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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
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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 MSC15A99 (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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