## Practice Final #1,2 Answers |
henry_lee |
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1. Eigenvalues: 2, -1 (mult. of 2) Using a diagonal matrix D with column vectors [-1,0,0], [0,-1, 0], and [0,0,2], one form of the basis matrix Q is [1,0,-1], [1,-1, 0] and [1,1,1] 2. Matrix A is diagonalizable; eigenvalues are 1,1,1; algebraic multiplicity of 3, geometric multiplicity of 1 Matrix B is diagonalizable, eigenvalues are 1,2,3,4,5 |

Cullen_Im: May 29, 2015, 1:03 a.m. |
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I believe that Matrix A in Question #2 is not diagonalizable since the sum of the geometric multiplicities does not equal the dimension of Matrix A. In other words, only one eigenvector is associated with the only eigenvalue which is "1", so there is only one basis in the eigenspace. Since Matrix A's dimension is 3, there needs to be three bases in the eigenspace in order for Matrix A to be diagonalizable. |

lilb: May 29, 2015, 6:40 p.m. |
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Just to understand the concept, what would the geometric multiplicity of the eigenvalue -1 in #1 be? I got the ker to be span{-y-z,y,z}, so would it's geometric multiplicity be 2? Why? |

Cullen_Im: May 29, 2015, 7:58 p.m. |
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The geometric multiplicity of the eigenvalue -1 in #1 is 2 since you get two eigenvectors from the one eigenvalue. The ker has two dimensions so there are two eigenvectors. |

henry_lee: May 29, 2015, 8:07 p.m. |
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Oh yeah, my mistake. Matrix A isn't diagonalizable, idk why I assumed since I got 3 eigenvalues that there would be a diagonal matrix D |

brookewenig: May 29, 2015, 8:23 p.m. |
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As a side note, 1 <= geometric multiplicity <= algebraic multiplicity. You can think of this as there is at least one eigenvector associated with each eigenvalue, but there can be no more eigenvectors than the algebraic multiplicity for that eigenvalue. |

missaelv95: May 29, 2015, 8:27 p.m. |
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i am with Cullen_Im however, another way to approach this would be the following manner... In Matrix A in Question#2 the characteristic polynomial is (1-λ)^3 which results in an eigenvalue of 1 In order for a matrix to be diagonalizable the algebraic multiplicity of each eigenvalue must equal its geometric multiplicity. The algebraic multiplicity in determined by the power, therefore almu(1)=3 The geometric multiplicity formula is geomu(λ)=n-rank(a-λI); therefore the geomu(1)= 3-2=1 therefore almu(1)≠geomu(1) therefore the matrix A is not diagonizable |

CarlHarmon: May 31, 2015, 8:10 a.m. |
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Also, for the second part of #2, since we get 5 distinct eigenvalues we know there are 5 distinct eigenvectors. |