Proving "almost all matrices over C are diagonalizable". - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T22:29:11Z http://mathoverflow.net/feeds/question/12657 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12657/proving-almost-all-matrices-over-c-are-diagonalizable Proving "almost all matrices over C are diagonalizable". Anweshi 2010-01-22T16:46:27Z 2013-01-31T20:00:45Z <p>This is an elementary question, but a little subtle so I hope it is suitable for MO.</p> <p>Let $T$ be an $n \times n$ square matrix over $\mathbb{C}$. </p> <p>The characteristic polynomial $T - \lambda I$ splits into linear factors like $T - \lambda_iI$, and we have the Jordan canonical form:</p> <p>$$J = \begin{bmatrix} J_1 \\ &amp; J_2 \\ &amp; &amp; \ddots \\ &amp; &amp; &amp; J_n \end{bmatrix}$$</p> <p>where each block $J_i$ corresponds to the eigenvalue $\lambda_i$ and is of the form</p> <p>$$J_i = \begin{bmatrix} \lambda_i &amp; 1 \\ &amp; \lambda_i &amp; \ddots \\ &amp; &amp; \ddots &amp; 1 \\ &amp; &amp; &amp; \lambda_i \end{bmatrix}$$</p> <p>and each $J_i$ has the property that $J_i - \lambda_i I$ is nilpotent, and in fact has kernel strictly smaller than $(J_i - \lambda_i I)^2$, which shows that none of these Jordan blocks fix any proper subspace of the subspace which they fix. Thus, Jordan canonical form gives the closest possible to a diagonal matrix. The elements in the superdiagonals of the Jordan blocks are the obstruction to diagonalization.</p> <p>So far, so good. What I want to prove is the assertion that "Almost all square matrices over $\mathbb{C}$ is diagonalizable". The measure on the space of matrices is obvious, since it can be identified with $\mathbb{C}^{n^2}$. How to prove, perhaps using the above Jordan canonical form explanation, that almost all matrices are like this?</p> <p>I am able to reason out the algebra part as above, but is finding difficulty in the analytic part. All I am able to manage is the following. The characteristic equation is of the form</p> <p>$$(x - \lambda_1)(x - \lambda_2) \cdots (x - \lambda_n)$$</p> <p>and in the space generated by the $\lambda_i$'s, the measure of the set in which it can happen that $\lambda_i = \lambda_j$ when $i \neq j$, is $0$: this set is a union of hyperplanes, each of measure $0$.</p> <p>But here I have cheated, I used only the characteristic equation instead of using the full matrix. How do I prove it rigorously?</p> http://mathoverflow.net/questions/12657/proving-almost-all-matrices-over-c-are-diagonalizable/12659#12659 Answer by Mariano Suárez-Alvarez for Proving "almost all matrices over C are diagonalizable". Mariano Suárez-Alvarez 2010-01-22T16:52:44Z 2010-01-22T16:52:44Z <p>The discriminant of the characteristic polynomial of a matrix depends polynomially on the coefficients of the matrix, and its vanishing detects precisely the existence of multiple eigenvalues. Therefore the set where the discriminant does not vanish is contained in the set of diagonalizable matrices.</p> <p>Now the set where a non-zero polynomial vanishes is very, very thin (in many senses: it does not contain open sets, it has zero Lebesgue measure, etc) so in consequence the set of diagonalizable matrices is correspondingly thick.</p> http://mathoverflow.net/questions/12657/proving-almost-all-matrices-over-c-are-diagonalizable/12660#12660 Answer by Emerton for Proving "almost all matrices over C are diagonalizable". Emerton 2010-01-22T16:52:55Z 2010-01-22T16:52:55Z <p>The map from $\mathbb C^{n^2}$ to the space of monic polynomials of degree $n$ which associates to a matrix $M$ its char. polynomial is the best kind of map you could imagine (algebraic, surjective, open, ... ). If a set in its source has positive measure, than so does its image.</p> <p>Now the set of polynomials with repeated roots is the zero locus of a non-trivial polynomial (the discriminant), and so has measure zero. Thus so does its preimage.</p> <p>In short, the space of matrices in ${\mathbb C}$ whose eigenvalues are distinct has full measure (i.e. its complement has measure zero). Any such matrix is diagonalizable (its Jordan Normal Form is a diagonalization).</p> http://mathoverflow.net/questions/12657/proving-almost-all-matrices-over-c-are-diagonalizable/12669#12669 Answer by Pete L. Clark for Proving "almost all matrices over C are diagonalizable". Pete L. Clark 2010-01-22T18:11:02Z 2010-01-23T09:57:03Z <p>The discriminant argument shows that for for $n \times n$ matrices over any field $k$, the Zariski closure of the set of non-diagonalizable matrices is proper in $\mathbb{A}^{n^2}$ -- an irreducible algebraic variety -- and therefore of smaller dimension. Being contained in a proper algebraic subset of affine or projective space is a very strong and useful way of saying that a set is "small" (except in the case that $k$ is finite!): in particular, its complement is Zariski dense.</p> <p>One can use this observation to reduce many theorems in linear algebra to the diagonalizable case, the idea being that any polynomial identity that holds on a Zariski-dense set of all $n \times n$ matrices must hold (by definition of the Zariski topology!) for all matrices. </p> <p>As a very simple example, one can immediately deduce that the characteristic polynomials $AB$ and $BA$ coincide, because if $A$ is invertible, the matrices are similar. </p> <p>With a bit more care, one can derive the entire theory of determinants and characteristic polynomials from such specialization arguments. In fact by purely algebraic means it is possible to reduce to the case of $k = \mathbb{R}$ (and thereby define the determinant in terms of change of volume, etc.).</p> http://mathoverflow.net/questions/12657/proving-almost-all-matrices-over-c-are-diagonalizable/12677#12677 Answer by Harald Hanche-Olsen for Proving "almost all matrices over C are diagonalizable". Harald Hanche-Olsen 2010-01-22T18:52:38Z 2010-01-23T01:26:25Z <p>In addition to the other answers, all of which are quite good, I offer a rather pedestrian observation: If you perturb the diagonal in each Jordan block of your given matrix $T$ so all the diagonal terms have different values, you end up with a matrix that has $n$ distinct eigenvalues and is hence diagonalizable. Such a perturbation can of course be as small as you wish.</p> <p>Edit: As gowers points out, you don't even need the Jordan form to do this, just the triangular form.</p> http://mathoverflow.net/questions/12657/proving-almost-all-matrices-over-c-are-diagonalizable/120455#120455 Answer by unknown (yahoo) for Proving "almost all matrices over C are diagonalizable". unknown (yahoo) 2013-01-31T20:00:45Z 2013-01-31T20:00:45Z <p>Dear Anweshi, a matrix is diagonalizable if only if it is a normal operator. That is, if and only if $A$ commutes with its adjoint ($AA^{+}=A^{+}A$). This equation is a restriction for a matrix $A$. Therefore, the set of diagonalizable matrices has null measure in the set of square matrices. That is, almost all complex matrices are not diagonalizable. Regards! -Dardo</p>