## Practice Final #3 |
Cullen_Im |
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To show that the eigenspace is a subspace of vector space "V": 1. The "0" vector is in the set of the eigenspace since multiplying each eigenbasis by the constant "0" will give you the direction (0, 0, ..., 0) with respect to the eigenbases. 2. if "v" and "w" are eigenvectors, the eigenspace is closed under addition if T(v + w) = "lambda"(v + w). T(v + w) = T(v) + T(w) and "lambda"(v + w) = "lambda"(v) + "lambda"(w) >>>>>>>> we know T(v) = "lambda"(v) and T(w) = "lambda"(w) so T(v) + T(w) = "lambda"(v) + "lambda"(w) and finally T(v + w) = "lambda"(v + w). 3. if "c" is a constant and "v" is an eigenvector, the eigenspace is closed under scaler multiplicaiton if T(cv) = "lambda"(cv). T(cv) = cT(v) and "lambda"(cv) = c"lambda"(v) >>>>>> since we know T(v) = "lambda"(v), cT(v) = c"lambda"(v) and finally T(cv) = "lambda"(cv). I believe a basis for P2(R) consisting of eigenvectors of T is [3/14, 3/14, 1] and [1,1,0] and that T is not diagonalizable since the dimension of the eigenspace of T is only 2. Please comment if there are errors/mistakes with this. |

Cullen_Im: May 29, 2015, 2:29 a.m. |
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I apologize for the cramped formatting as I can't figure out how to space between lines/make tabs |

alexwang0518: May 29, 2015, 8:49 p.m. |
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I think you need to find the eigenvalues and corresponding eigenvectors to the transformation formula turned into a matrix. as in( 6 3 1__-6 -5 -3__4 3 3). |

missaelv95: May 29, 2015, 10:42 p.m. |
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@alexwang0518 i was wondering how you arrived to that order for your eigenbasis, i got the same numbers but a different order (read top to bottom) in which my eigenvectors e_{1}$ = (1,3,6) e_{2}$ = (-3,-5,-6) and e_{3}$ = (3,3,4) |

missaelv95: May 29, 2015, 10:44 p.m. |
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excuse the formatting i have no idea how to use it but; e(1)= [[1,3,6]] e(2)=[[-3,-5,-6]] e(3)= [[3,3,4] in which these are the eigenvectors read from top to bottom |

lilb: May 30, 2015, 4:51 p.m. |
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did people get eigenvalues of -4, 2, -2? |

tedallen: May 30, 2015, 5:40 p.m. |
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I got the polynomial $λ^{3} -12λ - 16 = 0$, which yields λ = -2, 4 |

lilb: May 30, 2015, 6:22 p.m. |
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yea my bad. So is the eigenspace for 4 {1,1,1,} and the one for -2 { (-1,1,0) , (-2,0,1) }. And they have dimensions of 1 and 2, respectively? |

tedallen: May 30, 2015, 7:29 p.m. |
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hmm. For λ=-2 I got $\begin{matrix} 1\\1\\0 \end{matrix} and \begin{matrix} 0\\1\\1 \end{matrix}$, and for λ=4 I got $\begin{matrix} 1\\1\\2 \end{matrix}$ |

lilb: May 30, 2015, 9:05 p.m. |
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im probably doing something wrong, but you wanna check your work again? I can't find my mistake |

henry_lee: May 31, 2015, 3:26 a.m. |
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I also got 4, -2, -2 as eigenvalues |

henry_lee: May 31, 2015, 3:28 a.m. |
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Same eigenspaces as tedallen as well |

tanishaharlalka: May 31, 2015, 1:32 p.m. |
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I got my eigenvalues to be 4, -2, -2 as well. |

yulduzkhonbruin: June 1, 2015, 1:32 p.m. |
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For E(-2) I have span of {(1, 1, 0), (-1, 0, 1)} and for E(4) I have span of {(1, 1, 2)} both of which are the same answers as tedallen, but solved differently for E(-2), meaning the answer depends on what variable you set as a free variable. |

dannystapleton: June 1, 2015, 4:25 p.m. |
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For the proof: Must show that the Eigenspace is a subspace of V by proving the following three properties: 1) It is closed under addition. 2) It is closed under scalar multiplication. 3) The set contains the zero vector. |

dannystapleton: June 1, 2015, 5:25 p.m. |
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To find the Matrix given by the transformation, call it Matrix A, you have to see that the what the transformation is mapping from the std basis 1,x,$x^{2} . T(1) = 1 + 3x +6x^2. T(x) = -3 -5x -6x^2. T(x^2) = 3 + 3x + 4x^2. So your Matrix A will be composed of the three column vectors given by the transformation. First column: 1 3 6. Second column: -3 -5 -6. Third column: 3 3 4. |

Cullen_Im: June 3, 2015, 12:51 a.m. |
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I made calculation errors in finding a basis for P2(R) consisting of eigenvectors of T and finding the eigenspaces of T. The linear transformation T can be represented by multiplication by the matrix with columns [1, 3, 6], [-3, -5, -6], and [3, 3, 4]. The eigenvalues are found to be 4 (algebraic multiplicity 1) and -2 (algebraic multiplicity 2). A basis consisting of eigenvectors is (1, 1, 0), (1, 0, -1), and (1, 1, 2). The eigenspace for eigenvalue "-2" is the span of (1, 1, 0) and (1, 0, -1) and has dimension 2. The eigenspace for eigenvalue "4" is the span of (1, 1, 2) and has dimension 1. |

CatieImbery: June 5, 2015, 4:42 p.m. |
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David also mentioned in the review session that we should put the eigenvectors in terms of the basis. So for the eigenvalue of 4, you have [1,1,2] but the vector should be rewritten as 1+x+2x^2. |