Because the basis vectors are chosen arbitrarily, many different answers
are possible.
However, here is one way to go;
to diagonalize
take it as the representation of a transformation with respect to the
standard basis and look for
such that
that is, such that
and .
We are looking for scalars such that this equation
has solutions and , which are not both zero.
Rewrite that as a linear system
If then the first equation gives that , and then
the second equation gives that .
The case where both 's are zero is disallowed
so we can assume that .
Consider the bottom equation.
If then the first equation gives or .
The case is disallowed.
The other possibility for the bottom equation is that the numerator
of the fraction is zero.
The case gives a first equation of , and so
associated with we have
vectors whose first and second components are equal:
If then the first equation is
and so the associated vectors
are those whose first component is
twice their second:
This picture
shows how to get the diagonalization.
Comment.
This equation matches the definition under this renaming.
Problem 2
Diagonalize these upper triangular matrices.
Answer
Setting up
gives the two possibilities that and .
Following the possibility leads to the first equation
with the two cases that and that
.
Thus, under this first possibility, we find and the
associated vectors whose second component is zero, and whose
first component is free.
Following the other possibility leads to a first equation of
and so the vectors associated with this
solution have a second component that is four times their first
component.
The diagonalization is this.
The calculations are like those in the prior part.
The bottom equation
gives the two possibilities that and .
Following the possibility, and discarding the
case where both and are zero, gives
that , associated with vectors whose second component
is zero and whose first component is free.
The possibility gives a first equation of
and so the associated vectors have a
second component that is the negative of their first component.
We thus have this diagonalization.
This exercise is recommended for all readers.
Problem 3
What form do the powers of a diagonal matrix have?
Answer
For any integer ,
Problem 4
Give two same-sized diagonal matrices that are not similar.
Must any two different diagonal matrices come from different similarity
classes?
Answer
These two are not similar
because each is alone in its similarity class.
For the second half, these
are similar via the matrix that changes bases from
to
.
(Question.
Are two diagonal matrices similar if and only if their diagonal
entries are permutations of each other's?)
Problem 5
Give a nonsingular diagonal matrix.
Can a diagonal matrix ever be singular?
Answer
Contrast these two.
The first is nonsingular, the second is singular.
This exercise is recommended for all readers.
Problem 6
Show that the inverse of a diagonal matrix is the diagonal of
the inverses, if no element on that diagonal is zero.
What happens when a diagonal entry is zero?
Answer
To check that the inverse of a diagonal matrix is the diagonal
matrix of the inverses, just multiply.
(Showing that it is a left inverse is just as easy.)
If a diagonal entry is zero then the diagonal matrix is singular; it has a zero determinant.
is a bit jarring because for we must take the first matrix,
which is shown as an inverse, and for we take the inverse of the
first matrix, so that the two powers cancel and this matrix is
shown without a superscript .
Check that this nicer-appearing equation holds.
Is the previous item a coincidence?
Or can we always switch the and the ?
Answer
The check is easy.
It is a coincidence, in the sense that if
then need not equal .
Even in the case of a diagonal matrix , the condition that
does not imply that equals .
The matrices from Example 2.2 show this.
Problem 8
Show that the used to diagonalize in
Example 2.5 is not unique.
Answer
The columns of the matrix are chosen as the vectors associated with the 's. The exact choice, and the order of the choice was arbitrary. We could, for instance, get a different matrix by swapping the two columns.
Problem 9
Find a formula for the powers of this matrix
Hint: see Problem 3.
Answer
Diagonalizing and then taking powers of the diagonal matrix shows that
This exercise is recommended for all readers.
Problem 10
Diagonalize these.
Answer
Problem 11
We can ask how diagonalization interacts with the matrix operations.
Assume that are each diagonalizable.
Is diagonalizable for all scalars ?
What about ?
?
Answer
Yes, is diagonalizable by the final theorem of this
subsection.
No, need not be diagonalizable.
Intuitively, the problem arises when the two maps diagonalize with
respect to different bases (that is, when they are not
simultaneously diagonalizable).
Specifically, these two are diagonalizable but their sum is not:
(the second is already diagonal; for the first, see
Problem 10).
The sum is not diagonalizable because its square is the zero matrix.
The same intuition suggests that is not
be diagonalizable.
These two are diagonalizable but their product is not:
Show that matrices of this form are not diagonalizable.
Answer
If
then
so
The entries show that and the entries then show that . Since this means that . The entries show that and the entries then show that . Since this means that . But if both and are then is not invertible.
Problem 13
Show that each of these is diagonalizable.
Answer
Using the formula for the inverse of a
matrix gives this.
Now pick scalars so that
and and .
For example, these will do.
As above,
we are looking for scalars so that
and
and , no matter what values
, , and have.
For starters, we assume that , else the given matrix is
already diagonal.
We shall use that assumption because if we (arbitrarily) let
then we get
and the quadratic formula gives
(note that if , , and are real then these two
's are real as the discriminant is positive).
By the same token, if we (arbitrarily) let then
and we get here
(as above, if then this discriminant is positive
so a symmetric, real, matrix is similar to a real
diagonal matrix).
For a check we try , , .