= Trace and local volume change
{title2=$\det(I+tA)=1+t\operatorname{tr}A+O(t^2)$}
For any fixed real $n$-by-$n$ <matrix> $A$, as the real parameter $t$ tends to zero,
$$
\det(I+tA)=1+t\operatorname{tr}A+O(t^2).
$$
The <big O notation> means the remainder is bounded in <absolute value> by $Ct^2$ for some constant $C$ near zero. The <determinant> stays positive for sufficiently small $t$, so its <oriented volume> factor is also its ordinary <volume> factor. Thus the <trace> is the first-order fractional <volume> change, measured from the <identity matrix>.
A proof follows directly from <multilinearity of the determinant>. Each column of $I+tA$ is $e_j+tAe_j$. The constant term chooses all the <standard basis> columns and equals one. A term containing exactly one factor $t$ replaces just column $j$ by $Ae_j$; expanding that <vector> in the <standard basis> leaves only its $e_j$ component, $a_{jj}$, because every other choice repeats a column. Summing gives $\operatorname{tr}A$. Terms with two or more replacements have order $t^2$ or higher. This also proves the <derivative of the determinant> at the <identity matrix>.
This coefficient has a direct visual meaning. At the point $e_j$, the <linear map> $x\mapsto Ax$ has component $a_{jj}$ along direction $e_j$. In a small step $x\mapsto x+tAx$, these are the first-order changes of the square's or box's sides in their own directions. Summing them gives its fractional <area> or <volume> change; the off-diagonal entries initially change shape rather than contributing to this first-order total.
For the constant-coefficient <linear system of differential equations> $x'=Ax$, the exact <linear flow> is $x(t)=e^{tA}x(0)$, using the <matrix exponential>. Its <volume> factor is
$$
V(t)=\det(e^{tA})=e^{t\operatorname{tr}A},\qquad V(0)=1,
\qquad V'(t)=(\operatorname{tr}A)V(t).
$$
This is the <matrix exponential determinant identity>. One can derive it without <diagonalization>: set $E(t)=e^{tA}$, whose <matrix inverse> is $E(-t)$, and use $E'=AE$ in <Jacobi's formula> to get $V'=\det E\operatorname{tr}(E^{-1}AE)=(\operatorname{tr}A)V$. The scalar <differential equation> with initial value one gives the displayed <exponential function>. Thus positive <trace> causes <volume> expansion, negative <trace> causes contraction, and zero <trace> preserves <volume>, even when individual directions behave differently.
In two dimensions, compare four generators at $t=0.4$:
$$
\begin{array}{c|c|c|c}
A&\operatorname{tr}A&e^{0.4A}&\text{area factor}\\\hline
\operatorname{diag}(1,1)&2&\operatorname{diag}(e^{0.4},e^{0.4})&e^{0.8}\approx2.2255\\
\operatorname{diag}(1,-1)&0&\operatorname{diag}(e^{0.4},e^{-0.4})&1\\
\operatorname{diag}(-1,-1)&-2&\operatorname{diag}(e^{-0.4},e^{-0.4})&e^{-0.8}\approx0.4493\\
\begin{pmatrix}0&1\\0&0\end{pmatrix}&0&\begin{pmatrix}1&0.4\\0&1\end{pmatrix}&1
\end{array}
$$
The first and third <diagonal matrices> expand or contract both directions. The second expands horizontally and contracts vertically by reciprocal factors, making a rectangle of unchanged <area>. For the last <nilpotent matrix>, $A^2=0$, so its <matrix exponential> is exactly $I+tA$; it sends $(x,y)$ to $(x+ty,y)$, a <shear mapping>. Zero <trace> allows substantial shape change and does not mean the <flow map> is stationary.
\Image[/matrix-trace-local-volume.png]
{title=Four examples of a <linear system of differential equations>: <trace> controls <area> change while a saddle and a <shear mapping> preserve <area>}
The first-order statement and the exact <flow map> should be distinguished. For $A=\operatorname{diag}(1,-1)$, $\det(I+tA)=1-t^2$, whereas $\det(e^{tA})=1$ for all real $t$. Zero <trace> eliminates the first-order change in the approximate step, and it preserves <volume> exactly in the <matrix exponential> flow. The examples use an abstract parameter $t$ and coordinate <vectors>; no physical units are assumed.
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