3

Prove that for any nonsingular complex matrix $A$ and for any positive integer $k$, the equation $X^k = A$ has a solution.

Any tips or solution?

Ashtart
  • 123
  • 4
    What have you solved so far? – SPARSE Nov 24 '22 at 19:55
  • 1
    OP, the question itself is interesting, but it is still missing some things. What have you tried yourself? Also, we need context. Where did you get this problem? Is it a homework question or did it come up somewhere else. If it is homework, what is the course level. This is to e.g., give someone an idea of what techniques will give a useful answer. And in some instances it is to let them know whether this is a homework question where a solution is known, as opposed to an unsolved problem. – Mike Nov 27 '22 at 13:42
  • A proof is given in my answer to a related question. – Ramen Nii-chan Nov 27 '22 at 14:35

2 Answers2

2

If $A$ is normal, then $A$ is unitarily similar to a diagonal matrix, i.e. a unitary matrix $P$ exists s.t. $PAP^{-1}=D$ with $D$ diagonal. If so, then $$ PX^{k}P^{-1} = (PXP^{-1})^{k} = PAP^{-1} = D = \operatorname{diag}(\lambda_1,\ldots,\lambda_n) $$ If for any element $\lambda_i$ on the diagonal of $D$ we take one of its $k$-roots $\mu_i$, and then form $$ T = \operatorname{diag}(\mu_1,\ldots,\mu_n) $$ then $T^{k} = D = PX^{k}P^{-1}$ and then $$ X^{k} = P^{-1}T^{k}P = (P^{-1}TP)^{k} \implies X = P^{-1}TP. $$

Vincenzo Tibullo
  • 11,287
  • Can you explain the first line? I have never come across a unitary similar matrix. And I don't know why you assume A is in normal. – Ashtart Dec 11 '22 at 17:17
  • @Ashtart I assumed $A$ is normal, because only in this case it is diagonalizable and my subsequent reasoning work. On the other hand, I did not assume $A$ is non singular, because it is not required by my calculations. As regards to normal matrix and similarity, see https://en.wikipedia.org/wiki/Normal_matrix and https://en.wikipedia.org/wiki/Matrix_similarity – Vincenzo Tibullo Dec 11 '22 at 17:22
0

Here is more algebraic proof (as compared to my previous attempts which, admittedly, involved some hand waiving). If A is diagonal then this is easy i.e. $A= CD({\lambda}_1,{\lambda}_2,..)C^{-1}$ and one takes $X=CD({\lambda}_1^{1/k},{\lambda}_2^{1/k},..)C^{-1}$ . Now, $A$ has Jordan decomposition into blocks. Some of them are diagonal so this is not a problem due to the above. But some may be of the form "${\lambda}$ on main diagonal $1$ above". To handle this case, write such block in the form ${\lambda}(I + N)$ where $I$ is the identity and $N$ is nilpotent. Assuming the block size is $p$ we have $N^m = 0$ if $m>p$. Since the function $(1+x)^{1/k}$ is analytic in the neighborhood of $0$ we have $(1+x)^{1/k} = a_1 + a_2x + ...a_nx + ...$. Plugging in for $1$ the identity matrix and for $x$ the $N$ matrix one gets $(I + N) = (a_1I + a_2N + ... a_{p_1}N^{p-1} + ...)^k$. But the sum on the right is finte, i.e. $(I + N) = (a_1I + a_2N + ... a_{p_1}N^{p-1})^k$ Lastly, since $\lambda$ is not $0$ divide by $p$ root of it.

