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Change of basis matrix for $B_2\{a_1=(1,-1,0),a_2=(1,0,-1)\}$ to $B_1\{b_1=(0,1,-1),b_2=(1,1,-2)\}$, where $B_{1,2}$ are bases of $V=\{(x,y,z)|x+y+z=0\}$.

I haven't practiced that for a very long time and the definitions are a rather formal, total and loaded. I think I got the idea right, but I think I should have it checked:

I know that $b_1=-a_1+a_2,b_2=-a_1+2a_2$. Hence, for $P=\begin{pmatrix}-1&-1\\1&2\\ \end{pmatrix}$, $[b_1]_{B_1}=P[b_1]_{B_2}$ and the same goes for $b_2$. How do I arrive at the requested matrix from here? What about the standard basis? Is $B_2$ the standard basis here? I am confused, and could use some help.

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  • $\begingroup$ Basis in R3? Dim R3=3 , no two... $\endgroup$ – Martín Vacas Vignolo Mar 8 '16 at 15:19
  • $\begingroup$ I tried to do that, but I get non-invertible matrices... $\endgroup$ – Meitar Abarbanel Mar 8 '16 at 15:20
  • $\begingroup$ B1 and B2 aren't basis... $\endgroup$ – Martín Vacas Vignolo Mar 8 '16 at 15:21
  • $\begingroup$ I have been remiss in stating they are bases of a subspace V. Which is $\{x+y+z=0\}$, sorry. $\endgroup$ – Meitar Abarbanel Mar 8 '16 at 15:23
  • $\begingroup$ Duplicate of math.stackexchange.com/q/1688392/265466 $\endgroup$ – amd Mar 8 '16 at 19:15
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The confusion may go as follows: the vector space $V$ is basically 2 dimensional and if we do as usual $a_1 = \pmatrix{1 \\ 0}$, $a_2 = \pmatrix{0 \\ 1}$ then the matrix $P$ maps this basis to the basis $b_1 = \pmatrix{-1 \\ 1}$, $b_2 = \pmatrix{-1 \\ 2}$. So yes, $P$ is the correct change-of-base matrix in these terms, but it is NOT the matrix that maps $\pmatrix{1 \\ -1 \\ 0} \mapsto \pmatrix{0 \\ 1 \\ -1}$ and $\pmatrix{1 \\ 0 \\ -1} \mapsto \pmatrix{1 \\ 1 \\ 2}$, as obviously, that matrix needs to be $3 \times 3$.

To find the $3 \times 3$ matrix, try this: note that $V$ is set of vectors orthogonal to $\pmatrix{1 \\ 1 \\ 1}$, with the standard Euclid inner product. Start with a general $3 \times 3$ matrix having 9 unknown coefficients and the require it to map $a_1$ to $b_1$, $a_2$ to $b_2$ and to leave $\pmatrix{1 \\ 1 \\ 1}$ invariant. Unless I am wrong, such matrix would change the basis of $V$, leaving the orthogonal subspace in $\mathbb{R}^3$ invariant.

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  • $\begingroup$ Thank you Martin. Can you explain what is the importance of leaving $(1,1,1) $ invariant? How does one know to look at that? $\endgroup$ – Meitar Abarbanel Mar 9 '16 at 0:57
  • $\begingroup$ Well, basically you need to describe a transformation of 3 (linearly independent) vectors to obtain a unique linear mapping, that is the practical reason. I am not sure whether mapping it to $0$ vector or some other multiple of itself would work for you - it quite depends on what you want to do. E.g. mapping the vector to $0$ would, trivially, yield a noninvertible transformation. $\endgroup$ – Martin Plávala Mar 9 '16 at 8:18

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