Difference between revisions of "Diagonalizable"
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| − | An [[operator]] ''T'' on a finite-dimensional [[vector space]] ''V'' is ''' | + | An [[operator]] ''T'' on a finite-dimensional [[vector space]] ''V'' is '''diagonalizable''' if ''V'' has a [[basis]] of [[eigenvector]]s for ''T''. |
| − | ==Denseness of | + | ==Denseness of diagonalizable operators== |
The space of [[complex]] [[linear]] [[operator]]s on <math>\mathbb{C}^n</math> may be identified with the vector space <math>M_n(\mathbb{C})</math> of nxn [[matrices]] with complex [[coefficient]]s. As such, it inherits a natural structure as a [[topological space]]. | The space of [[complex]] [[linear]] [[operator]]s on <math>\mathbb{C}^n</math> may be identified with the vector space <math>M_n(\mathbb{C})</math> of nxn [[matrices]] with complex [[coefficient]]s. As such, it inherits a natural structure as a [[topological space]]. | ||
| − | Given this topology, the set of | + | Given this topology, the set of diagonalizeble functions is a [[dense]] subset of <math>M_n(\mathbb{C}^n)</math>. |
| − | We can prove this as follows: Every complex matrix <math>A</math> is [[conjugate]] to a matrix <math>B</math> in [[Jordan canonical form]]. One can then perturb the [[diagonal]] elements <math>b_{ii}</math> of <math>B</math> by arbitrarily small numbers <math>\epsilon_i</math> so that the diagonal elements <math>b_{ii}+\epsilon_i</math>of the perturbed matrix are distinct. But this implies that the perturbed matrix is | + | We can prove this as follows: Every complex matrix <math>A</math> is [[conjugate]] to a matrix <math>B</math> in [[Jordan canonical form]]. One can then perturb the [[diagonal]] elements <math>b_{ii}</math> of <math>B</math> by arbitrarily small numbers <math>\epsilon_i</math> so that the diagonal elements <math>b_{ii}+\epsilon_i</math>of the perturbed matrix are distinct. But this implies that the perturbed matrix is diagonalizable. Thus, we can find a diagonalizeable matrix arbitrarily close to a conjugate of <math>A</math>. But since conjugation is a length-preserving operation on the [[inner product]] space of complex matrices, this shows that <math>A</math> is arbitrarily close to a diagonalizable matrix. This completes the proof. |
[[Category:Mathematics]] | [[Category:Mathematics]] | ||
Revision as of 04:14, September 18, 2009
An operator T on a finite-dimensional vector space V is diagonalizable if V has a basis of eigenvectors for T.
Denseness of diagonalizable operators
The space of complex linear operators on <math>\mathbb{C}^n</math> may be identified with the vector space <math>M_n(\mathbb{C})</math> of nxn matrices with complex coefficients. As such, it inherits a natural structure as a topological space.
Given this topology, the set of diagonalizeble functions is a dense subset of <math>M_n(\mathbb{C}^n)</math>.
We can prove this as follows: Every complex matrix <math>A</math> is conjugate to a matrix <math>B</math> in Jordan canonical form. One can then perturb the diagonal elements <math>b_{ii}</math> of <math>B</math> by arbitrarily small numbers <math>\epsilon_i</math> so that the diagonal elements <math>b_{ii}+\epsilon_i</math>of the perturbed matrix are distinct. But this implies that the perturbed matrix is diagonalizable. Thus, we can find a diagonalizeable matrix arbitrarily close to a conjugate of <math>A</math>. But since conjugation is a length-preserving operation on the inner product space of complex matrices, this shows that <math>A</math> is arbitrarily close to a diagonalizable matrix. This completes the proof.