# Solving a matrix equation of four unknowns

Find all matrices $$A$$ that satisfy

$$A^2 - 4 A + 4I = \begin{pmatrix} 4 & 3 \\ 5 & 6 \end{pmatrix}.$$

I've tried solving this but I can only get a single equation constraining $$b$$ and $$c$$, i.e. $$b/c=3/5$$. I find it difficult to solve for the rest. Can anybody give me a hint? Thanks

• Please note that the answer to this question is not a unique matrix. You may see my Answer posted below. This is so because I find $\sqrt{B}$ by a consistent brute force method, where $B$ is the matrix on the left in the given question. Please discuss the non-uniqueness of $A$. – Dr Zafar Ahmed DSc Jun 19 at 18:44

Hints:

1. $$A^2 - 4A + 4I = (A - 2I)^2$$
2. The matrix on the right hand side is positive definite.
• I think it's pertinent to mention that taking the square roots of matrices isn't entirely trivial. You have to be careful because there are non-diagonal matrices whose squares are diagonal. – J_P Jun 19 at 9:02

Anurag A posted an answer but deleted it after I brought up some concerns. However, after some additional thought I think his argument can be completed.

First, diagonalise $$B$$ (the matrix on the RHS): $$B=PDP^{-1}$$. Now rewrite the equation: $$(A-2I)^2=PDP^{-1}\\ (P^{-1}(A-2I)P)^2=D\\ X^2=D$$ At this point, as $$B$$ has positive eigenvalues it is tempting to conclude that $$X=\sqrt{D}$$ where $$\sqrt{D}=\begin{bmatrix}\pm\sqrt{D_{11}} & 0\\0 & \pm\sqrt{D_{22}}\end{bmatrix}$$ However, this is not obvious since $$X^2=D$$ for some matrix $$X$$ does not imply that $$X$$ is diagonal; for example: $$\begin{bmatrix}1 & 1\\1 & -1\end{bmatrix}^2=\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}$$ But we can still make this work. If $$X=\begin{bmatrix}a & b\\c & d\end{bmatrix}$$ then $$X^2=\begin{bmatrix}a^2+bc & b(a+d)\\c(a+d) & d^2+bc\end{bmatrix}$$ If either $$b\neq0$$ or $$c\neq0$$, we can conclude that $$a+d=0$$, so $$a^2+bc=d^2+bc$$. Therefore, were $$X$$ not diagonal, the diagonal entries of $$D$$ would have to be the same, but it turns out they are not. So $$X$$ must be diagonal and indeed we can proceed from $$X=\sqrt{D}$$ for each of the $$4$$ options for $$\sqrt{D}$$.

• That works perfectly. – Anurag A Jun 19 at 9:09

Let $$X=A-2I$$, then we are solving $$X^2=B$$. Since $$X$$ is a $$2 \times 2$$ matrix, therefore by Cayley-Hamilton it satisfies $$X^2-(\text{tr}X)X+(\det X)I=0.$$ This implies $$(\text{tr}X)X=B+(\det X)I.$$ But $$X^2=B^ \implies \det X = \pm (\det B)=\pm 3.$$ Thus $$(\text{tr}X)X=B\pm 3I.$$ Let $$X=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$, then $$(a+d)\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix} 4\pm3 & 3 \\ 5 & 6\pm 3 \end{bmatrix}$$ From this we can conclude that $$(a+d)^2=(4 \pm 3)+(6 \pm 3)=16,4.$$ Thus $$\color{red}{\text{tr}X=\pm 4, \pm 2}$$ Now $$(\text{tr}X)X=B\pm 3I \implies \color{blue}{X=\frac{1}{\text{tr}X}(B\pm 3I)}$$

• If $X^2=D$, can we conclude that $X=\sqrt{D}$ for one of those choices of $\pm$? – J_P Jun 19 at 8:36
• @J_P for a diagonal matrix it works provided the square root of the entries lie in the field $\Bbb{F}$ over which the matrices are defined. – Anurag A Jun 19 at 8:39
• Which matrix are you saying must be diagonal, $X$ or $D$? There are examples of $2\times 2$ matrices $X$ which are not diagonal but whose squares are. Just take $d=-a$ where $a, d$ are the diagonal entries of $X$. – J_P Jun 19 at 8:46
• After you deleted your other answer, I thought about it some more and I think I managed to complete the argument, which I put into another answer. – J_P Jun 19 at 9:07
• @J_P Thanks. I realized that I did not have a complete argument to what I was claiming. That is why I tried a different method and after I deleted mine I saw your completion of the argument. – Anurag A Jun 19 at 9:09

Call the matrix on the left as $$B$$, your equation becomes $$(A-2I)^2=B~~~(1),$$ where $$B= \begin{pmatrix} 4 & 3 \\ 5 & 6 \end{pmatrix} ~~ \mbox{and let} ~~\sqrt{B}=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ This means $$A=2I + \sqrt{B}==~~~(2)$$. Now you have to find $$\sqrt{B}$$, Let $$B=\begin{pmatrix} a & b \\ c & d \end{pmatrix}^2=\begin{pmatrix} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc \end{pmatrix}=\begin{pmatrix} 4 & 3 \\ 5 & 6 \end{pmatrix}~~~~(3)$$ Now solve for $$a,b,c,d$$ the set of four equations $$a^2+bc=4~~~(i),~ b(a+d)=3~~~(ii), ~c(a+d)=5~~~(iii),~ d^2+bc=6~~~(iv)~~~(4)$$ Let $$c=k$$ then from (ii) and (iii) get $$b=3k/5.$$ From (i) and (ii) have $$a^2-d^2=-2 \Rightarrow (a-d)(a+d)=-2.$$ Next use (ii) to get $$a-d=-6k/5$$ and from (iii) we have $$(a+d)=5/k$$. Solve these two to get $$a=(5/k-6k/5)/2$$. Put this $$a$$ and $$b,c$$ in (i) to get $$(5/k-6k/3)^2/4+3k^2/5=4 ~~~~~~(5)$$ whose roots are $$k=\pm 5/2,~~ \pm \frac{5}{2 \sqrt{6}}~~~~~~~~~~(6)$$ Four $$2 \times 2$$ matrices are possible for $$\sqrt{B}$$. $$\sqrt{B}=\frac{\pm 1}{2} \begin{pmatrix} 1 & 3 \\ 5 & 3 \end{pmatrix} \mbox{and two more for other value of}~ k~~~~ (7)$$ The other two matrices do satisfy $$Trace(\sqrt{B})^2=Trace(B)$$ but they do not satisfy $$(\det |\sqrt{B}|)^2=\det |B|$$, hence they are rejected. from (2) you get $$A_1=\frac{1}{2}\begin{pmatrix} 5 & 3\\ 5 & 7 \end{pmatrix},~~~A_2=\frac{1}{2}\begin{pmatrix} 3 & -3 \\ -5 & 1 \end{pmatrix}.$$ It will be fun to check that these two matrices for A$will satisfy the original equation. Note that my answer though consistent will be different from that of the Answer of other methods. Also note that this question does not have a unique matrix as answer. • For matrices$P^2=Q$doesn't necessarily imply$P=\pm \sqrt{Q}\$. – Anurag A Jun 19 at 14:05
• @Anurag A See my Eq. (2), I have avoided it now. – Dr Zafar Ahmed DSc Jun 20 at 11:47