Trace, Determinant and Similarity

Definition The trace of a square matrix $A = [a_{ij}] \in \mathbb{F}^{n \times n}$ is the sum of its diagonal entries,

The trace is linear, with $\operatorname{tr}(A + B) = \operatorname{tr} A + \operatorname{tr} B$ and $\operatorname{tr}(cA) = c \operatorname{tr} A$, and $\operatorname{tr} A^{\mathsf{T}} = \operatorname{tr} A$. For columns $u, v \in \mathbb{F}^n$ the $i$th diagonal entry of $uv^{\mathsf{T}}$ is $u_i v_i$, so $\operatorname{tr}(uv^{\mathsf{T}}) = v^{\mathsf{T}}u$. Over $\mathbb{C}$ the matrix $uv^{*}$ has trace $v^{*}u$, which is a different number in general.

Definition Let $A \in \mathbb{F}^{m \times n}$ and $B \in \mathbb{F}^{n \times m}$. Then

even though $AB$ is $m \times m$ and $BA$ is $n \times n$. The proof is a finite double sum whose order can be reversed, using that scalars commute,

For three factors $A \in \mathbb{F}^{m \times n}$, $B \in \mathbb{F}^{n \times p}$ and $C \in \mathbb{F}^{p \times m}$, grouping two of them gives $\operatorname{tr}(ABC) = \operatorname{tr}(BCA) = \operatorname{tr}(CAB)$. Only these cyclic rotations are guaranteed to keep the trace. The identity is about traces and says nothing about $AB$ and $BA$ being equal.

Definition Matrices $A, B \in \mathbb{F}^{n \times n}$ are similar over $\mathbb{F}$ if

for some invertible $S \in \mathbb{F}^{n \times n}$. Similarity is an equivalence relation, and similar matrices are exactly the matrices of one operator in two bases. If $T \colon V \to V$ is linear on a finite-dimensional space $V$ with ordered bases $\mathcal{B}$ and $\mathcal{C}$, and $S = P_{\mathcal{B} \leftarrow \mathcal{C}}$ is the matrix whose columns are the $\mathcal{C}$ vectors written in $\mathcal{B}$ coordinates, then $P_{\mathcal{C} \leftarrow \mathcal{B}} = S^{-1}$ and

Conversely, every invertible $S$ arises this way, from the basis whose $\mathcal{B}$ coordinate columns are the columns of $S$.

Definition A function $f$ on $\mathbb{F}^{n \times n}$ is a similarity invariant if $f(S^{-1}AS) = f(A)$ for every $A \in \mathbb{F}^{n \times n}$ and every invertible $S \in \mathbb{F}^{n \times n}$. The trace and the determinant are similarity invariants. For the trace, apply the cyclic property to the two factors $S^{-1}A$ and $S$,

For the determinant, the product rule gives $\det(S^{-1}AS) = \det S^{-1} \det A \det S = \det A$, because $\det S^{-1} \det S = \det I = 1$. Neither proof needs $S$ to be orthogonal. The rank and the dimension of each eigenspace are similarity invariants too, since $\ker(\lambda I - S^{-1}AS) = S^{-1}\ker(\lambda I - A)$. So the trace and the determinant of an operator $T$ on a finite-dimensional space are well defined as those of $[T]_{\mathcal{B} \leftarrow \mathcal{B}}$ for any basis $\mathcal{B}$. Equal trace, determinant and characteristic polynomial are necessary for similarity, but for $n \ge 2$ they are not sufficient.

Definition The characteristic polynomial of $A \in \mathbb{F}^{n \times n}$ is

where $t$ is a scalar variable and $I$ is the $n \times n$ identity. In the permutation formula only the identity permutation contributes powers of $t$ above $t^{n-2}$, so $\chi_A$ is monic of degree $n$, the coefficient of $t^{n-1}$ is $-\operatorname{tr} A$, and the constant term is $\chi_A(0) = (-1)^n \det A$. For $n = 2$ this reads $\chi_A(t) = t^2 - (\operatorname{tr} A)\,t + \det A$. A scalar $\lambda \in \mathbb{F}$ is an eigenvalue of $A$, meaning $Av = \lambda v$ for some nonzero $v \in \mathbb{F}^n$, exactly when $\lambda I - A$ has a nonzero kernel, that is, exactly when $\chi_A(\lambda) = 0$. If $B = S^{-1}AS$, then

for every scalar $t$, so $\chi_B(t) = \chi_A(t)$ at every $t$. Two polynomials over $\mathbb{R}$ or $\mathbb{C}$ that agree at every scalar are equal, so $\chi_B = \chi_A$. Evaluating the finished polynomial $\chi_A$ at the matrix $A$ itself is a different question, settled later by the Cayley-Hamilton theorem.

Definition Let $A \in \mathbb{C}^{n \times n}$ and write $A = X + iY$ with $X, Y \in \mathbb{R}^{n \times n}$. Identify $\mathbb{C}^n$ with $\mathbb{R}^{2n}$ by sending $z = x + iy$, with $x, y \in \mathbb{R}^n$, to the real column $(x, y)$. Since $Az = (Xx - Yy) + i(Yx + Xy)$, multiplication by $A$ becomes the real linear map with matrix

and its determinant is

So multiplication by $A$ scales $2n$-dimensional real volume by $\lvert \det A \rvert^2$, and when $A$ is invertible it preserves the orientation of $\mathbb{R}^{2n}$. To prove it, work over $\mathbb{C}$ with the new variables $z = x + iy$ and $w = x - iy$. This change of variables is invertible, with $x = (z + w)/2$ and $y = (z - w)/(2i)$. In the variables $(z, w)$ the map sends $z$ to $Az$ and $w$ to $\overline{A}w$, so $R(A)$ is similar over $\mathbb{C}$ to the block diagonal matrix with blocks $A$ and $\overline{A}$. A block diagonal determinant is the product of the block determinants, and conjugation passes through the permutation formula, so

The permutation formula gives the same value for the real matrix $R(A)$ whether it is read over $\mathbb{R}$ or over $\mathbb{C}$, so this is also the real determinant. The case of singular $A$ is included, with factor $0$.