Let .
is said to be diagonalizable if there exist
and in such
that is a diagonal matrix and
.
Testing if a matrix is diagonalizable
is diagonalizable if and only if for every
eigenvalue of , the algebraic multiplicity of
is equal to the geometric multiplicity of .
An equivalent characterization is that
the sum of the geometric multiplicities of the eigenvalues of
is .
Examples
Let
Note that . Hence,
the only eigenvalue of is .
Now
The nullity of this matrix is , implying that
the geometric multiplicity is , not .
So is not diagonalizable.
Let
Note that .
So the eigenvalues are and .
The geometric multiplicity of is since the geometric multiplicity
cannot be and cannot exceed the algebraic multiplity of , which
is .
Now, note that
.
This matrix has rank since the second row is times
the first row and the third row is a row of 's.
Hence, the nullity of is , implying that
the geometric multiplicity of is .
So is diagonalizable as the sum of the geometric multiplities is .
Finding and
Now that we know the matrix in the second example above is diagonalizable,
how do we actually find and ?
First, observe that can be rewritten as
. Column of the left-hand side is given
by where denotes the column of
and column of the right-hand side is given by
. Since is a diagonal matrix,
.
Thus must be an eigenvector with eigenvalue .
Since we need to be invertible, we need linearly independent
eigenvectors. This is why we need the sum of the
geometric multiplicities to equal .
To form , first list all the eigenvalues
(including multiple appearances) in any order.
(Hence, an eigenvalue with algebraic
multiplicity will appear times in the list.)
Then form with this list of values as the diagonal.
For our example, the list could be .
So we can set
.
To form , first find a basis for each eigenspace.
To form the first column of , take a vector from the basis
for the eigenspace of the first diagonal entry of and remove
that vector from future inclusion into .
For each subsequent column , take a vector from the basis for
the eigenspace of the th diagonal entry of and
remove it from future includsion into .
For our example, we first need to find a basis for the eigenspace of
and a basis for the eigenspace of .
For the eigenvalue ,
we want to find a basis for the nullspace of
.
Row-reducing this matrix to RREF gives the matrix
.
Hence, all solutions to
are of the form
since is a free variable.
Thus, the single vector
forms a basis for the eigenspace.
For the eigenvalue ,
we want to find a basis for the nullspace of
.
Row-reducing this matrix gives
.
Hence, all solutions to
are of the form
since and
are free variables.
Hence, a basis for the nullspace is
.
To form the first column of , we take a vector from the basis
for the eigenspace of . There is only one such vector.
So is now
.
For the second column, we take a vector from the basis
for the eigenspace of . There are two vectors to choose from.
We choose .
So is now
.
And we have no choice but to take the remaining vector in the basis
for eigenspace of to fill the last column of .
Hence, .
Let .
You are given that is the only eigenvalue of .
Give a basis for the eigenspace of this eigenvalue.
Note that
We row-reduce this matrix to RREF:
Note that the second column is the only nonpivot column.
Thus, a basis for the eigenspace has exactly one vector and we can take any nonzero
vector in the nullspace of .
One such vector is .
Hence, a basis for the eigenspace is
.
Remark. Since we are given that 2 is the only eigenvalue and
the dimension of the eigenspace of this eigevalue is less than the total
number of columns of , we see that is not diagonalizable.