^{*3}\}$.

So I knew that if you let $P$ be the change of basis matrix from $\{u_i\}$ to $\{e_i\}$, then, the change of basis matrix from $\{u^{*i}\}$ to $\{e^{*i}\}$, would be $(P^{-1})^{\top}$. If the problem asks me the opposite way, from $\{e^{*i}\}$ to $\{u^{*i}\}$, the matrix would be $((P^{-1})^{\top})^{-1}=P^{\top}$. Thus, the solution should be $$(0,2,3)\begin{pmatrix} 1 & 2 & 0\\ 1 & 0 & 3\\ 1 & 0 & 0 \end{pmatrix}^{\top}=(4,9,0)$$.

The issue is that the solution says it's $$(0,2,3)\begin{pmatrix} 1 & 2 & 0\\ 1 & 0 & 3\\ 1 & 0 & 0 \end{pmatrix}=(5,0,6)$$.

Where am I mistaken?

  • $\begingroup$ What is your definition of a change of basis matrix? I think you should either be using column vectors and muliplying them by change of basis matrices on the left, or write the coordinates of $ u_i $ in the canonical basis in the rows of the change of basis matrix. $\endgroup$ Dec 2, 2022 at 10:28
  • $\begingroup$ Already solved the problem though I dont know why the solution I wrote on the comments was deleted... $\endgroup$
    – Conreu
    Dec 2, 2022 at 11:40
  • $\begingroup$ @JoanSGF you can't see it because that answer was deleted $\endgroup$
    – janmarqz
    Jan 3, 2023 at 14:41

1 Answer 1


Let us write \begin{eqnarray*} u_1&=&e_1+e_2+e_3\\ u_2&=&2e_1\\ u_3&=&3e_2 \end{eqnarray*} so we can associate the matrix $$B=\left( \begin{array}{ccc} 1&2&0\\ 1&0&3\\ 1&0&0\\ \end{array} \right).$$ That $B$ has determinant different from zero is due that those $u_i$ are linearly independent and they form a new basis. Then $B^{-1}$ exists, and in fact obeys $B^{-1}B=1\!\!1$, that is $$ \left( \begin{array}{ccc} 0&0&1\\ \frac{1}{2}&0&-\frac{1}{2}\\ 0&\frac{1}{3}&-\frac{1}{3}\\ \end{array} \right) \left( \begin{array}{ccc} 1&2&0\\ 1&0&3\\ 1&0&0\\ \end{array} \right)= \left( \begin{array}{ccc} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{array} \right).$$ Here, you can see how the rows of $B^{-1}$ behave as the duals $u^{*s}$ i.e. $$u^{*i}(u_j)=\delta^i_j.$$ as well as the canonical basis $e^{*i}(e_j)=\delta^i_j$. Then \begin{eqnarray*} u^{*1}&=&e^{*3}\\ u^{*2}&=&\frac{1}{2}e^{*1}-\frac{1}{2}e^{*3}\\ u^{*3}&=&\frac{1}{3}e^{*2}-\frac{1}{3}e^{*3} \end{eqnarray*} that immediately gives \begin{eqnarray*} e^{*1}&=&u^{*1}+2u^{*2}\\ e^{*2}&=&u^{*1}+3u^{*3}\\ e^{*3}&=&u^{*1}. \end{eqnarray*} Finally, subbing into you covector $w=2e^{*2}+3e^{*3}$, you'll get \begin{eqnarray*} w&=&2(u^{*1}+3u^{*3})+3u^{*1},\\ &=&5u^{*1}+6u^{*3}. \end{eqnarray*}


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