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Let $A=\{(1,0,5) (4, 5,5)(1,1,4)\}; B= \{(1,3,2)(-2,-1,1)(1,2,3) \}$. To find change of basis matrix.

$\boxed{ M_{A\to B} = M_{A\to e}M_{e\to B} } ---(1)\\= \begin{pmatrix} 1 & 4 & 1 \\ 0 & 5 & 1 \\ 5 & 5 & 4 \end{pmatrix}^{-1} \begin {pmatrix} 1 & -2 & 1 \\ 3 & -1 & 2 \\ 2 & 1 & 3 \end {pmatrix} = \boxed {\begin {pmatrix} -2 & -2 & -1 \\ 0 & -1 & 0 \\ 3 & 4 & 2 \end {pmatrix} } -----(2)$

Is equation (1) correct??? or it is $M_{A \to B} = M_{e \to B} M_{A \to e}$ as given in solution part of LINK:- How to construct change of basis matrix (I am confused here pls correct me if i am wrong...)

To reconfirm i did it in following way:-

to find coordinates of B interms of A

$(1,3,2)= a(1,0,5)+b(4,5,5)+c(1,1,4)=(a+4b+c,5b+c,5a+5b+4c)=[-2,0,3]_B$

Similarly $(-2,-1,4) \sim [-2,-1,4]_B \\ (1,2,3) \sim [-1,0,2]_B$

THis boils down to same above matrix by (2) enter image description here

Magneto
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  • There is also a typo, in your hypotesis it should be $M_{A\to B} = M_{e\to A}M_{B\to e}$ but how explained $M_{e\to A}M_{B\to e}=M_{B\to A}$. – user Jun 20 '18 at 10:45
  • Could you define what you mean by "change of basis matrix"? Because there is closely related "change of coordinate matrix", which is precisely the inverse of that. Maybe the confusion is there. – A.Γ. Jun 20 '18 at 11:03
  • @A.Γ. Sir i did not go deep. I am at linear transformations part of linear algebra. Change of basis matrix or change of coordinate matrix not same? pls elaborate... Change of basis matrix -- $[v]B = Q^{-1}{A \to B} [v]_A$ – Magneto Jun 20 '18 at 11:08
  • Your first way looks ok to me, the second is unclear, how did you get those matrices in "matrix method"? – A.Γ. Jun 20 '18 at 11:52
  • Got it. I misundersttod change of basis matrix. Re-read textbook again. Pls see my last below comment. – Magneto Jun 20 '18 at 12:38

1 Answers1

1

We need

$$M_{A\to B} = M_{e\to B}M_{A\to e}$$

that is

$$\begin {pmatrix} 1 & -2 & 1 \\ 3 & -1 & 2 \\ 2 & 1 & 3 \end {pmatrix}^{-1}\begin{pmatrix} 1 & 4 & 1 \\ 0 & 5 & 1 \\ 5 & 5 & 4 \end{pmatrix} $$

indeed the first matrix takes a vector in basis $A$ to the standard basis and the second takes a vector in the standard basis to the basis $B$, therefore the product matrix takes a vector from basis $A$ to basis $B$.

user
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  • can u pls have a look at part 2 or the other way i did to find change of basis matrix.

    To find change of basis matrix i followed -

    output basis = express in terms of input basis

     – Magneto Jun 20 '18 at 10:29
  • @anirudhb The matrix M you have found with tour method changes basis from B to A, indeed we have $M(1,0,0)=(-2,0,3)$ which are the components of $(1,3,2)$ in the basis $A$. – user Jun 20 '18 at 10:36
  • I understood matrix method. Actually i am bit confused about other way. For ex., @gimusi take $M_{e \to A}. \$$ (1,0,5) = 1e_1+0e_2+5e_3 \ $... like that i will express output basis vector interms of input basis vector. But i find using same way, output basis B vector = interms of input basis A vectors linear comination, i am gettting $M^{-1}$. Pls clarify. – Magneto Jun 20 '18 at 10:47
  • $M_{e \to A} \cdot (1,0,5)$ simply gives the components for $(1,0,5)$ in the basis $A$, I don't understand your doubts on that. Please explain better. – user Jun 20 '18 at 10:51
  • Sir i uploaded my worked out image. Pls have a look – Magneto Jun 20 '18 at 11:04
  • Also by your method you obtain a matrix which takes a vector in basis B and gives its components in basis A, thus $M_{B\to A}$ according to matrix method. – user Jun 20 '18 at 11:09
  • Pls see in uploaded image. I have changed in matrix representation – Magneto Jun 20 '18 at 11:11
  • what u r saying is correct. I did mistake in change of basis matrix. 1st feew lines in uploaded image - matrix i calculated is from A->standard basis. I mistook it as standard basis --> A. Now all problem got right. Took time to re-read textbook and understand. – Magneto Jun 20 '18 at 12:34
  • $(1,0,5 ) = a(1,3,2)+b(-2,-1,1)+c(1,2,3) =(a-2b+c, 3a-b+2c,2a+b+3c) \implies [a,b,c] = [-2,0,3]\ $ Similarly doing it igot $M_{A \to B} = \begin {pmatrix} {-2 & 0 &-1 \ 0 & -1 & 0 \ 3 & 2 &2 } \end{pmatrix}$ – Magneto Jun 20 '18 at 12:37