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Let $A$ be a real skew symmetric matrix. Prove that $I+A$ is non-singular, where $I$ is identity matrix.

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$A$ cannot have any non zero real eigenvalues : suppose $AX=\lambda X$ for some vector $X$ and real $\lambda$, then $$\lambda |X|^2=\langle AX|X\rangle=\langle X|A^*X\rangle=-\langle X|AX\rangle=-\lambda |X|^2.$$ Thus, either $\lambda=0$ or $X=0$. This in turn implies that $I+A$ has to be non singular (by injectivity). –  Olivier Bégassat Jul 20 '11 at 9:26
    
If $\mathbf A$ is skew-symmetric, $i\mathbf A$ is Hermitian and has real eigenvalues, so $\mathbf A$ will have pure imaginary eigenvalues or one zero eigenvalue for odd-order matrices. –  J. M. Jul 20 '11 at 11:30
    
Approach is not clear to me. Will you kindly explain clearly? –  user12290 Jul 22 '11 at 4:22

4 Answers 4

up vote 3 down vote accepted

Let $\lambda \neq 0$ be an eigenvalue of $A$ with eigenvector $x$. Then:

$$x^* A x = x^*(\lambda x) = \lambda x^* x$$

Where $x^*$ is the hermitian adjoint. Now, since $A$ is real, $A^* = A^T$ and we get $(x^* A x)^* = x^* A^T x = -x^* A x$. And since $(x^* A x)^* = (\lambda x^* x)^* = \lambda^* x^* x$. Putting these two equations together yields:

$$x^* A x = - \lambda^* x^* x$$

But since we have the same vector $x$, we have $-\lambda^* = \lambda$. Now, say $\lambda = a + ib$, so $-\lambda^* = -a + ib$. Thus we get $\lambda = ib$, i.e. $\lambda$ is pure imaginary.

Now, say we have an eigenvalue $\lambda = ib$ with eigenvector $x$, $Ax = \lambda x$. This implies $(Ax^*) = \lambda^* x^*$ and $(Ax^*) = x^* A^* = x^* A^T = - x^* A$, so $-x^* A = \lambda^* x^*$ Take the tranpose of both sides $(-x^* A)^T = -A^T \overline{x} = A \overline{x}$ and $(\lambda^* x^*)^T = \lambda^* \overline{x}$, where $\overline{x}$ is the complex conjugate of $x$. We have reached:

$$A \overline{x} = \lambda^* \overline{x}$$

Thus, $\lambda^* = -ib$ is an eigenvalue to $A$ with eigenvector $\overline{x}$.

So, all non-zero eigenvalues of a real skew symmetric matrix are pure imaginary and come in pairs $\lambda$ and $-\lambda$.

Now, let $\lambda$ be a (possibly zero) eigenvalue of $A$ with eigenvector $x$. From this eigenvalue we get an eigenvalue for $I + A$:

$$(I+A)x = Ix + Ax = x + \lambda x = (1+\lambda)x$$

since $\lambda$ is pure imaginary or zero, $1 + \lambda$ will always be non-zero. Since the determinant of a matrix is the product of its eigenvalues, we have that $\det(I+A) \neq 0$ and we can even deduce that $\det (I+A)$ is real and positive (since $(1 + ib)(1-ib) = 1 + b^2$). Hence $I+A$ is always invertible.

Just a note: If $A$ is $n \times n$ with $n$ odd, $A$ will always have a zero eigenvalue, since $$\det A = \det A^T = \det (-A) = (-1)^n \det A$$ if $n$ is odd we have $\det A = - \det A$ which implies $\det A = 0$, which implies that at least one eigenvalue is zero.

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As $A$ is skew symmetric, if $(A+I)x=0$, we have $0=x^T(A+I)x=x^TAx+\|x\|^2=\|x\|^2$, i.e. $x=0$. Hence $(A+I)$ is invertible.

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Hint 1: Computer $(I+A)(I+A^T)$.

Hint 2: Remember that for all real matrices $A$, the symmetric matrix $AA^T$ gives a positive semidefinite quadratic form.

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Looks like Olivier Bégassat has a simpler idea. I'll let this sit here, though. –  Jyrki Lahtonen Jul 20 '11 at 9:32

Here is a proof using the geometric meaning of a skew-symmetric matrix in 3D space.

For an arbitrary vector $a=[a_1,a_2,a_3]^T\in\mathbb{R}^3$, there exists a skew-symmetric matrix $$[a]_{\times}=\left( \begin{array}{ccc} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \\ \end{array} \right).$$

Conversely, given an arbitrary skew-symmetric matrix $A$, there is an associating vector $a$ such that $A=[a]_{\times}$.

Given a skew-symmetric matrix $A$ and a vector $x$, we have $Ax=a\times x$. Since $a\times x \perp kx, \forall k\ne 0$, it is impossible $Ax=kx, \forall k\ne 0$. Therefore, $kI+A, \forall k\ne 0$ is nonsingular.

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