Determine the coordinates of vector $v$ with respect to the base $u_1$, $u_2$ and $u_3$

Question

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?

Answer 1

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.

Answer 2

$\begin{bmatrix} 1&2&-2\\3&2&1\\2&2&-3 \end{bmatrix}\mathbf u = \mathbf v$ 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)\\ $