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trace of a matrix (Definition)

Definition
Let $ A=(a_{i,j})$ be a square matrix of order $ n$. The trace of the matrix is the sum of the main diagonal:

$ \operatorname{trace}(A)= \sum\limits _{i=1} ^{n} a_{i,i}$

Notation:
The trace of a matrix $ A$ is also commonly denoted as $ \operatorname{Tr}(A)$ or $ \operatorname{Tr}A$.

Properties:

  1. The trace is a linear transformation from the space of square matrices to the real numbers. In other words, if $ A$ and $ B$ are square matrices with real (or complex) entries, of same order and $ c$ is a scalar, then
    $\displaystyle \operatorname{trace}(A+B)$ $\displaystyle =$ $\displaystyle \operatorname{trace}(A)+ \operatorname{trace}(B),$  
    $\displaystyle \operatorname{trace}(cA)$ $\displaystyle =$ $\displaystyle c\cdot \operatorname{trace}(A).$  

  2. For the transpose and conjugate transpose, we have for any square matrix $ A$ with real (or complex) entries,
    $\displaystyle \operatorname{trace} (A^t)$ $\displaystyle =$ $\displaystyle \operatorname{trace} (A),$  
    $\displaystyle \operatorname{trace} (A^\ast)$ $\displaystyle =$ $\displaystyle \overline{\operatorname{trace} (A)}.$  

  3. If $ A$ and $ B$ are matrices such that $ AB$ is a square matrix, then
    $\displaystyle \operatorname{trace} (AB) = \operatorname{trace} (BA).$

    For this reason it is possible to define the trace of a linear transformation, as the choice of basis does not affect the trace. Thus, if $ A,B,C$ are matrices such that $ ABC$ is a square matrix, then

    $\displaystyle \operatorname{trace} (ABC) = \operatorname{trace} (CAB) = \operatorname{trace} (BCA).$
  4. If $ B$ is in invertible square matrix of same order as $ A$, then
    $\displaystyle \operatorname{trace} (A) = \operatorname{trace} (B^{-1}A B).$
    In other words, the trace of similar matrices are equal.
  5. Let $ A$ be a square matrix of order $ n$ with real (or complex) entries $ a_{ij}$. Then
    $\displaystyle \operatorname{trace} A^\ast A$ $\displaystyle =$ $\displaystyle \operatorname{trace} A A^\ast$  
      $\displaystyle =$ $\displaystyle \sum_{i,j=1}^n \vert a_{ij}\vert^2.$  

    Here $ ^\ast$ is the complex conjugate, and $ \vert\cdot\vert$ is the complex modulus. In particular, $ \operatorname{trace} A^\ast A\ge 0$ with equality if and only if $ A=0$. (See the Frobenius matrix norm.)
  6. Various inequalities for $ \operatorname{trace}$ are given in [2].

See the proof of properties of trace of a matrix.

Bibliography

1
The Trace of a Square Matrix. Paul Ehrlich, [online] http://www.math.ufl.edu/˜ehrlich/trace.html
2
Z.P. Yang, X.X. Feng, A note on the trace inequality for products of Hermitian matrix power, Journal of Inequalities in Pure and Applied Mathematics, Volume 3, Issue 5, 2002, Article 78, online.



"trace of a matrix" is owned by Daume. [ full author list (2) | owner history (1) ]
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See Also: Schur's inequality


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example of trace of a matrix (Example) by Daume
proof of properties of trace of a matrix (Proof) by Daume
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Cross-references: proof of properties of trace of a matrix, inequalities, Frobenius matrix norm, equality, complex modulus, complex conjugate, similar matrices, invertible, basis, conjugate transpose, transpose, scalar, complex, real numbers, linear transformation, properties, diagonal, sum, matrix, trace, order, square matrix
There are 3 references to this entry.

This is version 16 of trace of a matrix, born on 2001-11-16, modified 2006-07-19.
Object id is 930, canonical name is TraceOfAMatrix.
Accessed 39309 times total.

Classification:
AMS MSC: 15A99 (Linear and multilinear algebra; matrix theory :: Miscellaneous topics)
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Proof that Trace of a matrix equals sum of its eigenvalues? by sprocketboy on 2004-07-10 10:59:29
Does anyone know a general proof of the fact that the trace of a matrix equals the sum of its eigenvalues?

For the case that matrix $M is diagonalizable, this is fairly simple to prove. Assume $M is diagonalizable and form its diagonalization $D using matrix $V, which contains the eigenvectors of $M arranged in columns. Thus, we have:

$$ D = V^{-1} M V $$

where $D is a diagonal matrix with the eigenvalues of $M appearing on the diagonal of $D.

Now, use the following two facts:

$$ \operatorname{trace} (A^{-1}) = 1 / \operatorname{trace} (A).$$

and

$$ \operatorname{trace} (AB) = \operatorname{trace} (BA).$$

to show:

$$ \operatorname{trace} (D) = \operatorname{trace} (M).$$

and you are done.

What happens if $M is not diagonalizable?
[ reply | up ]
Reference by Logan on 2002-03-13 13:29:56
Please make the reference a hyperlink. For one thing, it is not encoded in latex properly, so the actual URL does not get displayed.

For example:

\htmlnormallink{\verb|http://www.math.ufl.edu/~ehrlich/trace.html|}{http://www.math.ufl.edu/~ehrlich/trace.html}

Or something like that. I haven't actually tried doing hyperlinks in latex2html.
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