# A Matrices Problem - Cayley-Hamilton or Bash?

Here's a cool problem I came across sometime back, and I haven't been able to solve it yet (let's hope that people at Math SE come up with interesting solutions for it!)

$A$ is a square matrix of order 2, with $|A| ≠ 0$ such that $|A + |A|adj(A)| = 0$, where $|A|$ and $adj(A)$ denote the determinant and adjoint/adjugate of matrix A, respectively. Find $|A - |A|adj(A)| = ?$

Is there any way to avoid a bash, by assuming elements of the matrix and then trying to compute the desired determinant? Would it be possible to proceed using Cayley Hamilton theorem?

Also, I think this result could possibly be generalised to a nxn square matrix. So, please help me find the value of the desired determinant, and also let's collectively investigate this problem for square matrices of higher order - could probably be an interesting generalisation, who knows?

Update: I just solved the problem for a 2x2 square matrix - the required determinant is 4, and the determinant of matrix A is 1. Now, the question that remains is, how do I solve it for an nxn square matrix?

• By adjoint do you mean adjugate? Sep 30 '18 at 3:03
• Yes! That's what it means. Sep 30 '18 at 3:14
• By "order 2", do you mean that $A^2 = 1$, or that $A$ is a $2 \times 2$ matrix? And just checking, does $\operatorname{adj}(A) = |A| A^{-1}$? Oct 1 '18 at 6:05
• 1. 2x2 Matrix 2. Yes Oct 1 '18 at 9:03
• Are you considering these to be matrices over the real numbers? I can find only two $2 \times 2$ matrices over the reals that satisfy the condition in your question. If you include complex numbers, there are infinite families of matrices, all with differing determinant, satisfying the conditions. Oct 1 '18 at 13:34

You are reading too much into the question. The result does not generalise well to higher dimensions. Let us consider the $$2\times2$$ case first. Presumably $$A$$ is real. Let $$d=|\det(A)|$$, the absolute value of $$\det(A)$$. The given condition can be rewritten as $$\det(A + d^2A^{-1}) = 0$$, which implies that $$\det(A^2 + d^2 I) = 0$$. This simply means that $$-d^2$$ is a negative eigenvalue of $$A^2$$.

Since $$\det(A^2)=d^2$$, when $$A$$ is $$2\times2$$, the other eigenvalue of $$A^2$$ has to be $$-1$$. Thus $$A^2$$ has two negative real eigenvalues $$-d^2$$ and $$-1$$. Therefore $$A$$ must have a conjugate pair of eigenvalues. Hence $$d=1$$ and also $$\det(A)=|\det(A)|$$. It follows that the characteristic polynomial of $$A^2$$ is $$p(x) = (x+d^2)(x+1) = (x+1)^2$$ and $$\det\left(A - \det(A)\operatorname{adj}(A)\right) =\det(A^{-1})\det(A^2 - d^2I) =\frac1d p(d^2) = 4.$$ This concludes the $$2\times2$$ case. When $$A$$ is at least $$3\times3$$, only what we've said in the first paragraph remains valid. The condition $$\det\left(A + \det(A)\operatorname{adj}(A)\right)$$ is true if and only if $$-d^2$$ is an eigenvalue of $$A^2$$, meaning that $$A$$ is similar to $$\left(\begin{array}{c|c}\pmatrix{0&-d\\ d&0}&\ast\\ \hline0&P^{-1}\end{array}\right)$$ for some real matrix $$P$$ with determinant $$\pm d$$ (unlike the $$2\times2$$ case, $$\det(A)$$ is not necessarily equal to $$|\det(A)|$$ here and its sign is controlled by $$P$$). Therefore \begin{align} \det\left(A - \det(A)\operatorname{adj}(A)\right) &=\det(A-d^2A^{-1})\\ &=4d^2\det\left(P^{-1} - d^2P\right)\\ &=4\det(P)\det\left(I - d^2P^2\right), \end{align} whose value can assume any positive, zero or negative real value. The answer by loup blanc is obtained by using a block-diagonal matrix $$A$$ with $$d=P=\frac1z$$.

Firstly, notice that the condition $$\operatorname{det}(A + \operatorname{det}(A) \operatorname{adj}(A)) = 0$$ is invariant under conjugation $$A \mapsto P A P^{-1}$$, and so it suffices to check one element for each similarity class. Each $$2 \times 2$$ real matrix is similar to either an upper triangular matrix, or a "complex number" matrix.

### Upper triangular case

$$A = \begin{pmatrix} \lambda_1 & t \\ 0 & \lambda_2 \end{pmatrix} \in \operatorname{Mat}_{2 \times 2}(\mathbb{R})$$ The condition $$\operatorname{det}(A) \neq 0$$ gives that $$\lambda_1$$ and $$\lambda_2$$ are nonzero. We compute $$A + (\operatorname{det} A)(\operatorname{adj} A) = \begin{pmatrix} \lambda_1 + \lambda_1 \lambda_2^2 & -t\lambda_1 \lambda_2 \\ 0 & \lambda_2 + \lambda_1^2 \lambda_2 \end{pmatrix}$$ which has determinant $$\lambda_1 \lambda_2 (1 + \lambda_1^2) (1 + \lambda_2^2)$$. This determinant being zero forces one of $$\lambda_1, \lambda_2$$ to be zero, which is not allowed. So there are no upper-triangular real matrices $$A$$ satisfying the conditions.

### Complex number case

$$A = \begin{pmatrix} a & -b \\ b & a \end{pmatrix} \in \operatorname{Mat}_{2 \times 2}(\mathbb{R})$$ In this case the matrix $$A$$ works like the complex number $$z = a + ib$$, and we have $$\operatorname{det}(A) = z \overline{z}$$ and $$\operatorname{adj}(A) = \overline{z}$$, where $$\overline{z} = a - ib$$ is the complex conjugate. So $$\operatorname{det}(A + (\operatorname{det} A)(\operatorname{adj} A)) = |z + z \overline{zz}|^2 = |z|^2 |1 + \overline{z}^2|^2 = 0$$ gives that $$z$$ must be a square root of $$-1$$, i.e. one of the two matrices $$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \quad \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$

### Conclusion

So the equation only has solutions when $$A$$ is an invertible $$2 \times 2$$ matrix with eigenvalues $$\pm i$$. In this case, it is easy to check (using either of the two matrices above) that $$\operatorname{det}(A - (\operatorname{det} A)(\operatorname{adj} A)) = 4$$.

I'm not going to do any of the cases with more dimensions, but perhaps this gives you a way to think about what might happen. For example, every $$3 \times 3$$ real matrix is similar either to an upper triangular matrix, or a "block upper-triangular" matrix where along the diagonal we have a real entry, and then a $$2 \times 2$$ "complex number" entry.

For $$n=3$$, the required result depends on the matrix $$A$$.

For example $$A=\begin{pmatrix}z&0&0\\0&0&1/z\\0&-1/z&0\end{pmatrix}$$ where $$z\in\mathbb{R}\setminus\{0\}$$.