Linear algebra problem

AngeloBenjamin0

New member
Hi, I don't know how to solve this problem. Any help will be helpful!
$$\displaystyle B=\{v_{1} ,v_{2} ,v_{3}\} \ B'=\{v_{1} -v_{2} ,v_{1} -v_{3} ,v_{3}\} .\ If\ [ u]_{B} =( 3,1,-2) \ then:\\ [ u]_{B'} =( 2,5,-2) \ or\ [ u]_{B'} =( 1,4,2) \ \ or\ [ u]_{B'} =( 3,1,-2) ,\ or\ another...$$

Subhotosh Khan

Super Moderator
Staff member
Hi, I don't know how to solve this problem. Any help will be helpful!
$$\displaystyle B=\{v_{1} ,v_{2} ,v_{3}\} \ B'=\{v_{1} -v_{2} ,v_{1} -v_{3} ,v_{3}\} .\ If\ [ u]_{B} =( 3,1,-2) \ then:\\ [ u]_{B'} =( 2,5,-2) \ or\ [ u]_{B'} =( 1,4,2) \ \ or\ [ u]_{B'} =( 3,1,-2) ,\ or\ another...$$
Please show us what you have tried and exactly where you are stuck.

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AngeloBenjamin0

New member
I think that $$\displaystyle [ u]_{B} : (3,1,-2) =a\cdot v_{1} +b\cdot v_{2} +c\cdot v_{3} \ ( base\ of\ B)\\ [ u]_{B'} : (3,1,-2) \ =\ a\cdot ( v_{1} -v_{2}) +b\cdot ( v_{1} -v_{3}) +c\cdot v_{3} \ ( base\ of\ B')$$
And the vectors of the base of B are the same of the base of B', but I don't know how to use that.

pka

Elite Member
$$\displaystyle B=\{v_{1} ,v_{2} ,v_{3}\} \ B'=\{v_{1} -v_{2} ,v_{1} -v_{3} ,v_{3}\} .\ If\ [ u]_{B} =( 3,1,-2) \ then:\\ [ u]_{B'} =( 2,5,-2) \ or\ [ u]_{B'} =( 1,4,2) \ \ or\ [ u]_{B'} =( 3,1,-2) ,\ or\ another...$$
This is a standard change of base problem. It should with a three by three matrix.
$$\left<\vec{v_1},\vec{v_2}.\vec{v_3}\right>$$$$\left[ {\begin{array}{*{20}{c}} 1&{ -1}&0 \\ 1&0&0 \\ 0&0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\overrightarrow {{v_1}} } \\ {\overrightarrow {{v_2}} } \\ {\overrightarrow {{v_3}} } \end{array}} \right] = \left\langle {\overrightarrow {{v_1}} - \overrightarrow {{v_2}} ,\overrightarrow {{v_1}} - \overrightarrow {{v_3}} ,\overrightarrow {{v_3}} } \right\rangle$$ [note LaTeX is not cooperating: that is a -1 in the first row second calcium]

pka

Elite Member
The matrix should be
$$\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0 \\ 1&0&{ - 1} \\ 0&0&1 \end{array}} \right] \left[ {\begin{array}{*{20}{c}} {\overrightarrow {{v_1}} } \\ {\overrightarrow {{v_2}} } \\ {\overrightarrow {{v_3}} } \end{array}} \right] = \left\langle {\overrightarrow {{v_1}} - \overrightarrow {{v_2}} ,\overrightarrow {{v_1}} - \overrightarrow {{v_3}} ,\overrightarrow {{v_3}} } \right\rangle$$

AngeloBenjamin0

New member
The matrix should be
$$\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0 \\ 1&0&{ - 1} \\ 0&0&1 \end{array}} \right] \left[ {\begin{array}{*{20}{c}} {\overrightarrow {{v_1}} } \\ {\overrightarrow {{v_2}} } \\ {\overrightarrow {{v_3}} } \end{array}} \right] = \left\langle {\overrightarrow {{v_1}} - \overrightarrow {{v_2}} ,\overrightarrow {{v_1}} - \overrightarrow {{v_3}} ,\overrightarrow {{v_3}} } \right\rangle$$
It shouldn't be equal to (3,1,-2)? Because I'm looking for the coordinates of u in base B'. Maybe I'm not understanding your answer.

pka

Elite Member
It shouldn't be equal to (3,1,-2)? Because I'm looking for the coordinates of u in base B'. Maybe I'm not understanding your answer.
I have taught linear algebra many times, but I have no idea what you are asking in this post.
If we know exactly what you are asking then maybe we can help.

AngeloBenjamin0

New member
I have taught linear algebra many times, but I have no idea what you are asking in this post.
If we know exactly what you are asking then maybe we can help.
I already post the exact question. The problem ask for the coordinates of the vector U in basis B'. The matrix that you write was the change of base matrix from basis B' to B? Then If I invert it, I'll get to the change of base matrix from basis B to B' ? Thanks.

pka

Elite Member
I already post the exact question. The problem ask for the coordinates of the vector U in basis B'. The matrix that you write was the change of base matrix from basis B' to B? Then If I invert it, I'll get to the change of base matrix from basis B to B' ?
You say "I already post the exact question." I have right to doubt you and I don't.
However, it does appear to me that it was written by a math-educator and not a mathematician.
Any $$D_3$$ vector space has three vector basis. Usually $$<1,0,0>,<0,1,0>,<0,0,1>$$.
From your post it seems that it calls for a basis such as $$<1,-1,0>,<1,0,-1>,<0,0,1>$$.
That is indeed a new basis for $$D_3$$. And I gave you a matrix that translates a point in the standard basis into a point in the transformed basis.
Now it seems that that is not at all what you understand the question to be. So please try to explain the meaning of the exact question.