0
$\begingroup$

Let \begin{align*} M =\begin{pmatrix} 2a&c&b&0\\b&a+d&0&b\\c&0&a+d&c\\0&c&b&2d \end{pmatrix}. \end{align*} When $M$ is invertible?

It is easy to compute the determinant and then solve $\det M=0$, but it's quite long and requires much manipulations.

So my question is can we find a way to write this matrix $M$ as a product of two $4$ by $4$ matrices, where one of them includes all the parameters as its entries and easier to compute the determinant, and the other is a matrix (not the identity one) with numbers entries?

(I came up with this since the matrix have a quite "nice" form, $M_{11}=a+a, M_{44} =d+d$ , $b$'s and $c$'s are symmetrical about the main diagonal.)

$\endgroup$
1
  • $\begingroup$ Instead of expressing $M$ as product of two $4×4$ matrices, it is easy to find required conditions through the equation $detM=0$ $\endgroup$
    – A learner
    Aug 30, 2020 at 9:01

1 Answer 1

0
$\begingroup$

I don't think what you suggests is a good way to follow. Your $M$ is a matrix representation of the linear operator $T:X\mapsto AX+XA$ on $M_2(\mathbb C)$. Since the invertibility of $T$ does not depend on the choice of basis, you can actually pick a basis of $\mathbb C^2$ such that $A$ becomes upper triangular (i.e. $c$ becomes $0$). Then $$ A=\pmatrix{a&b\\ 0&d}, \ M=\left(\begin{array}{cc|cc} 2a&0&b&0\\ b&a+d&0&b\\ \hline0&0&a+d&0\\ 0&0&b&2d \end{array}\right) $$ and it is easy to see that $\det(M)=4ad(a+d)^2=4\det(A)\operatorname{tr}(A)^2$.

Alternatively, note that $M=\pmatrix{A+aI&bI\\ cI&A+dI}$. Since the two sub-blocks in the bottom row commute, we have \begin{aligned} \det(M) &=\det\left((A+aI)(A+dI)-(bI)(cI)\right)\\ &=\det(A^2+\operatorname{tr}(A)A+\det(A)I)\\ &=\det(2\operatorname{tr}(A)A)\quad\text{(Cayley-Hamilton)}\\ &=4\operatorname{tr}(A)^2\det(A). \end{aligned}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.