Linear Algebra II HT23, Trace
Flashcards
What is the trace $\text{Tr}(A)$ of a square matrix $A$?
The sum of the diagonal entries.
What is the trace $\text{Tr}(T)$ of a linear transformation $T$?
The sum of the diagonal entries of a matrix $A$ representing $T$ with respect to some basis.
What’s a quick proof that the trace $\text{Tr}(T)$ of a linear transformation $T$ is the same for any choice of basis?
Consider $A, B$ both representing $T$ with respect to different bases
\[\begin{aligned} \text{Tr}(B) &= \text{Tr}(P^{-1} A P) \\\\ &= \text{Tr}(P^{-1}P A) \\\\ &= \text{Tr}(IA) \\\\ &= \text{Tr}(A) \end{aligned}\]Proofs
Bite-sized
Cyclic property of trace: $\mathrm{Tr}(AB) = $ $\mathrm{Tr}(BA)$ whenever both products are square. More generally, $\mathrm{Tr}(ABC) = \mathrm{Tr}(BCA) = \mathrm{Tr}(CAB)$ — but $\ne \mathrm{Tr}(ACB)$ in general (the cycle order matters).
Trace and Frobenius norm: for any $B \in \mathbb R^{m \times n}$, $\ \vert B\ \vert _ F^2 = $ $\mathrm{Tr}(B^\top B)$. This is the formula behind many trace-based bounds in NLA.
Why is $\mathrm{Tr}(T)$ basis-independent for a linear map $T : V \to V$?
For two matrices $A, B$ representing $T$ in different bases, $B = P^{-1} A P$ for the change-of-basis matrix $P$. Then $\mathrm{Tr}(B) = \mathrm{Tr}(P^{-1} A P) = \mathrm{Tr}(P P^{-1} A) = \mathrm{Tr}(A)$ by the cyclic property. So the trace is genuinely an invariant of the linear map, not of any particular matrix representation.