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.)

  • $\begingroup$ 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


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}


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