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# Kronecker product

Definition.
Let $A=(a_{{ij}})$ be a $n\times n$ matrix and
let $B$ be a $m\times m$ matrix. Then the
*Kronecker product* of $A$ and $B$ is
the $mn\times mn$ block matrix

$\displaystyle A\otimes B$ | $\displaystyle=$ | $\displaystyle\left(\begin{array}[]{ccc}a_{{11}}B&\cdots&a_{{1n}}B\\ \vdots&\ddots&\vdots\\ a_{{n1}}B&\cdots&a_{{nn}}B\\ \end{array}\right).$ |

The Kronecker product is also known as the *direct product*
or the *tensor product* [1].

1. The product is bilinear. If $k$ is a scalar, and $A,B$ and $C$ are square matrices, such that $B$ and $C$ are of the same order, then

$\displaystyle A\otimes(B+C)$ $\displaystyle=$ $\displaystyle A\otimes B+A\otimes C,$ $\displaystyle(B+C)\otimes A$ $\displaystyle=$ $\displaystyle B\otimes A+C\otimes A,$ $\displaystyle k(A\otimes B)$ $\displaystyle=$ $\displaystyle(kA)\otimes B=A\otimes(kB).$ 2. If $A,B,C,D$ are square matrices such that the products $AC$ and $BD$ exist, then $(A\otimes B)(C\otimes D)$ exists and

$\displaystyle(A\otimes B)(C\otimes D)$ $\displaystyle=$ $\displaystyle AC\otimes BD.$ If $A$ and $B$ are invertible matrices, then

$\displaystyle(A\otimes B)^{{-1}}$ $\displaystyle=$ $\displaystyle A^{{-1}}\otimes B^{{-1}}.$ 3. If $A$ and $B$ are square matrices, then for the transpose ($A^{T}$) we have

$\displaystyle(A\otimes B)^{{T}}$ $\displaystyle=$ $\displaystyle A^{{T}}\otimes B^{{T}}.$ 4. Let $A$ and $B$ be square matrices of orders $n$ and $m$, respectively. If $\{\lambda_{i}|i=1,\ldots,n\}$ are the eigenvalues of $A$ and $\{\mu_{j}|j=1,\ldots,m\}$ are the eigenvalues of $B$, then $\{\lambda_{i}\mu_{j}|i=1,\ldots,n,\,j=1,\ldots,m\}$ are the eigenvalues of $A\otimes B$. Also,

$\displaystyle\det(A\otimes B)$ $\displaystyle=$ $\displaystyle(\det A)^{m}(\det B)^{n},$ $\displaystyle\mathop{\mathrm{rank}}(A\otimes B)$ $\displaystyle=$ $\displaystyle\mathop{\mathrm{rank}}A\,\mathop{\mathrm{rank}}B,$ $\displaystyle\mathop{\mathrm{trace}}(A\otimes B)$ $\displaystyle=$ $\displaystyle\mathop{\mathrm{trace}}A\,\mathop{\mathrm{trace}}B,$

# References

- 1
H. Eves,
*Elementary Matrix Theory*, Dover publications, 1980. - 2
T. Kailath, A.H. Sayed, B. Hassibi,
*Linear estimation*, Prentice Hall, 2000

## Mathematics Subject Classification

15-00*no label found*

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## Comments

## invariants of the tensor product

The entry for "Kronecker Product" or alternatively "Tensor Product"

shows formulas for the trace, rank, and determinant of the product

in terms of those for its factors.

Are there corresponding formulas for the other invariants, and in

particular, can the characteristic equation of the product be

related to the characteristic equations of its factors?

At worst, I suppose they could be deduced by knowing all the roots.

- hvm

## Re: invariants of the tensor product

this is possible, if not all that illuminating. recall that the $k$th coefficient of the characteristic polynomial of A is $(-1)^k {\rm tr}(\wedge^k A)$. Thus, for $A\otimes B$ we get ${\rm tr}(\wedge^n A\otimes B)={\rm tr}(\wedge^n A)+{\rm tr}(\wedge^{n-1} A\otimes B)+\cdots={\rm tr}(\wedge^n A)+{\rm tr}(\wedge^{n-1} A){\rm tr}(B)+\cdots$.

## Re: invariants of the tensor product

Allright, why not try to make it more illuminating? Those

wedgies are determinants, the trace takes sums, and the final

form looks like a convolution. But I'm suspicious of anything

that starts off with something depending only on A; the

determinant of the tensor product doesn't look like that,

although the trace does. Call those wedgies, which are the

symmatric functions of the roots, sigma-k. Then Sigma-2 (for

the tensor product) would be sigma-2-A + sigma-1-A * sigma-1-B

+ sigma-2-B. Is that correct?

Is it possible to run this in Mathematica(TM) and get a human-

readable result?