Maths Item of the Month – August 2012
Linear matrices
(
)
(
)
for all natural numbers n.
Can you find other matrices, M, such that all the elements of Mn are linear functions of n
(where n is a natural number)?
Solution
We can begin by checking the example. This will help with our understanding of the
question.
Finding low powers of a matrix is relatively straightforward; we just multiply the matrix by
itself n times. However, this is very time-consuming for larger powers, and impossible in the
general case.
n
Instead we use the equation Mn = P.D .P-1, where D is the diagonalised form of M.
(
) where λ represents the eigenvalues of M.
This means that we only need to find the power of the diagonalised matrix, which is much
easier.
Using det(M  λI)=0,
|(
Collecting,
|(
Using det(M) = ad  bc,
So
λ
(
)
(
)|
)|
)
(This is the characteristic polynomial)
.
This means we cannot construct a complete diagonalised form; in this case we need to use
the “Jordan” form. (Page 2 of http://www.math.niu.edu/~fletcher/jnf.pdf explains how to find
the Diagonal or Jordan form of a 2x2 matrix, depending on the characteristic polynomial.)
For the example matrix, the Jordan form is
(
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).
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So, we see that
(
)
We could, at this point, calculate S and S-1, however this is not necessary to show that the
statement is true. All we need to know is that S will be constant (i.e. not in terms of n).
As the matrices S and S-1 are both independent of n, we only need to show that the elements
of Jn are linear functions of n. This is much more straightforward.
Try for n = 1,
(
)
(
)
then for n = 2,
(
)
(
)
for n = 3,
(
)
(
)
We can continue checking values, but already we see the pattern emerging, and in doing the
calculations you will see why this pattern will hold for any value of n.
(
for n = n,
)
(
)
So we see that the elements will be linear for the example matrix, and that this is defined by
the Jordan/Diagonal form of the matrix.
We should examine what conditions are necessary for a general Jordan/Diagonal form to
give this result. We will divide the matrices into 3 cases.
Case 1:
A normal matrix with 2 eigenvalues, i.e. one which can be diagonalised. For this case the
matrix D will be of the form
(
).
Try for n = 1,
(
)
(
)
then for n = 2,
(
)
(
)
for n = 3,
(
)
(
)
leading to the general
form,
(
)
(
)
So we see that any 2x2 matrix that is diagonalisable will have exponential functions of n as
its elements.
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Case 2:
(
A matrix, M, with one eigenvalue, λ, where
).
In this case the matrix is already in its Jordan form. We can quickly see that this case gives
much the same result as the first, the only difference being that λ1
= λ2.
Case 3:
A matrix, M, with one eigenvalue, λ, where M ≠ λI.
(
In this case the Jordan form is as follows;
).
Try for n = 1,
(
)
(
)
then for n = 2,
(
)
(
for n = 3,
(
)
(
)
for n = 4,
(
)
(
)
leading to the general form,
(
)
(
)
)
For most values of λ this will also give exponential functions of n. The only value of λ for
which this does not happen, is 1. In this case, the powers of n take no effect, leaving the
linear function of n:
(
)
(
)
(
)
So the 2x2 matrix, Mn, will have linear functions of n as it’s elements provided that M has
one eigenvalue, λ = 1, and M is not the identity matrix.
Now, we must figure out what matrices fulfil these criteria. We derive our first equation from
the condition λ = 1.
(
We define our general matrix as
given that λ = 1, det(M  λI) = 0,
collecting,
|(
)
)
|
(
)|
|
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(
calculate determinant,
)(
)
simplify,
rearrange,
( )
substitute det(M) = ad – bc,
(1)
The next equation is derived from the condition that there is only one eigenvalue.
given that det(M  λI) = 0,
)
|(
collecting,
|
calculate determinant,
(
(
)|
|
)(
)
simplify,
rearrange
(
)
substitute det(M) = ad – bc,
(
)
only one root; b  4ac = 0,
(
)
substitute (1),
(
)
2
( )
( )
(
)
expand,
(
this factorises,
)
(2)
square root and rearrange,
So we now have the matrix
(
). Now we must eliminate c.
using (1) and (2),
( )
simplify,
( )
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(
)
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using det(M) = ad – bc,
(
substitute (2),
)
rearrange,
n
So a matrix, M, for which all the elements of M are linear functions of n, is always of the
form:
(
)
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