Orthogonality of skew-symmetric matrix

Question

Let $S$ be a real skew-symmetric matrix. Prove that $I-S$ and $I+S$ are orthogonal, where $I$ is $n \times n$ identity matrix.

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My approach: I am not sure whether the question is asking to prove I-S and I+S orthogonal individually or together. In either case I can't get the answer.

In first case it should be (I-S)'(I-S)=(I-S^2) which I don't think is equal to I.

In second case if we were to prove both are orthogonal to each other in my view we need to check X'Y=0 i.e. (I-S)'(I+S)=0. Which means (I+S)^2=0.

Can some one throw light on this question as in what the examiner is willing to ask and what is the solution.

Thanks.

Answer 1

The statement is false in general. Take $S = \begin{pmatrix} 0 & - 1 \\ 1 & 0 \end{pmatrix}$. Then

$$I + S = \begin{pmatrix} 1 & - 1 \\ 1 & 1 \end{pmatrix}, \qquad I - S = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}, $$

and both $I + S$ and $I - S$ are not orthogonal. Note that there is a result that $(I - S)(I + S)^{-1}$ is orthogonal, see here.

Answer 2

The two matrices clearly aren’t orthogonal matrices in general, since $S^TS$ doesn’t vanish just because $S$ is skew-symmetric. So, you’re probably meant to show that the inner product of $I+S$ and $I-S$ vanishes, but to do that you have to actually use an inner product. The inner product of two vectors must product a scalar, so $(I-S)^T(I+S)$ or some other product of these two matrices and their transposes isn’t the right thing to use: the product of two matrices is another matrix, not a scalar.

No doubt an inner product of square matrices was defined somewhere in the material leading up to this exercise, so you should go back over that and see what it is. However, a common one for real matrices is $\langle A,B\rangle = \operatorname{tr}(B^TA)$, which corresponds directly to the dot product of elements of $\mathbb R^n$. So, using this inner product, you would have to show that $\langle I-S,I+S\rangle = \operatorname{tr}\left((I+S)^T(I-S)\right) = 0$ when $S$ is skew-symmetric. (It turns out that this is false for this particular inner product, as you can verify by direct calculation with $2\times2$ matrices, so some other inner product must be intended.)