Lets assume that we want to find the base change matrix from an arbitrary basis $\mathbb{B_u}$ to an ON-basis $\mathbb{B_v}$ in $\mathbb{R^3}$

i.e we want to find $A\vec{z}_{\mathbb{B_u}} = \vec{z}_{\mathbb{B_v}}$

When transforming from an arbitrary basis to a ON-basis the base exchange matrix $A$ looks like this:

$$A = \left( \begin{array}{3} v_1 \cdot u_1, v_2 \cdot u_1, v_3 \cdot u_1 \\ v_1 \cdot u_2, v_2 \cdot u_2, v_3 \cdot u_3 \\ v_1 \cdot u_3, v_2 \cdot u_3, v_3 \cdot u_3 \end{array}\right ) $$

I am trying to understand why it looks like this.

Any vector $\vec{z}$ in $\mathbb{B_v}$ can be written as such:

$\vec{z} = (v_1 \cdot z)v_1 + (v_2 \cdot z)v_2 + (v_3 \cdot z)v_3$

where $v_i \cdot z$ is the coefficient of $\vec{z}$ in $v_i$.

Any vector $\vec{u_1}, \vec{u_2}, \vec{u_3}$ in $\mathbb{B_v}$ can be written as such:

$\vec{u_1} = (v_1 \cdot u_1)v_1 + (v_2 \cdot u_1)v_2 + (v_3 \cdot u_1)v_3$

$\vec{u_2} = (v_1 \cdot u_2)v_1 + (v_2 \cdot u_2)v_2 + (v_3 \cdot u_2)v_3$

$\vec{u_3} = (v_1 \cdot u_3)v_1 + (v_2 \cdot u_3)v_2 + (v_3 \cdot u_3)v_3$

here we have expressed the coefficient matrix for $u_1, u_2, u_3$ in $\mathbb{B_v}$

$A = \left( \begin{array}{3} v_1 \cdot u_1, v_2 \cdot u_1, v_3 \cdot u_1 \\ v_1 \cdot u_2, v_2 \cdot u_2, v_3 \cdot u_3 \\ v_1 \cdot u_3, v_2 \cdot u_3, v_3 \cdot u_3 \end{array}\right ) $

Where the trouble begins: Lets assume we have a vector $\vec{z}$ in $\mathbb{B_u}$:

$\vec{z} = a_1u_1 + a_2u_2 + a_3u_3$

We want to write it in $\mathbb{B_v}$:

$\vec{z} = b_1v_1 + b_2v_2 + b_3v_3$

This means we have to map from $\mathbb{B_u}$ to $\mathbb{B_v}$

$A\vec{z} = A(a_1u_1 + a_2u_2 + a_3u_3) = b_1v_1 + b_2v_2 + b_3v_3$

Does that mean that we simply have to do the multiplication with the matrix $A$ that was developed above? Can you please explain to me why that is so?

Thank you!


Going from $\mathbb{B_u}$ to $\mathbb{B_v}$ your matrix should be $A=$ Mtx$_{\mathbb{B_v},\mathbb{B_u}}(\boldsymbol \iota)=[[u_1]_\mathbb{B_v},[u_2]_\mathbb{B_v},[u_3]_\mathbb{B_v}],$ which is then actually the transpose of your matrix $A$.

Now take $z = a_1u_1 + a_2u_2 + a_3u_3$. Then \begin{eqnarray} z &=& (v_1 \cdot z)v_1 + (v_2 \cdot z)v_2 + (v_3 \cdot z)v_3\\ &=& (v_1 \cdot (a_1u_1 + a_2u_2 + a_3u_3))v_1 + (v_2 \cdot (a_1u_1 + a_2u_2 + a_3u_3))v_2 \\ && \quad \quad + (v_3 \cdot (a_1u_1 + a_2u_2 + a_3u_3))v_3 \\ &=& (a_1(v_1 \cdot u_1) + a_2(v_1 \cdot u_2) + a_3(v_1 \cdot u_3))v_1+(a_1(v_2 \cdot u_1) + a_2(v_2 \cdot u_2) + a_3(v_2 \cdot u_3))v_2 \\ && \quad \quad +(a_1(v_3 \cdot u_1) + a_2(v_3 \cdot u_2) + a_3(v_3 \cdot u_3))v_3, \end{eqnarray} which is exactly the same as multiplying $\vec{z}_{\mathbb{B_u}}$ on the left by $A$...except, just be careful to mix vectors with their representation as column vectors. Above I have calculated the vector form, but as a column vector with respect to $\mathbb{B_v}$ we than have: \begin{equation} \begin{bmatrix}a_1(v_1 \cdot u_1) + a_2(v_1 \cdot u_2) + a_3(v_1 \cdot u_3) \\ a_1(v_2 \cdot u_1) + a_2(v_2 \cdot u_2) + a_3(v_2 \cdot u_3) \\ a_1(v_3 \cdot u_1) + a_2(v_3 \cdot u_2) + a_3(v_3 \cdot u_3)\end{bmatrix},\end{equation} which is the coefficients of each of the vectors $v_1,v_2,v_3$ ordered by row as per definition of coordinate vectors.

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  • $\begingroup$ Thank you for your answer Christiaan! $\endgroup$ – Lukas Arvidsson May 16 '14 at 6:33
  • $\begingroup$ @LukasArvidsson always a pleasure! $\endgroup$ – Christiaan Hattingh May 16 '14 at 6:38
  • $\begingroup$ After looking at your answer some more, I am a bit curious about the $a_1, a_2, a_3$. In my book the base change matrix is said to be: $\left( \begin{array}{3} v_1 \cdot u_1, v_1 \cdot u_1, v_1 \cdot u_1 \\ v_2 \cdot u_2, v_2 \cdot u_2, v_2 \cdot u_3 \\ v_3 \cdot u_3, v_3 \cdot u_3, v_3 \cdot u_3 \end{array}\right )$. But from your calculation the addition of $a_1, a_2, a_3$ seems logical. Any idea what my book presents it without the $a_j$ coefficients? $\endgroup$ – Lukas Arvidsson May 18 '14 at 15:14
  • $\begingroup$ Hi @LukasArvidsson. no the change of base matrix will not have the $a_i$. What I have calculated there is just $A\vec{z}_{\mathbb{B_u}}$ $\endgroup$ – Christiaan Hattingh May 18 '14 at 15:19
  • $\begingroup$ Thank you for your comment! It makes sense and I see what you mean. $\endgroup$ – Lukas Arvidsson May 18 '14 at 15:39

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