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Exercise: Let M be a square complex matrix satisfying M = M^T, and furthermore assume that M is not singular. Prove that there exists a matrix K such that K^2 =
by ogogmad 4y ago
Exercise: Let M be a square complex matrix satisfying M = M^T, and furthermore assume that M is not singular. Prove that there exists a matrix K such that K^2 = M and K = K^T.
Non-solution: You are essentially being asked to show that given an invertible complex-symmetric matrix, you can find a complex-symmetric square root. You might be tempted to use the Taylor series of sqrt(z) centred at some z_0 to obtain K, but this won't work if M's eigenvalues can't be fit inside a half-plane not include the complex number 0. An example of this is when M is
⎡1 0 ⎤
⎢ ⎥
⎣0 -1⎦
Another non-solution: You might attempt to extend the spectral theorem to complex-symmetric matrices, but the only such extension I know of is complicated, and results in a block-diagonal (but not necessarily diagonal) canonical form.
Solution: To actually extend the sqrt function from the complex numbers to matrices, use the Jordan canonical form of M. Then you only need to take the sqrt's of each Jordan block. Note that this sqrt won't exist if a Jordan block is non-zero and nilpotent, so we require M to be non-singular. This obtains K. Finally, observe that there exists a polynomial p such that p(M) = K (which can be obtained via Hermite interpolation), and therefore K^T = p(M)^T = p(M^T) = p(M) = K.
- WoahNoun 4y agoThere's a geometric solution as well! Since M is invertible and symmetric. It acts on R^N by independently scaling the coordinates of an orthogonal basis. Eg, M = O^T*A*O where O is orthogonal and A is a diagonal matrix. Let sqrt(A) be the diagonal matrix with entries equal to the square root of the corresponding entry in A. Then K = O^T*sqrt(A)*O is symmetric and K^2 = (O^T*sqrt(A)*O)*(O^T*sqrt(A)*O). Since O is orthogonal O^T*O = I, and K^2 = O^T*A*O = M.
- ogogmad 4y agoThat doesn't work. It's complex symmetric, but not real symmetric or Hermitian, so you can't use the spectral theorem. This statement here: It acts on R^N by independently scaling the coordinates of an orthogonal basis is precisely the spectral theorem, which you can't use here. [edit] Also, the corresponding claim for Hermitian (and real symmetric) matrices isn't true either. If an eigenvalue of a Hermitian matrix H is negative, then no square root of it can be Hermitian (or real symmetric).
- WoahNoun 4y agoAh I missed the complex part. Clearly a complex matrix doesn't act on R^N by scaling the coordinates.