📂Matrix Algebra

Prove that the trace of powers of a diagonalizable matrix is equal to the sum of powers of its eigenvalues

Theorem

Suppose a diagonalizable matrix $A \in \mathbb{C}^{n \times n}$ and a natural number $k \in \mathbb{N}$ are given. Let the eigenvalue of $A$ be $\lambda_{1} , \cdots , \lambda_{n}$, then the following holds. $$\operatorname{tr} A^{k} = \sum_{i=1}^{n} \lambda_{i}^{k}$$ Here, $\operatorname{tr}$ is the trace.

Explanation

Although it is not quite a corollary, it is useful to note that when $A \in \mathbb{R}^{n \times n}$ is a symmetric matrix, the trace of $A^{2}$ equals the sum of the squares of the elements of $A$. $$\sum_{i=1}^{n} \sum_{j=1}^{n} A_{ij}^{2} = \operatorname{tr} A A^{T} = \tr A^{2} = \sum_{i=1}^{n} \lambda_{i}^{2}$$ In fact, this is used in the proof of Craig’s theorem.

Proof ¹

Let $Q$ be a unitary matrix and $\diag \left( \lambda_{1} , \cdots , \lambda_{n} \right)$ be a diagonal matrix, then let $A = Q \Lambda Q^{\ast}$.

Cyclic property of the trace: $$\operatorname{tr} (ABC) = \operatorname{tr} (BCA) = \operatorname{tr} (CAB)$$

$$\begin{align*} \operatorname{tr} A^{k} =& \operatorname{tr} Q \Lambda Q^{\ast} \cdots Q \Lambda Q^{\ast} \\ =& \operatorname{tr} Q \Lambda^{k} Q^{\ast} \\ =& \operatorname{tr} Q^{\ast} Q \Lambda^{k} \\ =& \operatorname{tr} I \Lambda^{k} \\ =& \sum_{i=1}^{n} \lambda_{i}^{k} \end{align*}$$

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