I have two exercises, $(1)$ and $(2)$. In exercise $(1)$ I was given a basis $e_1,e_2,e_3$ (no given coordinates) and $v=(1,4,3)$, then given the relationship between this basis and a new basis in which I was supposed to determine the coordinates of $v$.

\begin{align*} e^{'}_1&=e_1+e_2+e_3\\ e^{'}_2&=e_3\\ e^{'}_3&=e_1+e_3.\\ \end{align*}

Since $v=e_1+4e_2+3e_3$, I did this by solving for $e_1,e_2,e_3$ and then multiplying the results with original coordinates, like this:

$$ \underbrace{e^{'}_3-e^{'}_2}_{e_1}+4(\underbrace{e^{'}_1-e^{'}_3}_{e_2})+3\underbrace{e^{'}_2}_{e_3}=4e^{'}_1+2e^{'}_2-3e^{'}_3. $$

This is the correct answer and I have no issues with this exercise, I believe I have understood it. However, in a similar exercise, $(2)$, I am given the four vectors

\begin{align*} u_1&=(1,2,-2)\\ u_2&=(3,2,1)\\ u_3&=(2,2,-3)\\ v&=(1,4,-8). \end{align*}

I am supposed to determine the coordinates of $v$ with respect to the basis $u_1, u_2,u_3$.

In this exercise $(2)$, however, not only do I get the wrong answer using the same method as in exercise $(1)$, but the suggested solution is completely different. The solution is supposed to be finding coefficients $c_1,c_2,c_3$ to make the $x,y,z$ coordinates of my $u_1,u_2,u_3$ equal to the $x,y,z$ of the vector $v$, like this:

$$ c_1 \begin{bmatrix} 1\\ 2\\ -2\\ \end{bmatrix} + c_2 \begin{bmatrix} 3\\ 2\\ 1\\ \end{bmatrix} + c_3 \begin{bmatrix} 2\\ 2\\ -3\\ \end{bmatrix} = \begin{bmatrix} 1\\ 4\\ -8\\ \end{bmatrix} $$

So my question is, why would I use this method here, instead of the method in exercise $(1)$? What is the difference between the two that I'm missing? Are the two methods perhaps somehow equivalent, just that I have made a miscalculation somewhere?


Since you stated that you don't want to use matrices, here is a spelled out explanation.

I guess you can use both 'methods' in both exercises. Let's find the solution for (2) with your way.

First, we should represent $e_1,e_2,e_3$ in the basis $u_1,u_2,u_3$. From the coordinates of $u_1,u_2,u_3$ we get $$\begin{align*}u_1\phantom{+u_2+u_3}&=\phantom{1}e_1+2e_2-2e_3\\ \phantom{u_1+}u_2\phantom{+u_3}&=3e_1+2e_2+\phantom{1}e_3\\ \phantom{u_1+u_2+}u_3&=2e_1+2e_2-3e_3\end{align*}$$ To obtain the coordinates of $e_i$ with respect to the $u_i$, solve this system of equations for the $e_i$.

Comparing this with Doug M's answer, we are in fact now calculating the inverse matrix in his second line.

We obtain $$\begin{matrix} -\frac 45u_1 &+& \frac 15u_2 &+& \frac 35u_3&=&e_1&&&\\ \phantom{-}\frac{11}{10} u_1 &+& \frac 1{10} u_2 &-& \frac 7{10} u_3&=&&e_2&\\ \phantom{-}\frac 15u_1 &+& \frac 15 u_2 &-& \frac 25 u_3&=&&&e_3 \end{matrix}$$ Since we know that $v=e_1+4e_2-8e_3$, we obtain $$\begin{align*}v&=\underbrace{-\frac 45 u_1 + \frac 15 u_2 + \frac 35 u_3}_{e_1} + 4\left(\underbrace{\frac {11}{10} u_1 + \frac 1{10} u_2 - \frac 7{10} u_3}_{e_2}\right) -8\left(\underbrace{\frac 15 u_1 + \frac 15 u_2 - \frac 25 u_3}_{e_3}\right) \\ &=\frac{10}5 u_1-\frac55u_2+\frac55u_3 = 2u_1-u_2+u_3\end{align*}$$

Now for method 2:

The equation $$ c_1 \begin{bmatrix} 1\\ 2\\ -2\\ \end{bmatrix} + c_2 \begin{bmatrix} 3\\ 2\\ 1\\ \end{bmatrix} + c_3 \begin{bmatrix} 2\\ 2\\ -3\\ \end{bmatrix} = \begin{bmatrix} 1\\ 4\\ -8\\ \end{bmatrix} $$ describes exactly the representation of $v$ in the basis $u_1,u_2,u_3$. If we solve the system of equations, we obtain the coordinates $c_1,c_2,c_3$ of $v$ with respect to $u_1,u_2,u_3$, which are $c_1=2, c_2=-1, c_3=1$ (just plug in $c_1,c_2,c_3$ into the equation).

Let's apply this method to (1). We search for $\lambda_1,\lambda_2,\lambda_3$ such that $$ \lambda_1 \begin{bmatrix} 1\\ 1\\ 1\\ \end{bmatrix} + \lambda_2 \begin{bmatrix} 0\\ 0\\ 1\\ \end{bmatrix} + \lambda_3 \begin{bmatrix} 1\\ 0\\ 1\\ \end{bmatrix} = \begin{bmatrix} 1\\ 4\\ 3\\ \end{bmatrix} $$The solution is $\lambda_1=4, \lambda_2=2, \lambda_3=-3$, exactly what you got.

So both 'methods' yield the same results. What is the difference? Method (1) yields a way to calculate the coordinates of an arbitrary vector with respect to the basis $e'_1,e'_2,e'_3$. If I give you another vector, say $v'=\begin{pmatrix}13\\3\\7\end{pmatrix}$, you can easily calculate its coordinates by writing $v=13e_1+3e_2+7e_3$ and plugging in your representation of $e'_i$ with respect to $e_i$.

This is the same as multiplying the inverse matrix of Doug M. to the vector $v'$. The computational investment is only expanding the multiplication. But we had to calculate the inverse matrix before, which is a lot of work. You had to find the representation of $e_i$ with respect to the $e'_i$s, which is the same

Method (2) computes the coordinates directly. This is faster since you don't have to find the coordinates of all $e_i$ with respect to the $e'_i$s first. But if I give you another vector $v'$, you have to start over from scratch and solve the system of linear equations again.

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$\begin{bmatrix} 1&2&-2\\3&2&1\\2&2&-3 \end{bmatrix}\mathbf u = \mathbf v$ will transform a vector in the basis $U$ to the standard basis.

$\begin{bmatrix} 1&2&-2\\3&2&1\\2&2&-3 \end{bmatrix}^{-1}\mathbf v = \mathbf u$ transforms a vector from the standard basis to the basis for $U.$

Then you will need to use algebra to show that:

$-\frac 45 u_1 + \frac 15 u_2 + \frac 35 u_3 = (1,0,0)\\ \frac 65 u_1 + \frac 1{10} u_2 - \frac 7{10} u_3 = (0,1,0)\\ \frac 15 u_1 + \frac 15 u_2 - \frac 25 u_3 = (0,0,1)\\ $

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  • $\begingroup$ I think I've seen that somewhere but unfortunately it doesn't really help answering my question since I'm not using matrices at this point, and since I'm trying to figure out the differences between exercises $(1)$ and $(2)$. $\endgroup$ – Chisq Jul 30 '18 at 17:34

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