From A First Course in Linear Algebra
Version 1.06
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/
The matrix trace is a function that sends square matrices to scalars. In some
ways it is reminiscent of the determinant. And like the determinant it has many
useful and surprising properties.
Definition MT
Matrix Trace
Suppose is an
matrix. Then the
matrix trace of ,
, is the sum of the
diagonal entries of .
Symbolically,
The next three proofs make for excellent practice. In some books they would be left as exercises for the reader as they are all “trivial” in the sense they do not rely on anything but the definition of the matrix trace.
Theorem MTL
Matrix Trace is Linear
Suppose
and are
matrices. Then
. Furthermore,
if is a complex
number then .
Proof These properties are exactly those required for a linear transformation. To prove these results we just manipulate sums,
The second part is as straightforward as the first,
Theorem MTSRM
The Matrix Trace is Symmetric with Respect to Multiplication
Suppose
and are
matrices.
Then .
Proof
Theorem MTIST
Matrix Trace is Invariant Under Similarity Transformations
Suppose
and are
matrices and
is invertible.
Then .
Proof Invariant means constant under some operation. In this case the operation is a similarity transformation. A lengthy exercise (but possibly a educational one) would be to prove this result without referencing theorem MTSRM. But here we will,
Now we could define the trace of a linear transformation as the trace of any matrix representation of the transformation. Would this definition be well-defined, that is will two different representations of the same linear transformation always have the same trace? Why? (Think Theorem SCB.) We will now prove one of the most interesting and surprising results about the trace.
Theorem MTSE
Matrix Trace is the Sum of the Eigenvalues
Suppose that is a square
matrix of size with
distinct eigenvalues .
Then
Proof It is amazing that the eigenvalues would have anything to do with the sum of the diagonal entries. Our proof will rely on double counting. We will demonstrate two different ways of counting the same thing therefore proving equality. Our object of interest is the coefficient of in the characteristic polynomial of (Definition CP), which will be denoted . From the proof of Theorem NEM we know that:
First we want to prove that is equal to and to do this we will use a straight forward counting argument. Induction can be used here as well (try it), but the intuitive approach is a much stronger technique. Let’s imagine creating each term one by one from the extended product. How do we do this? From each we pick either a or a . But we are only interested in the terms that have . As , we have factors of the form . Then to get terms with we need to pick ’s in every , except one. Since we have linear factors there are ways to do this, namely each eigenvalue represented as many times as it’s algebraic multiplicity. Now we have to take into account the sign of each term. As we pick ’s and one (which has a negative sign in the linear factors) we get a then we have to take into account the in the characteristic polynomial. Thus is the sum of these terms which is .
Now we will now show that is also equal to . For this we will proceed by induction on the size of . If is a square matrix then and . With our base case in hand let’s assume is a by matrix. By Definition CP
First let’s consider the maximum degree of when . For polynomials, the degree of , denoted , is the highest power of in the expression . A well known result of this definition is: if then (can you prove this?). Now has degree of zero when . Furthermore has rows, one of which has entries of all degree zero as column is removed. The other rows have one entry with degree one and rest have degree zero. Then by problem 4 (reference this), the maximum degree of is . So these terms will not affect the coefficient of . Now we are free to focus all of our attention on the term . As is a maxtrix the induction hypothesis tells us that has a coefficient of for . We also note that the proof of Theorem NEM tells us that the leading coefficent is . Then,
Expanding the product shows (the coeficient of ) to be
Now we have counted in two ways proving that
Then as ,
Problems
1 Prove there are no square matrices
and
such
that .
2 If then
prove is a
subspace of
and determine it’s dimension.
3 Assume
is a matrix.
Prove .
4 Assume is a
matrix with polynomial
entries. Define
to be the maximum degree of the entries in row
. Then
. (Hint:
If ,
then )
5 If
is a
matrix, the matrix exponential is defined as
Prove that .
Solutions 4 We will proceed by induction. If is a matrix then clearly . Now assume is a matrix then by Theorem DER,
Let’s consider the degree of the term, . By definition of the function , . We use our induction hypothesis to examine the other part of the product which tells us that
Furthermore by definition of (Definition SM) row of matrix contains all the entries of the corresponding row in then,
So,
Then Using the property that if then ,
As is arbitrary the degree of all terms in the determinant are so bounded. Finally using the fact that if then we have .