Difference between revisions of "Trace of a square matrix"
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\newcommand{\Tr}{\mathop{\mathrm{Tr}}} | \newcommand{\Tr}{\mathop{\mathrm{Tr}}} | ||
\newcommand{\Sp}{\mathop{\mathrm{Sp}}} | \newcommand{\Sp}{\mathop{\mathrm{Sp}}} | ||
| + | \newcommand{\End}{\mathop{\mathrm{End}}} | ||
$ | $ | ||
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\tr A = \sum_{i=0}^n a_{ii} | \tr A = \sum_{i=0}^n a_{ii} | ||
$$ | $$ | ||
| − | Let $A$ be a square matrix of order $n$ over a [[ | + | Let $A$ be a square matrix of order $n$ over a [[field]] $k$. The trace of $A$ coincides with the sum of the roots of the [[characteristic polynomial]] of $A$. If $k$ is a field of characteristic 0, then the $n$ traces $\tr A, \ldots \tr A^n$ uniquely determine the characteristic polynomial of $A$. In particular, $A$ is nilpotent if and only if $\tr A^m = 0$ for all $m=1,\ldots,n$. |
If $A$ and $B$ are square matrices of the same order over $k$, and $\alpha,\beta \in k$, then | If $A$ and $B$ are square matrices of the same order over $k$, and $\alpha,\beta \in k$, then | ||
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$$ | $$ | ||
while if $\det B \neq 0$, | while if $\det B \neq 0$, | ||
| − | $$ | + | $$\label{eq:a1} |
\tr(BAB^{-1}) = \tr A. | \tr(BAB^{-1}) = \tr A. | ||
$$ | $$ | ||
The trace of the [[Tensor product|tensor (Kronecker) product]] of square matrices over a field is equal to the product of the traces of the factors. | The trace of the [[Tensor product|tensor (Kronecker) product]] of square matrices over a field is equal to the product of the traces of the factors. | ||
| − | The trace of a product of matrices $A \in \mathbb{R}^{n \times m}$, $B \in \mathbb{R}^{m \times n}$ with a resulting square matrix is equal to the sum over all components of a [[Hadamard | + | The trace of a product of matrices $A \in \mathbb{R}^{n \times m}$, $B \in \mathbb{R}^{m \times n}$ with a resulting square matrix is equal to the sum over all components of a [[Hadamard product]] of $A$ and $B^T$: |
$$ | $$ | ||
\tr(AB) = \sum_{i=1}^n \sum_{j=1}^n (A \circ B^T)_{i,j}. | \tr(AB) = \sum_{i=1}^n \sum_{j=1}^n (A \circ B^T)_{i,j}. | ||
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|valign="top"|{{Ref|Ga}}||valign="top"| F.R. [F.R. Gantmakher] Gantmacher, "The theory of matrices", '''1''', Chelsea, reprint (1959) (Translated from Russian) | |valign="top"|{{Ref|Ga}}||valign="top"| F.R. [F.R. Gantmakher] Gantmacher, "The theory of matrices", '''1''', Chelsea, reprint (1959) (Translated from Russian) | ||
| + | |- | ||
| + | |} | ||
| + | |||
| + | ====Comment==== | ||
| + | The trace of an endomorphism $\alpha$ of a finite-dimensional vector space $V$ over the field $k$ may be defined as the trace of any matrix representing it with respect to a given basis for $V$. Since the trace is invariant for [[similar matrices]] (equation 1), this is well-defined. It may be defined in a basis-independent way from the sequence | ||
| + | $$ | ||
| + | \End(V) \leftrightarrow V^* \otimes V \rightarrow k \ . | ||
| + | $$ | ||
| + | |||
| + | ====References==== | ||
| + | |||
| + | {| | ||
| + | |- | ||
| + | |valign="top"|{{Ref|Bo}}||valign="top"| N. Bourbaki, "Algebra", '''1''', Chap.1-3, Springer (1989) §4.3 | ||
|- | |- | ||
|} | |} | ||
Latest revision as of 19:21, 11 November 2023
2020 Mathematics Subject Classification: Primary: 15A15 [MSN][ZBL] $ \newcommand{\tr}{\mathop{\mathrm{tr}}} \newcommand{\Tr}{\mathop{\mathrm{Tr}}} \newcommand{\Sp}{\mathop{\mathrm{Sp}}} \newcommand{\End}{\mathop{\mathrm{End}}} $
The sum of the entries on the main diagonal of this matrix. The trace of a matrix $A = [a_{ij}]$ is denoted by $\tr A$, $\Tr A$ or $\Sp A$: $$ \tr A = \sum_{i=0}^n a_{ii} $$ Let $A$ be a square matrix of order $n$ over a field $k$. The trace of $A$ coincides with the sum of the roots of the characteristic polynomial of $A$. If $k$ is a field of characteristic 0, then the $n$ traces $\tr A, \ldots \tr A^n$ uniquely determine the characteristic polynomial of $A$. In particular, $A$ is nilpotent if and only if $\tr A^m = 0$ for all $m=1,\ldots,n$.
If $A$ and $B$ are square matrices of the same order over $k$, and $\alpha,\beta \in k$, then $$ \tr(\alpha A + \beta B) = \alpha \tr A + \beta \tr B, \quad \tr AB = \tr BA, $$ while if $\det B \neq 0$, $$\label{eq:a1} \tr(BAB^{-1}) = \tr A. $$ The trace of the tensor (Kronecker) product of square matrices over a field is equal to the product of the traces of the factors.
The trace of a product of matrices $A \in \mathbb{R}^{n \times m}$, $B \in \mathbb{R}^{m \times n}$ with a resulting square matrix is equal to the sum over all components of a Hadamard product of $A$ and $B^T$: $$ \tr(AB) = \sum_{i=1}^n \sum_{j=1}^n (A \circ B^T)_{i,j}. $$
References
| [Co] | P.M. Cohn, "Algebra", 1, Wiley (1982) pp. 336 |
| [Ga] | F.R. [F.R. Gantmakher] Gantmacher, "The theory of matrices", 1, Chelsea, reprint (1959) (Translated from Russian) |
Comment
The trace of an endomorphism $\alpha$ of a finite-dimensional vector space $V$ over the field $k$ may be defined as the trace of any matrix representing it with respect to a given basis for $V$. Since the trace is invariant for similar matrices (equation 1), this is well-defined. It may be defined in a basis-independent way from the sequence $$ \End(V) \leftrightarrow V^* \otimes V \rightarrow k \ . $$
References
| [Bo] | N. Bourbaki, "Algebra", 1, Chap.1-3, Springer (1989) §4.3 |
Trace of a square matrix. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Trace_of_a_square_matrix&oldid=31319