## 2009年5月11日 星期一

### Some linear algebra

These days some of my friends discussed some linear algebra problems with me. I found them quite interesting, so let me post some of them here.

1. If A and B are square matrices of the same size, then the spectrums of AB and BA are the same. More precisely, the generalized eigenvalues of AB and BA over the complex numbers occur with the same multiplicity. This is clear when A is invertible, because then AB and BA are similar. If A is not invertible, then we can approximate A by invertible matrices. The spectrum of AB is just the zeroes of det(xI-AB) (counted with multiplicities). The claim then follows easily. This generalizes the fact that the traces and determinants of AB and BA are the same.

2. If A and B are real square matrices of the same size and if there is a complex matrix U such that $A = U^{-1} B U$, then there is a real matrix M such that $A = M^{-1} B M$. In other words, if two real matrices are similar over the complex numbers, then they are similar over the reals. Proof: Look at the equation MA = BM, where M is a square matrix of the same size as A and B. This is a linear system of equations in the entries of M with real coefficients. Hence if S_c is the complex vector space of complex matrices that satisfy this equation and S_r is the real vector space of real matrices that satisfy the same equation, then the complex dimension of S_c is the same as the real dimension of S_r. Now by assumption this dimension is at least 1. We want to find a real matrix in S_r that is invertible. Let $M_1, \dots, M_k$ be a basis of S_r over the reals. Then this is also a basis of S_c over the complex. By assumption there is a linear combination $z_1M_1 + \dots + z_kM_k$ over the complex numbers whose determinant is non-zero. Hence the polynomial $det(z_1M_1 + \dots + z_kM_k)$ in $z_1, \dots, z_k$ does not vanish identically over $\mathbb{C}^k$. It follows that this polynomial cannot vanish identically over $\mathbb{R}^k$. Hence there is a real linear combination of $M_1, \dots, M_k$ that is invertible. This finishes the proof.

3. This is more or less for my own benefit, to recall the following proof of the well-known fact that any normal matrix is diagonalizable over C. Key: Assume A is normal, i.e. A commutes with A*. Then if x is an eigenvector of A, x is also an eigenvector of A*. In fact if Ax = tx, then
$0 = \langle(A-tI)x, (A-tI)x\rangle = \langle(A^*-\overline{t} I)x, (A^*-\overline{t} I) x\rangle$
when A commutes with A*. This allows us to proceed by induction on the dimension of the vector space: first pick any eigenvector x of A. Then its orthogonal complement is preserved by A, because x is also an eigenvector of A*. Hence we can reduce to one dimension lower, and by induction this completes the proof.

#### 1 則留言:

Alex 提到...

Nice post Polam! In (1), you prove that AB and BA have the same characteristic polynomials by approximating A with a sequence of invertible matrices. I think there is another way of proving this without using limit. Simply observe that the characteristic polynomial of A is
\det(A-tI)=\sum_{i=1}^n(-1)^i\text{tr}(\bigwedge^{n-i}A)t^i
where \bigwedge^i A:\bigwedge^i C\to\bigwedge^i C is given by
\bigwedge^i A(z_1\wedge\cdots\wedge z_i)=Az_1\wedge\cdots\wedge Az_i

and \bigwedge^i(AB)=\bigwedge^i A\bigwedge^i B. So the equality of the characteristic polynomials of AB and BA boils down to the fact that tr(CD)=tr(DC).