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Testing a method with the use of C.-H. theorem for finding square roots of real $ 2 \times 2 $ matrices I have noticed that some matrices probably don't have their square roots with real and complex entries.

An example the matrix $A= \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$.

However is it at all a proof that it is impossible to extend somehow the field of entries in order to satisfy equation $B^2=A$ similar to the situation when many years ago solution of $a^2=-1$ seemed to be impossible to solve for real numbers hence imaginary numbers $i$ were introduced ?

Is it possible to devise such numbers (...quaternions? octonions ? or others..) that $B^2=A$ would be however satisfied ?

Additionally, when we are sure in general case for $n \times n$ matrices that a square root exists if we are free to vastly extend a field?

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Two partial answers to your question make a full answer!

Let $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$.

  • Answer 1: The set of matrices form a ring (rings are sets where you have algebraic operations of addition and multiplication). In abstract algebra, one learns about ring extensions, in other words, you can construct a bigger ring which contains the ring of matrices, but your given matrix $A$ has a square root. This is easier to do in a commutative ring, but matrix multiplication is not commutative.

    In this case, let $M_{2,\mathbb{R}}$ be the set of $2\times 2$ matrices with coefficients in $\mathbb{R}$, and we consider elements which are sums of matrices times $B$. In other words, you have sums whose terms are like $$ BM_1BM_2BM_3B\cdots M_kB $$ with or without the leading $B$'s. The one extra condition is that $B^2=A$.

    This gets complicated, but does include a square root of $B$. The problem is that $B$ is not a matrix, it's just an extra element in the ring that acts like the square root $A$.

  • Answer 2: Suppose that $B$ must be a matrix (and we're working in a field of characteristic $0$). Then, we have the situation $$ \begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}a^2+bc&ab+bd\\ac+cd&bc+d^2\end{bmatrix}. $$ Let's start with the lower left corner. We have that $c(a+d)=0$, so either $c=0$ or $a=-d$.

    • Let's start with the case where $c=0$. In this case, the matrix on the RHS simplifies to $$ \begin{bmatrix}a^2&ab+bd\\0&d^2\end{bmatrix} $$ Since the upper left and lower right corners are also $0$, $a^2=0$ and $d^2=0$, so $a=0$ and $d=0$. But this makes the upper right corner $0$ as well, a contradiction.

    • Suppose now that $a=-d$, but then the upper right corner is $b(a+d)=0$, which is also not possible.

    Therefore, if $xy=0$ implies that $x=0$ or $y=0$, then there is no way to write $A$ as a square of a matrix no matter what field you work with (assuming in our field that $0\not=1$).

Concluding remark: If you work with matrices over a commutative ring which has zero divisors (so not an integral domain), then it may be possible to find a matrix which is the square root of $A$.

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  • $\begingroup$ "acts like the square root A", but not the matrix? What else can be said about this element? Can we make on it some operations as if it would be matrix? $\endgroup$ – Widawensen Jun 1 '17 at 11:37
  • $\begingroup$ I'm finding your questions confusing, could you please elaborate. $\endgroup$ – Michael Burr Jun 1 '17 at 11:47
  • $\begingroup$ it concerns the last sentence of answer 1 part. I suppose we can add and multiply this element $B$ with matrices of the proposed ring but can we somehow devise for him also matrix-like constructs, for example, as inverse or determinant ? $\endgroup$ – Widawensen Jun 1 '17 at 11:54
  • $\begingroup$ $B$ would not have an inverse because $B^2=A$, which is not invertible, but you can define a determinant for $B$ using the desired property that $\det(CD)=\det(C)\det(D)$. Technically, one would need to check that the determinant homomorphism is also a homomorphism on this ring, but it will work. $\endgroup$ – Michael Burr Jun 1 '17 at 11:59

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