**Kapil H. Paranjape**

I have frequently used the above question to gauge the solidity of the mathematical training of another person. The incorrect responses vary between:

- An unhesitating ``Yes''.
- An unhesitating ``No'' with no counter-example forthcoming.
- A ``Yes'' after some thought.
- ``Yes, over the field of complex numbers''.

To begin with a matrix *A* is said to be ``diagonalisable'' if it is a
square matrix and there is an invertible matrix *B* such that
*B*^{-1}*AB* is a diagonal matrix. So we need restrict ourselves only to
square matrices for the purposes of our discussion. Now it is clear
that if
{*e*_{1},..., *e*_{n}} is the standard basis for the vector space
on which *A* acts, then
{*Be*_{1},..., *Be*_{n}} is another basis, which
has the property that each of the basis vectors is an * eigenvector*; moreover, the diagonal entries of *B*^{-1}*AB* are just
the eigenvalues of *A*. Thus an equivalent definition of
diagonalisability is be that there be a basis consisting of
eigenvectors.

If *A* is a symmetric matrix then one can show that there is an
orthogonal basis of eigenvectors and thus we obtain the
diagonalisability of symmetric matrices. One can argue similarly for
some other classes of matrices. This is perhaps what people
who give the third answer are thinking of. However, there is an
important class of matrices of which none except the zero matrix is
diagonalisable (see below).

Now it is clear that a matrix like
cannot have eigenvalues over the field of real numbers; hence it isn't
diagonalisable over the real numbers. However, it *is*
diagonalisable over the field of complex numbers with eigenvectors
*e*_{1}±^{ . }*e*_{2}. More generally, one could argue as
follows. From the second definition and the usual definition of
eigenvectors we know that we should be looking at the roots of the
Characteristic Polynomial
*P*_{A}(*t*) = det(*A* - *tI*). Now if we are working
over the field of complex numbers we have all roots of *P*_{A}(*t*). Thus
we have eigenvectors corresponding to every eigenvalue. This is
perhaps the reasoning of people who give the fourth answer. However,
the problem is that if an eigenvalue occurs with multiplicity greater
than one it is not clear that we can find two or more (linearly
independent) eigenvectors.

Of course the correct answer to the main question is ``No'' because of
the phenomenon of nilpotent matrices. A matrix such as
has 0 as its only eigenvalue but it is not the zero matrix and thus it
cannot be diagonalisable. It is clear that if *N* is nilpotent matrix
(i. e. *N*^{k} = 0 for some *k*) then it is diagonalisable if and only
*N* = 0. In fact, the role of nilpotent matrices is
made more precise by the ``Jordan decomposition theorem'' which says
that any matrix *A* over the real numbers (or more generally over a
perfect field) can be written as *A* = *S* + *N* where *S* and *N* commute,
*N* is nilpotent and *S* is diagonalisable over the complex numbers
(respectively over an algebraic extension of the ground field).

Now we can approach the question from a different viewpoint and ask
for conditions which ensure that the nilpotent matrix *N* is zero. One
way to do this is to ensure that *P*_{A}(*t*) has distinct roots. In this
case we clearly have a basis of eigenvectors. Let
*f* (*t*) = *t*^{n} + *c*_{1}*t*^{n - 1} + ... + *c*_{n} be any polynomial. There is a (universal)
polynomial
*D*_{n}(*C*_{1},..., *C*_{n}) called the discriminant with the
property that
*D*_{n}(*c*_{1},..., *c*_{n}) = 0 if and only if *f* (*t*) has
multiple roots (e. g. for *n* = 2 we have
*D*_{2} = *C*_{1}^{2} - 4*C*_{2}). Now

the coefficients of the characteristic polynomial of a matrixThus, treating the entries of the matrix as variables we get a (universal) polynomial in the matrix entries ofAare polynomials in the matrix entries ofA.

- Every matrix is not diagonalisable. Take for example non-zero nilpotent matrices. The Jordan decomposition tells us how close a given matrix can come to diagonalisability.
- If we choose our matrix ``randomly'' (in a uniform distribution) from within a bounded region, then it will turn out to be diagonalisable over the complex numbers with probability 1 (in other words ``always'').

Kapil Hari Paranjape 2002-11-21