If $A$ is a skew-symmetric Matrix then prove that $I-A$ is non singular, where $I$ is the identity Matrix of the order equal to that of $A$.
I know that this question can be done by using the concept of eigenvalues. But I want a solution that does not involve that. I have one solution but it is like very difficult to think.
My approach
We see that the matrix $(I-A)(I+A)^{-1}$ is orthogonal. Hence we can say that $\frac{\det(I-A)}{\det(I+A)}=1$ so $\det(I-A)\ne0$
This solution is nice but hard to think.
Second approach (Solution given in book)
Consider a homogeneous system of equation $(I-A)X=0$ $\implies AX=X$
Now, $$(AX)^{T}=X^T$$ $$\implies -X^TA=X^T$$ Now post multiplying by $X$ gives, $$X^TX=0$$ now the book says $\implies X=0$ so $(I-A)X$ has only trivial solution. And hence $\det(I-A)\ne0$
I cannot understand how they concluded that $X$ is $0$. How did they just cancel $X^T?$
I want you guys to give nice solutions of this problem and if possible pls explain the last step in the second solution.
Any help is greatly appreciated.
The statement in question is false if the elements of $A$ are taken from a general field. E.g. the complex matrix $$ I_2-\pmatrix{0&i\\ -i&0}=\pmatrix{1&-i\\ i&1}=\pmatrix{1\\ i}\pmatrix{1&-i} $$ is singular.
The author of your book has probably implicitly assumed that the underlying field is real. First, observe that when $A$ is a skew-symmetric matrix over a field of characteristic $\ne2$, we must have $X^TAX=0$ for every vector $X$, because \begin{align*} X^TAX &=(X^TAX)^T\quad\text{(as $X^TAX$ is a $1\times1$ matrix)}\\ &=X^TA^TX^T\\ &=-X^TAX^T.\\ \end{align*} (We actually also have $X^TAX=0$ in a field of characteristic $2$, if the definition of skew-symmetric matrix requires that the matrix to have a zero diagonal.)
In particular, $X^TAX=0$ when the field is $\mathbb R$. Now, if $(I-A)X=0$, then $X=AX$. In turn, $\sum_ix_i^2=X^TX=X^TAX=0$. Since all $x_i$s are real numbers, each of them must be zero. Hence $X$ is necessarily zero and $I-A$ is nonsingular.