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Chapter 4: Problem 11

A matrix \(M \in M_{n \times n}(C)\) is called skew-symmetric if \(M^{t}=-M\). Prove that if \(M\) is skew-symmetric and \(n\) is odd, then \(M\) is not invertible. What happens if \(n\) is even?

Short Answer

Expert verified
If the matrix M is skew-symmetric and n is odd, then the determinant of M is zero, which implies that the matrix is not invertible. However, when n is even, we cannot make a definitive statement regarding the invertibility of skew-symmetric matrices, as there are both invertible and non-invertible examples.

Step by step solution

01

Recall the definition of an invertible matrix

To prove that a matrix is not invertible, we need to show that it has no inverse. One property of invertible matrices is that their determinant is non-zero, while for non-invertible matrices, the determinant is zero. We will focus on this property to prove the given statement.
02

Compute the determinant of the matrix

We have a skew-symmetric matrix M, with \(M^t = -M\). To relate the determinant with the symmetry properties of M, we need to compute the determinant of \(M^t\). Let's first take the determinant of both sides of the equation \(M^t = -M\): \[\det(M^t) = \det(-M)\]
03

Use the properties of the determinant

One property of the determinant is that the determinant of a transpose is equal to the determinant of the original matrix: \[\det(M^t) = \det(M)\] Another property of the determinant is that the determinant of a matrix multiplied by a scalar (constant) is the product of the scalar raised to the power of n and the determinant of the matrix: \[\det(-M) = (-1)^n \det(M)\] Now we can replace the determinants in our original equation, \( \det(M^t) = \det(-M) \), using these properties: \[\det(M) = (-1)^n \det(M)\]
04

Utilize the odd/even properties of n

We are given that n is odd, so n = (2k + 1) for some integer k. We can plug this into our equation: \[\det(M) = (-1)^{2k + 1} \det(M)\] This simplifies to: \[\det(M) = -\det(M)\]
05

Prove the non-invertibility of M

Since \( \det(M) = -\det(M) \), we can add \(\det(M)\) to both sides, yielding: \(2 \det(M) = 0\) which gives us \(\det(M) = 0\). Since the determinant of the matrix M is zero, the matrix is not invertible when n is odd.
06

Discuss the case when n is even

When n is even, we cannot conclude that the matrix is not invertible, and in fact, there exist skew-symmetric, invertible matrices with even dimensions. For example, consider the 2x2 skew-symmetric matrix: \[ M= \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \] This matrix is skew-symmetric and has a determinant equal to 1, implying that it is invertible. However, in general, there is no definitive statement on the invertibility of skew-symmetric matrices when n is even, as there can be both invertible and non-invertible examples.

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Most popular questions from this chapter

Write the polynomial \(f(t)=a t^{2}+b t+c\) as a linear combination of the polynomials \(p_{1}=(t-1)^{2}\), \(p_{2}=t-1, p_{3}=1 .\left[\text { Thus, } p_{1}, p_{2}, p_{3} \text { span the space } \mathbf{P}_{2}(t) \text { of polynomials of degree } \leq 2 .\right]\).

Determine whether the following polynomials \(u, v, w\) in \(\mathbf{P}(t)\) are linearly dependent or independent: (a) \(u=t^{3}-4 t^{2}+3 t+3, \quad v=t^{3}+2 t^{2}+4 t-1, \quad w=2 t^{3}-t^{2}-3 t+5\); (b) \(u=t^{3}-5 t^{2}-2 t+3, v=t^{3}-4 t^{2}-3 t+4, w=2 t^{3}-17 t^{2}-7 t+9\).

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