# Eigenvalues of Real Skew-Symmetric Matrix are Zero or Purely Imaginary and the Rank is Even

## Problem 269

Let $A$ be a real skew-symmetric matrix, that is, $A^{\trans}=-A$.

Then prove the following statements.

**(a)** Each eigenvalue of the real skew-symmetric matrix $A$ is either $0$ or a purely imaginary number.

**(b)** The rank of $A$ is even.

Contents

## Proof.

### (a) Each eigenvalue of the real skew-symmetric matrix $A$ is either $0$ or a purely imaginary number.

Let $\lambda$ be an eigenvalue of $A$ and let $\mathbf{x}$ be an eigenvector corresponding to the eigenvalue $\lambda$. That is, we have

\[A\mathbf{x}=\lambda \mathbf{x}.\]
Multiplying by $\bar{\mathbf{x}}^{\trans}$ from the left, we have

\begin{align*}

\bar{\mathbf{x}}^{\trans}A\mathbf{x}=\lambda \bar{\mathbf{x}}^{\trans} \mathbf{x}=\lambda ||\mathbf{x}||^2. \tag{*}

\end{align*}

Note that the left hand side $\bar{\mathbf{x}}^{\trans}A\mathbf{x}$ is the dot (inner) product of $\bar{\mathbf{x}}$ and $A\mathbf{x}$. Since the dot product is commutative, we have

\begin{align*}

&\text{The left hand side of (*)}\\

&=\bar{\mathbf{x}}^{\trans}A\mathbf{x}=(A\mathbf{x})^{\trans}\bar{\mathbf{x}}\\

&=x^{\trans}A^{\trans}\bar{\mathbf{x}}.

\end{align*}

Since $A$ is skew-symmetric, we have $A^{\trans}=-A$. Substituting this into the above equality, we have

\begin{align*}

&\text{The left hand side of (*)}\\

&=x^{\trans}A^{\trans}\bar{\mathbf{x}}=-\mathbf{x}^{\trans}A\bar{\mathbf{x}}

\end{align*}

Taking conjugate of $A\mathbf{x}=\lambda\mathbf{x}$ and use the fact that $A$ is real, we have

\[A\bar{\mathbf{x}}=\bar{\lambda}\bar{\mathbf{x}}.\]

Thus, we have

\begin{align*}

&\text{The left hand side of (*)}\\

&=-\mathbf{x}^{\trans}A\bar{\mathbf{x}}\\

&=-\mathbf{x}^{\trans}\bar{\lambda}\bar{\mathbf{x}}=-\bar{\lambda}||\mathbf{x}||^2.

\end{align*}

Therefore comparing the left and right hand sides of (*) yields

\[-\bar{\lambda}||\mathbf{x}||^2=\lambda ||\mathbf{x}||^2.\]
Since $\mathbf{x}$ is an eigenvector, it is nonzero by definition. Thus $||\mathbf{x}||\neq 0$.

Hence we have

\[-\bar{\lambda}=\lambda,\] and this implies that $\lambda$ is either $0$ or purely imaginary number.

(To see this, let $\lambda=a+ib$, where $a, b\in \R$. Then

\[-\bar{\lambda}=-a+ib=a+ib=\lambda\] implies $a=0$, thus $\lambda=bi$.)

Remark: Another similar proof is to take conjugate of (*).

### (b) The rank of $A$ is even

From part (a), we know that the eigenvalues of $A$ are $0$ or purely imaginary.

Thus if $\lambda$ is a purely imaginary eigenvalue of $A$, then its conjugate $\bar{\lambda}=-\lambda$ is also an eigenvalue of $A$ since $A$ is a real matrix.

Thus, nonzero eigenvalues come in pairs $\lambda, -\lambda$ (and their algebraic multiplicities are the same).

Let

\[\lambda_1, -\lambda_1, \lambda_2, -\lambda_2, \dots, \lambda_k, -\lambda_k\]
be nonzero eigenvalues of $A$.

Since a real skew-symmetric matrix is normal, it is diagonalizable (by a unitary matrix).

Thus there exists an invertible matrix $P$ such that

\begin{align*}

&P^{-1}AP=\\

&\diag\begin{bmatrix}

\lambda_1 & -\lambda_1 & \lambda_2 & -\lambda_2 & \dots &\lambda_k &-\lambda_k & 0 &\dots 0

\end{bmatrix},

\end{align*} where $\diag[a_1,a_2,\dots, a_n]$ denotes the $n\times n$ matrix whose diagonal entries are $a_1, a_2, \dots, a_n$ and all the off diagonal entries are zero.

Since $P$ is an invertible matrix, the rank of $A$ is the same as the rank of the diagonal matrix on the right-hand side, which is easily seen to be $2k$.

Thus the rank of $A$ is $2k$, and we have proved that the rank of the real skew-symmetric matrix $A$ is even.

## Related Question.

As an application of this problem, try the following problem.

**Problem 1**. If $A$ is a skew-symmetric matrix, then $I+A$ and $I-A$ are nonsingular matrices, where $I$ is the identity matrix.

For a proof, see the post↴

If $A$ is a Skew-Symmetric Matrix, then $I+A$ is Nonsingular and $(I-A)(I+A)^{-1}$ is Orthogonal.

**Problem 2**. Prove that the determinant of an $n\times n$ skew-symmetric matrix is $0$ if $n$ is odd.

For a proof, check out the post ↴

The Determinant of a Skew-Symmetric Matrix is Zero.

### More Eigenvalue and Eigenvector Problems

Problems about eigenvalues and eigenvectors are collected on the page:

Add to solve later

Sponsored Links

## 3 Responses

[…] of a skew-symmetric matrix are either $0$ or purely imaginary numbers. (See the post “Eigenvalues of Real Skew-Symmetric Matrix are Zero or Purely Imaginary and the Rank is Even” for a proof of this […]

[…] For a proof, see the post “Eigenvalues of Real Skew-Symmetric Matrix are Zero or Purely Imaginary and the Rank is Even“. […]

[…] each eigenvalue of a skew-symmetric matrix is either $0$ or purely imaginary. (See the post “Eigenvalues of Real Skew-Symmetric Matrix are Zero or Purely Imaginary and the Rank is Even“.) Also if the degree of the matrix is odd, it has $0$ as an eigenvalue. (See the post […]