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

  • 11
    $\begingroup$ $A$ 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). $\endgroup$ Jul 20 '11 at 9:26
  • $\begingroup$ 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. $\endgroup$ Jul 20 '11 at 11:30
  • $\begingroup$ Approach is not clear to me. Will you kindly explain clearly? $\endgroup$
    – user12290
    Jul 22 '11 at 4:22

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.


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.

  • $\begingroup$ Ok, the eigenvalues of A are purely imaginary. But from this info does it not follow that A isn't diagonalizable, since A is a real matrix? And isn't the property : "the determinant of a matrix is the product of its eigenvalues" appliable only for a diagonizable matrix? $\endgroup$ Dec 7 '16 at 23:55
  • $\begingroup$ @math.h, no, it is not only applicable for diagonizable matrices. $\endgroup$
    – Calle
    Dec 8 '16 at 7:19

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.


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.

  • $\begingroup$ Looks like Olivier Bégassat has a simpler idea. I'll let this sit here, though. $\endgroup$ Jul 20 '11 at 9:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.