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From a problem set I'm working on: (Edit 04/11 - I fudged a sign in my matrix...)

Let $A(t) \in M_3(\mathbb{R})$ be defined: $$ A(t) = \left( \begin{array}{crc} 1 & 2 & 0 \\ 0 & -1 & 0 \\ t-1 & -2 & t \end{array} \right).$$

For which $t$ does there exist a $B \in M_3(\mathbb{R})$ such that $B^2 = A$?

In a previous part of the problem, I showed that $A(t)$ could be diagonalized into a real diagonal matrix for all $t \in \mathbb{R}$, with eigenvalues $1,-1,t$.

A few things I've thought of:

  • The matrix is not positive-semidefinite, so the general form of the square root does not work. (Is positive-definiteness a necessary condition for the existence of a square root?)
  • Since $A = B^2$, then $\det(B^2) = (\det B)^2 = \det A$. So $\det A \geq 0$ for there to be a real-valued square root, forcing $t \leq 0$ to be necessary.
  • My professor suggested that, since $B^2$ fits the characteristic polynomial of $A$, $\mu_A(x) = (x-1)(x+1)(x-a)$, then the minimal polynomial of $B$ must divide $\mu_A(x^2) = (x^2-1)(x^2+1)(x^2-a) = (x-1)(x+1)(x^2+1)(x^2-a)$. Examining the possible minimal polynomials, one can find the rational canonical form, square it, and check whether the eigenvalues match. This probably could get me the right answer, but I am fairly sure that there is an alternative to a "proof by exhaustion".
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You should include into the formulation of the problem that $t\in\mathbb R$ and B has real values. (You did not write this explicitly, but from what you tried to get the solutions, I am guessing you want to work in reals.) –  Martin Sleziak Apr 11 '11 at 6:18
    
About your first bullet point: You probably meant positive-semidefiniteness? That's not a necessary condition for the existence of a square root -- a counterexample is $$\left(\begin{array}{cc}1&0\\2c&1\end{array}\right)$$ with $|c|>1$, with square root $$\left(\begin{array}{cc}1&0\\c&1\end{array}\right)\;.$$ –  joriki Apr 11 '11 at 7:23
    
About your second bullet point: You probably meant $t\le0$? –  joriki Apr 11 '11 at 7:27
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$B$ cannot be real, as its eigenvalues must be $\pm i$, $\pm 1$, $\pm\sqrt{t}$ (one choice for each $\pm\mathit{something}$). If $B$ were real then its eigenvalues would have to be real or in complex conjugate pairs - your only chance would be for $t=-1$ (I didn't check, so I don't know whether then a real $B$ exists). –  user8268 Apr 11 '11 at 8:11
    
@Martin and @joriki: Thank you, I corrected the problem statement. –  Michael Chen Apr 11 '11 at 14:56

2 Answers 2

up vote 8 down vote accepted

Assume that there exists a real number $t$ and a real matrix $B$ such that $A(t)=B^2$.

Note that $-1$ is an eigenvalue of $A(t)$, hence $A(t)+I=(B-\mathrm{i}I)(B+\mathrm{i}I)$ is singular. This implies that $B-\mathrm{i}I$ or $B+\mathrm{i}I$ is singular. Since $B$ is real valued, this means that both $B-\mathrm{i}I$ and $B+\mathrm{i}I$ are singular. Likewise, $1$ is an eigenvalue of $A(t)$, hence $A(t)-I=(B-I)(B+I)$ is singular. This implies that $B-I$ or $B+I$ is singular. Hence the eigenvalues of $B$ are $\{\mathrm{i},-\mathrm{i},1\}$ or $\{\mathrm{i},-\mathrm{i},-1\}$.

In both cases, $B$ has three distinct eigenvalues hence $B$ is diagonalizable on $\mathbb{C}$. This implies that the eigenvalues of $A(t)$ are $-1$ (twice) and $1$ (once) and that $A(t)$ is diagonalizable as well. Hence $t=-1$. We now look at the matrix $A(-1)$.

One can check that $A(-1)$ is diagonalizable hence $A(-1)$ is similar to a diagonal matrix with diagonal $(1,-1,-1)$. Both $I_1$ (the $1\times1$ matrix with coefficient $1$) and $-I_2$ (the $2\times2$ diagonal matrix with diagonal coefficients $-1$) have square roots: take $I_1$ for $I_1$ and the rotation matrix $\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$ for $-I_2$. Hence $A(-1)$ is a square.

Finally $A(t)$ is a square if and only if $t=-1$.

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I believe I follow. I was a little confused because I showed in a previous part that $A(t)$ is diagonalizable in $M_2(\mathbb{R})$ for all real $t$, and then I realized I missed a sign in my original problem! So I'm trying to continue where you left off. –  Michael Chen Apr 12 '11 at 2:31
    
If $A(-1)$ is diagonalizable, then I can construct a matrix $B$ with eigenvalues $1,i,-i$ such that $B^2 = D$, where $D = P^{-1}AP$ is the diagonalization of $A$. If I use change of basis, then I can find $PBP^{-1}$ and see if that is real. Is that the way to go about doing it? –  Michael Chen Apr 12 '11 at 2:36
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No that is not. The matrix $M=A(-1)$ is defective because the algebraic multiplicity of the eigenvalue $-1$ is $2$ while its geometric multiplicity is $1$. See en.wikipedia.org/wiki/Generalized_eigenspace –  Did Apr 12 '11 at 5:22
    
In the corrected matrix (where the 3,2 entry has value -2, not 2 -- my fault), I believe $A(-1)$ does have a full basis of eigenvectors: $(0,0,1)$ and $(1,-1,0)$ associated with $\lambda = -1$, and $(1,0,1)$ associated with $\lambda = 1$. –  Michael Chen Apr 12 '11 at 5:56
    
See edit. $ $ $ $ –  Did Apr 12 '11 at 6:08

A common exercise is to first diagonalize a matrix with positive eigenvalues, and then find a "square root" for the matrix as you have been asked to do. This is trivial since if $A=PDP^{-1}$ and $B=P\sqrt{D}P^{-1}$ then $A=B^2$ where $\sqrt{D}$, of course, denotes the diagional matrix D after taking the square root of it's entries (which is why we need positive eigenvalues if you are working with real matricies). Thus your question follows from what you have already determained about diagionalizing the matrix.

A not so trivial result is to prove that a matrix with positive eigenvalues has a square root regardless of whether or not it can be diagonalized.

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If I understand you correctly, then the triangular Pascal-matrix is an example for this. You can easily define a squareroot but if you try to diagonalize you get zero-eigenvectors. –  Gottfried Helms Apr 11 '11 at 8:08
    
so if I understand you correctly, there is no square root precisely because the eigenvalues are not nonnegative? –  Michael Chen Apr 11 '11 at 14:53
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This answer is irrelevant: the eigenvalues are not positive. –  Plop Apr 11 '11 at 15:49
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Any invertible matrix has a square root over the complex numbers. A real matrix having a negative eigenvalue with odd multiplicity (or more generally an odd number of Jordan blocks of some size) has no real square root. Some non-invertible matrices have no square root. –  Robert Israel Apr 12 '11 at 2:46

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