What is the coordinate vector of x relative to B?

Question

I've been working on this simple coordinate system problem for at least 15 minutes now, trying to figure out how my answer is incorrect.

Here is the question; it's in the "true-or-false" format (you can ignore their answer for now):

And here is my work, in which I am unable to find an error:

This is the part of the textbook (David C. Lay's Linear Algebra and Its Applications) that I reference:

It seems to me that their solution contradicts the definition given in the textbook, but maybe I'm simply not seeing it or doing the problem incorrectly.

Any help is appreciated. Thanks.

Answer 1

There is no contradition. When we write $[x]_\mathcal{B},$ we are referring to the coordinates of a vector $x$ with respect to basis $\mathcal{B}.$ In your case, the vector of interest is $x = \begin{pmatrix} 3 \\ 5 \end{pmatrix}.$ The coordinates of $x$ with respect to $\mathcal{B} = \left\{ \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \end{pmatrix} \right\}$ is indeed $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ because $\begin{pmatrix} 3 \\ 5 \end{pmatrix} = 1 \cdot \begin{pmatrix} 1 \\ 1 \end{pmatrix} + 2 \cdot \begin{pmatrix} 1 \\ 2 \end{pmatrix}.$ What you calculated was $\left[ \begin{pmatrix} 1 \\ 2 \end{pmatrix} \right]_\mathcal{B}.$

Answer 2

What's confusing is that without knowing the definition, there are two intuitive ways ways to "guess" $\begin{pmatrix} 3\\5 \end{pmatrix}_\mathcal{B}$. The one is $3b_1 + 5b_2$. The other (the correct way) is that under the standard basis, the vector is $\begin{pmatrix} 3\\5 \end{pmatrix}$, what is it in terms of $b_1$ and $b_2$? An easy way to remember the latter is correct is that vectors, by default, are always assumed to be under the standard basis. So $\begin{pmatrix} 3\\5 \end{pmatrix}_\mathcal{B}$ is a function that plugs in the vector $\begin{pmatrix} 3\\5 \end{pmatrix}=\begin{pmatrix} 3\\5 \end{pmatrix}_E$.