Calculating eigenvalues of a matrix written using Kronecker products

Question

If $\lambda_1$, $\lambda_2$ and $\lambda_3$ are the eigenvalues of the matrix $\mathbf{Q}$=$S\left(\mathbf{n}\otimes\mathbf{n} \right)$+$P\left(\mathbf{m}\otimes\mathbf{m} \right)-1/3\left(S+P\right)\mathbf{I}$, with real scalars S and P, $\mathbf{n}$ and $\mathbf{m}$ are unit vectors, and $\mathbf{n}\cdot\mathbf{m}=0$, how do I calculateS and P as a function of the eigenvalues, i.e., $S=S\left(\lambda_1, \lambda_2, \lambda_3\right)$ and $P=P\left(\lambda_1, \lambda_2, \lambda_3\right)$?

Are $\mathbf{n}$ and $\mathbf{m}$ its eigenvectors? If so, which are their eigenvalues?

The Eqs. (17) to (19) in https://arxiv.org/abs/1409.3542 say that $\lambda_1=\left(2S-P\right)/3$, $\lambda_2=-\left(S+P\right)/3$, and $\lambda_3=\left(2P-S\right)/3$.

Answer 1

Here $\otimes$ does not correspond to Kronecker product, but to the outer product (see below). Let us see what are the consequences.

Let $m,n$ be unit orthogonal vectors of $\mathbb{R^3}$ that will be assimilated to column vectors;

Let us stop here in order to recall that the product

• $u^Tv$ of a line vector $u^T$ by a column vector $v$ is a number which is called their dot product (or scalar product, or in tensor terms, the inner product),

$$u^Tv=\begin{pmatrix} a&b&c \end{pmatrix}\begin{pmatrix} d\\e\\f \end{pmatrix}=ae+be+cf$$

whereas the product:

• $uv^T$ of a column vector $u$ by a line vector $v^T$ which is a matrix

$$uv^T=\begin{pmatrix} a\\b\\c \end{pmatrix}\begin{pmatrix} d&e&f \end{pmatrix}=\begin{pmatrix} ad&ae&af \\ bd&be&bf \\ cd&ce&cf\end{pmatrix}$$

called their outer product. Such a matrix has rank one.

The fact that $m$ and $n$ are unit vectors and orthogonal can be expressed using matrix-vector notations like this:

$$\begin{cases}m^Tm&=&n^Tn&=&1\\n^Tm&=&m^Tn&=&0\end{cases}.\tag{1}$$

Let : $$Q := Snn^T+Pmm^T−\frac13(S+P)I_3$$

In order to check that $n$ is an eigenvector of $Q$, we apply $Q$ to $n$, getting

$$Qn = Snn^Tn+Pmm^Tn−\frac13(S+P)I_3n\tag{2}$$

Using (1), (2) becomes:

$$Qn = Sn\underbrace{(n^Tn)}_1+Pm\underbrace{(m^Tn)}_0−\frac13(S+P)n$$

$$Qn = Sn + 0 −\frac13(S+P)n=(S-\frac13(S+P))n$$

i.e.,

$$Qn = \underbrace{(\frac13(2S-P))}_{\lambda}n$$

we find the awaited eigenvalue.

Same computation for $Qm=\lambda' m$ giving $\lambda'=(\frac13(2P-S)).$

(we can in fact avoid this calculation, because the definition of $Q$ is symmetrical with respect to variables $S$ and $P$).

Answer 2

$ \def\LR#1{\left(#1\right)} \def\fracLR#1#2{\left(\frac{#1}{#2}\right)} $In addition to $(m,n)$ there is a third unit vector $$p = \frac{m\times n}{\|m\times n\|}$$ Together these three unit vectors form a basis for ${\mathbb R}^{3}$ and allow one to write the identity tensor as $$I=m\otimes m + n\otimes n+p\otimes p$$ Substituting this into the expression for the $Q$ matrix yields $$\eqalign{ Q &= Pm\otimes m + Sn\otimes n - \fracLR{P+S}{3}\LR{m\otimes m+n\otimes n+p\otimes p} \\ &= \fracLR{2P-S}{3}m\otimes m + \fracLR{2S-P}{3}n\otimes n - \fracLR{P+S}{3}{p\otimes p} \\ }$$ Since the three basis vectors are mutually orthogonal, each is an eigenvector of $Q$ and the associated eigenvalue is the coefficient in parentheses.