Square root of Matrix $A=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$

Question

Find Square root of Matrix $A=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$ without using concept of Eigen Values and Eigen Vectors

I assumed its square root as $$B=\begin{bmatrix} a &b \\ c&d \end{bmatrix}$$

hence

$$B^2=A$$ $\implies$

$$\begin{bmatrix} a^2+bc &b(a+d) \\ c(a+d)&d^2+bc \end{bmatrix}=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$$

So

$$a^2+bc=1 \tag{1}$$

$$b(a+d)=2 \tag{2}$$

$$c(a+d)=3 \tag{3}$$

$$d^2+bc=4 \tag{4}$$ From $(2)$ and $(3)$ we get

$$\frac{b}{c}=\frac{2}{3}$$

Let $b=2k $ and $c=3k$ Then

$$a^2=1-6k^2$$ and $$d^2=4-6k^2$$ So

$$B=\begin{bmatrix} \sqrt{1-6k^2} &2k\\ 3k&\sqrt{4-6k^2} \end{bmatrix}$$

So

$$B^2=\begin{bmatrix} 1 &2k \left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right) \\ 3k \left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right)&4 \end{bmatrix}=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$$

Now we have to solve for $k$ using equation

$$ \left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right)=\frac{1}{k} \tag{5}$$ Also

$$\left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right) \times \left(\sqrt{4-6k^2}-\sqrt{1-6k^2}\right)=3$$ So from $(5)$

$$\left(\sqrt{4-6k^2}-\sqrt{1-6k^2}\right)=3k \tag{6}$$

Subtracting $(6)$ from $(5)$ we get

$$2\sqrt{1-6k^2}=\frac{1}{k}-3k$$ Squaring Both sides we get

$$33k^4-10k^2+1=0$$ we get

$$k^2=\frac{5 \pm i \sqrt{8}}{33}$$ From this we get $k$. is there any other simpler way to find ?

Answer 1

@ Umesh shankar, this is exactly what should not be done (if you want to make progress)!!

Note that $A^2=trace(A)A-\det(A)I=5A+2I$. Since $B^2=A$ we deduce that $A,B$ commute and since $A$ is not a scalar matrix, $B$ is in the form $B=aI+bA$. Then $(aI+BA)^2=A$, that is $a^2I+b^2(5A+2I)+2abA=A$, that implies

$$a^2+2b^2=0\;,\;5b^2+2ab=1.$$

We deduce $33b^4-10b^2+1=0$,....... and, finally, the four solutions.

EDIT.

Recall that the determinant and the trace of a matrix do not change after a change of basis.

Proposition 1. If $A\in M_2$ is not a scalar matrix, then $I,A$ is a basis of the vector space $E=\{X|AX=XA\}$.

Proof. With a change of basis, we may assume that $A=\begin{pmatrix}0&a\\1&b\end{pmatrix}$ (there is $u$ s.t. $u,Au$ is a basis). Let $X=\begin{pmatrix}x&y\\z&t\end{pmatrix}$. Then $AX=XA$ is equivalent to $y=az,x+bz=t$, that is $2$ linear independent linear relations; thus $dim(E)=4-2=2$.

Proposition 2. If $A\in M_2$ is not a scalar matrix, then $A^2=trace(A)A-\det(A)I$.

Proof. As above, we may assume that $A=\begin{pmatrix}0&a\\1&b\end{pmatrix}$. $A^2=\begin{pmatrix}a&ab\\b&a+b^2\end{pmatrix}=bA+aI=trace(A)A-\det(A)I$.

Answer 2

Just variation of the loup blanc's answer...

Interesting it would be also to check whether it is possible to use

$B^2-\text{trace}(B)B+\det(B)I=0$ ?

Indeed $B^2+\det(B)I= A\pm\sqrt{\det(A)}I=\text{trace}(B)B=tB $

If we name $ C=A\pm\sqrt{\det(A)}I$ ( $C$ can be directly calculated) then $\text{trace}(C/t)=t$, hence $t^2=\text{trace}(C )$.

Also $4$ solutions.

Answer 3

Eliminate $bc$ from the first equation (the uppercase letters are the RHS),

$$a^2+\frac{BC}{(a+d)^2}=A,$$ and from the first and the last, $$d^2-a^2=D-A.$$

Let $p:=(d+a)/2,m:=(d-a)/2$, giving

$$(p-m)^2+\frac{BC}{4p^2}=A,\\mp=4(D-A)$$

and multiplying by $4p^2$,

$$4(p^2-mp)^2+BC=4(p^2-4(D-A))^2+BC=4Ap^2.$$

This biquadratic equation will give the values of $p$, then $m, a$ and $d$, followed by $b$ and $c$.

Answer 4

Square roots suck, so I'll do this with polynomials instead. Any matrix $B$ with $B^2=A$ must commute with $A$. Since there exist vectors $e$ such that $(e,Ae)$ form a basis of $\Bbb R^2$ (in fact the first standard basis vector will do, as will almost any other vector), any matrix commuting with $A$ must be a polynomial in$~A$, as shown for instance elementarily in this answer.

Since $A^2=2I_2+5A$ (check it) any polynomial in$~A$ can be rewritten to a polynomial of degree${}<2$ in$~A$. So let $B=sI_2+tA$ then $B^2=s^2I_2+2stA+t^2A^2=(s^2+2t^2)I_2+(2st+5t^2)A$. You need that to be equal to $A$ and since $I_2$ and $A$ are linearly independent, it will hold if and only if $s^2+2t^2=0$ and $2st+5t^2=1$. Clearly there are no solutions over $\Bbb R$, so if the question was to find a real square root of$~A$, the answer is there are none.

Working over the complex numbers, we can however continue. We must have $t\neq0$, and then the first equation says $(s/t)^2=-2$. First take $s/t=\sqrt2\,\mathbf i$, then the second equation becomes $\sqrt8\,\mathbf it^2+5t^2=1$ or $t^2=\frac{5-\sqrt8\,\mathbf i}{33}$. Taking complex square roots gives you two solutions. Taking instead $s/t=-\sqrt2\,\mathbf i$ one gets $t^2=\frac{5+\sqrt8\,\mathbf i}{33}$ with again two solutions. These appear to be the ones you found in the question. I don't think it will clarify anything to detail these solutions any further (it just becomes a mess, though it will work if you persevere).