I know that we can use the diagonalization of a matrix:

$$\hspace{9.4cm} A = V \Lambda V ^{-1} \hspace{9cm} (1)$$

(where $V$ is the matrix of eigenvectors and $\Lambda$ is the (diagonal) matrix of eigenvalues) to find the $k$'th power of a matrix using the formula:

$$A^k = V \Lambda ^k V ^{-1}$$

It is pretty easy to show that this holds by just multiplying $(1)$ $k$ times.

But can we use the same formula for $k < 1$? For example if I would want to find $A ^ {1/2}$ could I do something like:

$$A ^ {1/2} = V \Lambda ^{1/2} V ^{-1}$$

where each $\lambda^{*}$ (lambda star) from the diagonal of $\Lambda ^{1/2}$ would simply be the square roots of the $\lambda$'s (lambda) from the original eigenvector matrix $\Lambda$. Am I allowed to do that? Why/ Why not?


You can certainly find a square root of the matrix $ A $ this way assuming that the eigenvalues of $ A $ are positive (assuming we're working over $ \mathbb R $), but it may not be the only square root of $ A $ if you only consider a single matrix $ V $ which diagonalizes $ A $. In addition, $ A $ may have a square root despite the fact that its eigenvalues are not positive, in which case this method will fail to find the square root. For instance, if $ R(x) $ represents the rotation matrix by $ x $ radians in $ 2 $ dimensions, then $ R(\pi/2)^2 = R(\pi) $, but $ R(\pi) = -I $ while $ R(\pi/2) $ has no real eigenvalues, and so isn't diagonalizable over $ \mathbb R $ at all.

If the ground field is algebraically closed, then this second problem doesn't arise, and you can assume that any possible square root $ A^{1/2} $ can be put into Jordan normal form, in which case in characteristic $ 0 $ (over a field like $ \mathbb C $, for example) it's easy to show a matrix in Jordan normal form which squares to a diagonal matrix is itself diagonal, and this applies to all higher roots as well. In other words, any square root $ A^{1/2} $ of a diagonalizable matrix $ A $ over an algebraically closed ground field of characteristic $ 0 $ can be represented as $ V \Lambda^{1/2} V^{-1} $, where $ A = V \Lambda V^{-1} $. Of course, different values of $ V $ may still give different square roots - for instance, the identity matrix has many square roots in general, corresponding to involutions of your vector space, but all of them can be obtained by this method.

For irrational powers one has to first define what these even mean, and the usual context in which it makes sense to talk about these is in a small enough neighborhood of the identity matrix in which the $ \exp $ and $ \log $ maps make sense and are well defined. In that case, the formula $ A^r = \exp(r \log(A)) $ defines $ A^r $ for all real exponents $ r $, and in such a situation it's easy to see the diagonalization method will work to compute both the matrix logarithm and the matrix exponential, so the simple formula $ A^r = V \Lambda^r V^{-1} $ is appropriate.

  • $\begingroup$ What about the case where $k = 1/3$ instead of $k = 1/2$? Then I wouldn't have the restriction that the eigenvalues of $A$ have to be positive since we're finding the third root, so then I can just carry on, right? $\endgroup$ – user592938 May 10 '20 at 20:33
  • $\begingroup$ @user1502 You can't just carry on. In that case your method will always give some cube root of the matrix, but as I said it may not be the only one. Consider, for example, that the identity matrix in two dimensions has the cube root $ R(2 \pi/3) $, which has no real eigenvalues. $\endgroup$ – Ege Erdil May 10 '20 at 20:40

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