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Let's we're in vector space $E$ of dimension 3, and we have the basis $B_0 = (e1,e2,e3)$. I have another basis $B = (e1+e2, e1+e3, e2+e3)$. The problem asks me to give the coordinates of $e1,e2,e3$ "in the basis $B$". I'm not exactly sure what this means. I'm assuming this is what a "change of basis" is, but I'm not too sure.

I know that, if I create a matrix in which the column vectors are the vectors of $B$, then we have the "change of basis matrix", and right-multplying that matrix with a set of coordinates of a vector in basis $B$ with get me the coordinates in the canonical base, right? However, I'm not sure how to apply this knowledge to this particular problem. Any help is appreciated, thanks!

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Let $$B = (e_1+e_2, e_1+e_3, e_2+e_3) = (u_1, u_2, u_3)$$

You need to write $e_1,e_2,e_3$ as linear combination of $u_1, u_2, u_3$

For example $e_1 = (1/2)(u_1+u_2-u_3)$, therefore, in $B$ we have $e_1 = (1/2, 1/2, -1/2)$

Similarly for $e_2 $and $e_3 $ you find the components in new basis $B$.

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  • $\begingroup$ @MohammadRiazi-Karmani Thanks for your answer. There's something I'm not sure about, however. So, we want to find $a,b,c \in R$ such that $a(e_1 + e_2) + b(e_1 + e_3) + c(e_2 + e_3) = (e_1, e_2, e_3)$, correct? So this gives us $ae_1 + ae_2 + be_1 + be_3 + ce_2 + ce_3 = (e_1, e_2, e_3)$. From here, I don't know what to do; am I doing something wrong? Thanks. $\endgroup$ – iaskdumbstuff May 13 '18 at 1:39
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    $\begingroup$ Let $ae_1 + ae_2 + be_1 + be_3 + ce_2 + ce_3 = e_1$ and find coefficients. Similarly for$ e_2$ and$ e_3.$ $\endgroup$ – Mohammad Riazi-Kermani May 13 '18 at 1:44
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The coordinates are just the coefficients of the linear combination. Just find out how to express $B_0$ with $B_1$.

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Here the systematic way for basis transformations using the inverse of a matrix.

  • The transformation matrix describing vectors in $B$ as vectors in $B_0$ is given: $$A = \begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix}$$
  • The transformation matrix describing vectors in $B_0$ as vectors in $B$ is the inverse $A^{-1}$: $$A^{-1} = \frac{1}{2}\begin{pmatrix} 1 & 1 & -1 \\ 1 & -1 & 1 \\ -1 & 1 & 1 \end{pmatrix}$$ So, for example, if $B= (e1+e2, e1+e3, e2+e3) = (b_1,b_2,b_3)$, then $$e_1 = \frac{1}{2}(b_1+b_2-b_3) \mbox{ etc.}$$
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