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Let $V$ be a finite-dimensional vector space, $T\colon V \to V$ a linear map and $W \subset V$ a $T$-invariant subspace (i.e., $T(W) \subset W$). Then there is a well-defined induced quotient map $\overline{T}\colon V/W \to V/W$.

Now I recall having seen the following determinant formula: $\det(T) = \det(\overline{T}) \det(T|_W)$, where $T|_W\colon W \to W$ is the restriction of $T$ to the subspace $W$ (though I do not remember the proof anymore).

Is there some similar formula relating the traces of $T$, $\overline{T}$ and $T|_W$?

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The idea is the same. Choose a basis $\mathfrak{B}$ of $W$, and extend it to a basis $\mathfrak{B} \cup \mathfrak{C}$ of $V$, where $\mathfrak{C} \subseteq V \setminus W$. With respect to this basis, the matrix of $T$ will be of the block form $$ \begin{bmatrix} B & A\\ 0 & C \end{bmatrix}, $$ where $A, B, C$ are suitably-sized matrices, with $B$ and $C$ square matrices.

Now $B$ represents the action of $T$ on $W$ with respect to $\mathfrak{B}$, and $C$ the action of $T$ on $V/W$, with respect to the basis given by the image of $\mathfrak{C}$. Hence the determinant formula, and the analogue $$ \operatorname{tr}(T) = \operatorname{tr}(T|_W) + \operatorname{tr}(\overline{T}). $$ for the trace.

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  • $\begingroup$ Thanks! Does this also hold true in the infinite-dimensional case (under the assumption that $T$ is traceable)? $\endgroup$
    – AlexE
    Apr 1, 2013 at 12:21

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