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Let $A$ be a square matrix and $I$ the $k \times k$ identity matrix. Then the identity $$ \det(A \otimes I) = \det(A)^k,$$ holds as can be seen from a general result on the determinant of block matrices.

Since the structure of $A \otimes I$ is quite beautiful I am wondering

Can someone find a short proof of the above equality?

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  • $\begingroup$ isn't it the Kroenecker product (and not Hadamard)? Because with the Hadamard product, this result is wrong. However with the Kroenecker product, you can show that the eigenvalues of $A \circ I$ and $A$ coincides (up to multiplicity). $\endgroup$ – Surb Nov 20 '14 at 20:56
  • $\begingroup$ @Surb Err you're right I mean the Kronecker product! $\endgroup$ – Jernej Nov 20 '14 at 21:01
  • $\begingroup$ The simplest proof proceeds as follows: Let $K$ be your base ring. Let $n$ be the size of $A$. There is an isomorphism $K^n \otimes K^k \to \bigoplus\limits_{i=1}^k K^n$ of $K$-modules, and under this isomorphism the endomorphism $A \otimes I$ of $K^n \otimes K^k$ corresponds to the endomorphism $\bigoplus\limits_{i=1}^k A$ of $\bigoplus\limits_{i=1}^k K^n$ (where we identify matrices with endomorphisms in a hopefully clear way). Now, you know what $\det\left(\bigoplus\limits_{i=1}^k A\right)$ is? (What is $\det\left(A\oplus B\right)$ ?) $\endgroup$ – darij grinberg Sep 19 '15 at 13:05
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Consider the canonique basis $e_1,\ldots,e_k$ of $\Bbb R^k$. Furthermore let $\lambda_1,\ldots,\lambda_n$ be the eigenvalues of $A \in \Bbb R^{n\times n}$ and $v_1,\ldots,v_n$ their associated eigenvectors. Then $v_i \otimes e_j$ is an eigenvector of $A \otimes I$ associated to the eigenvalue $\lambda_i$ for every $j = 1,\ldots,k$. Since $v_i \otimes e_l$ and $v_i \otimes e_j$ and linearly independent for every $l \neq j$. It is clear that the eigenvalues of $A \otimes I$ are exactly $\underbrace{\lambda_1,\ldots,\lambda_1}_{k \text{ times}},\ldots,\underbrace{\lambda_n,\ldots,\lambda_n}_{k \text{ times}}$ it follows that $\det(A\otimes I) = \det(A)^k$.

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  • $\begingroup$ $n$ linearly independent eigenvectors don't always exist. In the general case you would have to extend the base field to an algebraic closure (no, $\mathbb{R}$ is not enough), WLOG assume that the matrix is diagonalizable (that, or find a trigonalization), etc. All in all, not a particularly short proof. $\endgroup$ – darij grinberg Sep 19 '15 at 13:02

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