## Possible Solution to #9 |
TLindberg |
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Hey just wanted to see if my logic and calculations for number 9 make sense and find the correct answer. First I try to find the Identity matrix from \beta to \gamma. This is a 3 x 3 matrix with each column representing a transformation from a basis vector in \beta to a basis vector in \gamma. I got this as: \begin{pmatrix} 1 & 2 & 2 \\ 0 & 3 & 1 \\ 2 & 2 & 0 \end{pmatrix} Next I calculate the inverse of this matrix because we need to find the identity matrix from \gamma to \beta. I augment the matrix above with the identity matrix and then row reduce it to a matrix that looks like this: \begin{pmatrix} 1/5 & -2/5 & 2/5 \\ -1/5 & 4/15 & 1/10 \\ 3/5 & -1/5 & -3/10 \end{pmatrix} After this I calculate the the linear transformation from \gamma to \gamma by multiplying the matrixes in this order. (Identity matrix from \beta to \gamma) * (linear transformation matrix from \beta to \beta) * (identity matrix from \gamma to \beta) After calculating this (which is pretty lengthy) I get the matrix: \begin{pmatrix} 27/5 & -43/15 & 29/5 \\ 17/5 & -3/5 & 19/5 \\ 16/5 & -2 & 17/5 \end{pmatrix} The above matrix is the answer I got to the question. Please correct me if any of my calculations or understandings are incorrect. |

Jackhe: May 5, 2015, 12:46 a.m. |
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that's what I also tried to do, but I'm not sure about the answer |

Sam_Lai: May 5, 2015, 1:01 a.m. |
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This was the exact same logic that I used to solve the problem, but I interpreted the given matrix to be from $\gamma\$ to $\beta$. Here's my reason for interpreting it that way: Imagine if I had a basis {(1,2), (0,1)}. My basis with respect to the standard basis {(1,0), (0,1)} is $\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$ This means that if I take my (1,0) and multiply it by this transformation matrix, I get (1,2) which is my coordinates in the standard basis. My basis with respect to the standard takes my coordinates to the standard. So if the vectors of my friend's basis $\gamma$ with respect to my basis $\beta$ is given, shouldn't it be a transformation matrix from $\gamma$ to $\beta$? Or is there something wrong with my interpretation? |

Sam_Lai: May 5, 2015, 1:02 a.m. |
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Wow, something messed up big time there. I'm so sorry. Hopefully it's still comprehensible. |

Cullen_Im: May 5, 2015, 5:32 a.m. |
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@TLindberg: I believe the first step necessary is to find an identity matrix from gamma to beta and NOT beta to gamma. Because the given linear transformation is from beta to beta, the vector needs to be first transformed into the beta basis. I similarly found an inverse to the identity matrix but the only calculation difference I see in your inverse identity matrix is row 2, column 2. I got "2/5" instead of your "4/15." I then multiplied the matrices in the order: (identity matrix from gamma to beta) * (linear transformation matrix from beta to beta) * (identity matrix from beta to gamma). I got the final answer as a matrix of linear transformation from gamma to gamma with row 1: 3, 33/5, 21/5; row 2: 5/2, 52/5, 19/5; row 3: -3/2, -41/5, -12/5 |

akaur: May 5, 2015, 11:46 a.m. |
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I got a very similar identity matrix, so I guess one of us made a calculation mistake somewhere (probably me). Isn't the given identity matrix from gamma to beta and the one we find is from beta to gamma? |

julianna_burke: May 5, 2015, 2:55 p.m. |
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I did same process but got a different answer. In the identity matrix I got 3/5 instead of your 4/15 and then my final answer was 1st row: {3, -11/5, 1} 2nd row: {5/2, 52/5, 27/10} 3rd row {-3/2, -41/5, 9/2} This could very well be wrong but it is similar to someone else's answer in a different discussion except our third column differs |

julianna_burke: May 5, 2015, 3:06 p.m. |
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Nevermind I found my mistake! I got the same answer as @Cullen_Im |

elaine_lei: May 5, 2015, 3:58 p.m. |
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I did the problem twice and got the same answer as @Cullen_Im |

TLindberg: May 5, 2015, 5:32 p.m. |
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@Cullen_Im: When composing matrices together its similar to a composition of functions. Therefore, the identity matrix from gamma to beta goes on the right because it is the "first" matrix to act on the vector in the gamma basis. After this, the vector is now in terms of the beta basis and can transformed with the linear transformation from beta to beta. After calculating the transformation, we need to switch the coordinates back to the gamma basis so we apply the identity matrix from beta to gamma, causing it to be on the very left. ------------------ A possibly easier way to imagine it is like this: f(g(h(x))). In this representation h(x) is the identity matrix from gamma to beta, g(x) is the linear transformation matrix from beta to beta, and f(x) is the identity matrix from beta to gamma. -------------- This ends with the function appearing backwards and instead reading right to left; however, the matrices can still be multiplied left to right because the matrices are associative and the multiplication can be done in any order, as long as the overall order of the three matrices is fixed. --------------- Please tell me if this makes sense. I was a bit confused by it at first too because it makes sense to form an equation from left to right in my mind, but according to the lecture notes and my TA I'm pretty sure this is the right way to go about it. |

TLindberg: May 5, 2015, 6:50 p.m. |
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Oh, I might see what you were trying to correct me on. I interpreted the first matrix as being from beta to gamma, but instead it should be gamma to beta. By switching the values in those two matrices but still multiplying it in the same order I specified above it should be the correct answer. I probably calculated the inverse incorrectly, so I'll just leave it at that and try to post a solution I get later tonight. |

angela: May 5, 2015, 8:31 p.m. |
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The explanation @TLindberg posted was super helpful! After multiplying everything out, I got $\begin{bmatrix} 27/5 & 1/5 & 29/5 \\ 17/5 & 11/5 & 19/5 \\ 16/5 & -2/5 & 17/5 \end{bmatrix}$ |

wllmskn: May 5, 2015, 9:02 p.m. |
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@angela That's what I got too! |

LizvetteV: May 6, 2015, 2:43 p.m. |
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okay so my answer is way off, but I got 3 33/5 21/5 5/2 52/5 19/5 -3/2 -41/5 -12/5 But it might be because the identity matrix i got from beta to gamma was: 1/5 -2/5 2/5 -1/5 2/5 1/10 3/5 -1/5 -3/10 Because i still multiplied the identity matrix of beta to gamma * (T) from beta to beta * Identity matrix of gamma to beta |

KianJerKoh: May 7, 2015, 2:51 p.m. |
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@Cullem_Im Yes it is from gamma to beta, because the output "with respect to our basis" means that the output should be in beta. Again the solutions that both David and my TA gave are from $\gamma$ to $\beta$, first, then we find the inverse. |