Salcio
  • 2,515
  • 2
    Why is $X_\epsilon$ bounded? – user1551 Nov 25 '22 at 05:59
  • Look at the formula for $X_{\epsilon}$ this is the product of $C_{\epsilon}$, $D$, and $C_{\epsilon}^{-1}$. If each one is bounded then the product is bounded. Now, $D$ is bounded. So is $C_{\epsilon}$ since it consist of eigenvalues of $A_{\epsilon}$. Lastly, $C_{\epsilon}^{-1}$ is bounded by the formula for inverse matrix, that is it contains some expression of entries of the original matrix dived by determinant. And determinant is separated from $0$. – Salcio Nov 25 '22 at 13:07
  • Is this the only way? I understand the first part, but not with epsilon. – Ashtart Nov 25 '22 at 15:21
  • 7
    Sorry, I don't find your reasoning convincing. E.g. for $A=\pmatrix{1&1\ 0&1}$ and $A_\epsilon=\pmatrix{1&1\ 0&1-\epsilon}$, we have $$ A_\epsilon=C_\epsilon D_\epsilon C_\epsilon^{-1} =\pmatrix{1&-\epsilon^{-1}\ 0&1}\pmatrix{1&0\ 0&1-\epsilon}\pmatrix{1&\epsilon^{-1}\ 0&1}. $$ Although $X_\epsilon=C_\epsilon D_\epsilon^{1/k} C_\epsilon^{-1}$ is bounded in this example, $C_\epsilon$ and $C_\epsilon^{-1}$ are unbounded when $\epsilon\to0$. – user1551 Nov 25 '22 at 15:28
  • @user1551 - well, one can always normalize eigenvectors i.e. assume that the have length 1 say. What you are doing is, you take an eigenvector $x$ and them claim that ${\epsilon}^{-1}x$ is also eigenvector. Everything cancels because $\epsilon$ comes back in inverse but this is redundant. Though, a good point. – Salcio Nov 25 '22 at 21:14
  • 1
    This is not as easy as you might think. If you scale a column of $C_\epsilon$ by a factor $k$, you must scale the corresponding row of $C_\epsilon^{-1}$ by $k^{-1}$. While you can bound $C_\epsilon$ by normalisation, it is unclear why you may bound $C_\epsilon$ and $C_\epsilon^{-1}$ simultaneously. E.g. the $C_\epsilon$ in my previous comment becomes $\pmatrix{1&-\epsilon^{-1}/\sqrt{1+\epsilon^{-2}}\ 0&1/\sqrt{1+\epsilon^{-2}}}$ after normalising its columns, but $C_\epsilon$ then becomes $\pmatrix{1&\epsilon^{-1}\ 0&\sqrt{1+\epsilon^{-2}}}$, which remains unbounded when $\epsilon\to0$. – user1551 Nov 26 '22 at 05:21
  • I think it is not that difficult to see ... determinant of $C$ is no-zero, say its absolute value is bigger than $\delta > 0$. Also, entries of $C_{\epsilon}$ converge to entries of $C$ so does determinant. In words, determinant of $C_{\epsilon}$ is bigger than ${\delta}/2$ for $\epsilon$ small enough. It means that entries of $C_{\epsilon}^{-1}$ are bounded since they come from operations on entries of $C_{\epsilon}$ divided by determinant of $C_{\epsilon}$. – Salcio Nov 26 '22 at 23:43
  • What is $C$? If $A$ is not diagonalisable, there is no $C$ for $C_\epsilon$ to converge to! Please take a look at my counterexample again. Both $C_\epsilon=\pmatrix{1&-\epsilon^{-1}\ 0&1}$ and $C_\epsilon^{-1}=\pmatrix{1&\epsilon^{-1}\ 0&1}$ have unit determinants, but both of them diverge when $\epsilon\to0$. – user1551 Nov 27 '22 at 05:36
  • $C$ is the matrix of normalized eigenvectors of $A$. Sorry for this. – Salcio Nov 27 '22 at 13:23
  • @Salcio $C$ does not exist if $A$ is not diagonalizable. (You don't need to mess with $C_\epsilon$ in the first place if $A$ is diagonalizable.) – Ramen Nii-chan Nov 27 '22 at 14:43
  • Added explanation of why entries of $X_{\epsilon}$ are bounded to the proof. – Salcio Nov 27 '22 at 15:08
  • I'm not convinced. I wonder if a canonical form approach would help alleviate the issues here. – Cameron L. Williams Nov 27 '22 at 15:44
  • 1
    My counterexample has already shown that your explanation does not work. When your theory does not fit reality, please look at the reality. – user1551 Nov 27 '22 at 16:12