# Are all matrices diagonalisable?

**Kapil H. Paranjape**

**http://www.imsc.res.in/~kapil/papers/matrix/**

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''.

Linear Algebra forms the heart of ``modern'' mathematics. It is thus disappointing to see the number of students confused about such an important issue. This note is an attempt to show the ramifications of this question and provide a detailed answer.

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 matrixAare polynomials in the matrix entries ofA.

Thus, treating the entries of the matrix as variables we get a
(universal) polynomial in the matrix entries of *A* with the property that
the characteristic polynomial of *A* has multiple roots if
and only if vanishes at *A*. It is not difficult to show that
is not the identically zero polynomial; moreover,
the zero set of a non-zero polynomial is always a ``thin set'' in
the sense of measure theory. In other words, if we bound the
matrix entries of *A* and choose them randomly (but
uniformly) within these bounds then *with probability 1*
we will find a diagonalisable matrix. Thus my own answer to the
question posed above is two-fold:

- 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