From A First Course in Linear Algebra
Version 2.23
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/
This section contributed by Andy Zimmer.
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 T
Trace
Suppose is a square
matrix of size .
Then the trace of ,
, is the sum of the
diagonal entries of .
Symbolically,
(This definition contains Notation T.)
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 TL
Trace is Linear
Suppose and
are square
matrices of size .
Then .
Furthermore, if ,
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 TSRM
Trace is Symmetric with Respect to Multiplication
Suppose and
are square
matrices of size .
Then .
Proof
Theorem TIST
Trace is Invariant Under Similarity Transformations
Suppose and
are square
matrices of size
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 TSRM. 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 TSE
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 have,
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 result in to the power . 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 factor) we get a factor of . Then we have to take into account the in the characteristic polynomial. Thus is the sum of these terms,
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 square matrix of size . 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 zero when . Furthermore has rows, one of which has all of its entries of degree zero, since column is removed. The other rows have one entry with degree one and the remainder of degree zero. Then by Exercise T.T30, 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 matrix 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 coefficient of is . Then,
Expanding the product shows (the coefficient of ) to be
With two expressions for , we have our result,
T10 Prove there are no square matrices
and
such
that .
Contributed by Andy Zimmer
T12 Assume is a
square matrix of size
matrix. Prove .
Contributed by Andy Zimmer
T20 If then
prove is a
subspace of
and determine it’s dimension.
Contributed by Andy Zimmer
T30 Assume is a
matrix with polynomial
entries. Define
to be the maximum degree of the entries in row
. Then
. (Hint:
If , then
.)
Contributed by Andy Zimmer Solution [2363]
T40 If is a square matrix, the matrix exponential is defined as
Prove that .
(You might want to give some thought to the convergence of the infinite sum as
well.)
Contributed by Andy Zimmer
T30 Contributed by Andy Zimmer Statement [2361]
We will proceed by induction. If
is a square matrix of size 1, then clearly
. Now assume
is a square
matrix of size
then by Theorem DER,
Let’s consider the degree of 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