# Change of basis for dual bases

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?

• 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. Dec 2, 2022 at 10:28
• Already solved the problem though I dont know why the solution I wrote on the comments was deleted... Dec 2, 2022 at 11:40
• @JoanSGF you can't see it because that answer was deleted Jan 3, 2023 at 14:41

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 $$\begin{eqnarray*} w&=&2(u^{*1}+3u^{*3})+3u^{*1},\\ &=&5u^{*1}+6u^{*3}. \end{eqnarray*}$$