The Trace of a Matrix

The Trace of a Matrix#

The trace of a square matrix is the sum of its diagonal entries. For \(A\in\mathbb{R}^{n\times n}\) or \(A\in\mathbb{C}^{n\times n}\),

\[\operatorname{tr}(A)=\sum_{i=1}^n a_{ii}.\]

For example,

\[\begin{split} A=\begin{pmatrix} 3 & 0 & 1 \\ 5 & -2 & 9 \\ 4 & 6 & 8 \end{pmatrix} \quad\Longrightarrow\quad \operatorname{tr}(A)=3+(-2)+8=9. \end{split}\]

In particular, \(\operatorname{tr}(I_n)=n\). Only diagonal entries contribute to the trace, so it takes \(n-1\) additions to compute it.

Properties of the Trace#

Linearity#

For square matrices \(A,B\) of the same size and scalars \(\alpha,\beta\),

\[ \operatorname{tr}(\alpha A+\beta B) =\alpha\operatorname{tr}(A)+\beta\operatorname{tr}(B). \]

Proof. The diagonal entries of \(\alpha A+\beta B\) are \(\alpha a_{ii}+\beta b_{ii}\). Thus

\[\begin{split} \begin{aligned} \operatorname{tr}(\alpha A+\beta B) &=\sum_{i=1}^n(\alpha a_{ii}+\beta b_{ii}) \\ &=\alpha\sum_{i=1}^n a_{ii}+\beta\sum_{i=1}^n b_{ii}. \end{aligned} \end{split}\]

Cyclic Property#

If \(A\) is \(m\times n\) and \(B\) is \(n\times m\), then

\[\operatorname{tr}(AB)=\operatorname{tr}(BA).\]

The factors may be rectangular: \(AB\) is \(m\times m\), while \(BA\) is \(n\times n\).

Proof. Expand the diagonal entries of each product and interchange the finite sums:

\[\begin{split} \begin{aligned} \operatorname{tr}(AB) &=\sum_{i=1}^m\sum_{j=1}^n a_{ij}b_{ji} \\ &=\sum_{j=1}^n\sum_{i=1}^m b_{ji}a_{ij} \\ &=\operatorname{tr}(BA). \end{aligned} \end{split}\]

For three factors of compatible sizes, grouping them into two factors gives

\[ \operatorname{tr}(ABC) =\operatorname{tr}(BCA) =\operatorname{tr}(CAB). \]

For example, apply the two-factor identity to \(A\) and \(BC\) to move \(A\) from the front to the back. The same argument works for any number of factors: moving the first factor to the end preserves the trace.

An arbitrary reordering need not preserve the trace. For example, let

\[\begin{split} A=\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad B=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad C=\begin{pmatrix}0&0\\1&0\end{pmatrix}. \end{split}\]

Then \(ABC=A\) and \(ACB=0\), so \(\operatorname{tr}(ABC)=1\) while \(\operatorname{tr}(ACB)=0\).

The trace is also not multiplicative: in general, \(\operatorname{tr}(AB)\neq\operatorname{tr}(A)\operatorname{tr}(B)\). Taking \(A=B=I_2\) gives \(2\) on the left and \(4\) on the right.

Transpose and Conjugate Transpose#

For a square matrix \(A\),

\[ \operatorname{tr}(A^T)=\operatorname{tr}(A),\qquad \operatorname{tr}(A^H)=\overline{\operatorname{tr}(A)}. \]

Here, \(A^H\) denotes the conjugate transpose and the overline denotes complex conjugation.

Proof. Transposition leaves the diagonal entries unchanged. The conjugate transpose replaces each diagonal entry \(a_{ii}\) by \(\overline{a_{ii}}\), so its trace is the complex conjugate of their sum.

Invariance Under a Change of Basis#

If \(S\) is invertible, then

\[\operatorname{tr}(S^{-1}AS)=\operatorname{tr}(A).\]

Thus similar matrices have the same trace. Although their individual diagonal entries may differ, the trace does not depend on the basis used to represent the linear transformation.

Proof. Use the cyclic property to move \(S^{-1}\) to the end:

\[\begin{split} \begin{aligned} \operatorname{tr}(S^{-1}AS) &=\operatorname{tr}(ASS^{-1}) \\ &=\operatorname{tr}(A). \end{aligned} \end{split}\]

The later section on applications of eigenvalues uses this property to relate the trace to the sum of the eigenvalues.

Useful Identities#

Outer Products#

For column vectors \(u,v\in\mathbb{R}^n\),

\[\operatorname{tr}(uv^T)=\sum_{i=1}^n u_iv_i=v^Tu.\]

The diagonal entries of \(uv^T\) are \(u_iv_i\), which gives the identity directly. For complex vectors, the corresponding identity with conjugation is

\[\operatorname{tr}(uv^H)=\sum_{i=1}^n u_i\overline{v_i}=v^Hu.\]

The Frobenius Norm#

For a possibly rectangular matrix \(A\in\mathbb{C}^{m\times n}\),

\[\begin{split} \begin{aligned} \operatorname{tr}(A^HA) &=\sum_{j=1}^n\sum_{i=1}^m |a_{ij}|^2 \\ &=\|A\|_F^2. \end{aligned} \end{split}\]

Indeed, diagonal entry \(j\) of \(A^HA\) is the squared norm of column \(j\) of \(A\). Summing over the columns gives the squared Frobenius norm. For a real matrix, this reduces to \(\operatorname{tr}(A^TA)=\|A\|_F^2\).

Interpretation: Divergence of a Linear Vector Field#

Let \(A\in\mathbb{R}^{n\times n}\) and consider the vector field \(F(x)=Ax\). The divergence is the sum of the partial derivatives of each component with respect to its matching coordinate:

\[ \nabla\cdot F=\sum_{i=1}^n\frac{\partial F_i}{\partial x_i}. \]

Since \(F_i(x)=\sum_{j=1}^n a_{ij}x_j\), we have \(\partial F_i/\partial x_i=a_{ii}\). Therefore,

\[\nabla\cdot F=\operatorname{tr}(A).\]

If particles move with velocity \(F(x)\), so that \(\dot{x}=Ax\), the divergence gives the instantaneous rate of volume change per unit volume:

  • A positive trace means that a volume of moving particles grows.

  • A negative trace means that it shrinks.

  • A zero trace means that volume is preserved; the flow is divergence-free.

This concerns volume, rather than expansion or contraction in every direction. For example,

\[\begin{split}A=\begin{pmatrix}2&0\\0&-1\end{pmatrix}\end{split}\]

has positive trace, but the motion \(\dot{x}=Ax\) expands in the first coordinate and contracts in the second. If the first diagonal entry is changed to \(1\), the trace becomes zero: expansion in one coordinate and contraction in the other preserve area while changing shape.

These statements describe motion under the velocity field \(F(x)=Ax\). For a single application of the linear map \(x\mapsto Ax\), the volume scaling factor is \(|\det(A)|\), as discussed in the preceding section.