## Geometric Multiplicity vs Algebraic Multiplicity |
alexwang0518 |
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What is the difference? I know that if you are solving for eigenvalues and you get something like: (λ-1)^3 The eigenvalue will be 1 with an algebraic multiplicity of 3. But how do you find geometric multiplicity? |

brookewenig: May 30, 2015, 5:07 p.m. |
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The geometric multiplicity is the number of eigenvectors that you get associated with that eigenvalue. For example, plug in 1 for λ in your matrix, and see how many eigenvectors you get. |

tedallen: May 30, 2015, 7:36 p.m. |
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for example, $\begin{matrix} 1&1&-1\\0&0&0\\0&0&0\end{matrix}$ will give you two eigenvectors, therefore having a geometric multiplicity of 2. whereas gm would be 1 for something like $\begin{matrix} 1&2&-1\\0&1&3\\0&0&0\end{matrix}$. |

tanishaharlalka: May 31, 2015, 1:20 p.m. |
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An easier way that I found how to do it was to calculate the the dimension of the kernel when you plug in the eigenvalue. For example, for #2A, when you plug in the eigenvalue 1, and row reduce, you get the dim(ker(A)) to equal 1, which gives you the geometric multiplicity. Hence, the algebraic multiplicity does not equal the geometric multiplicity. |

dannystapleton: May 31, 2015, 3:52 p.m. |
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Geometric multiplicity is equal to the number of eigenvectors that you get from determining one specific eigenvalue. If the algebraic multiplicity is equal to the geomtertic multiplicity for all eigenvalues, the Matrix is diagonalizable. |