Proving Invertibility of Matrices

Question

Given that $A$ is an $n×n$ skew symmetric matrix $I$ being the $n×n$ identity matrix, prove that $A − I$ and $A + I$ are invertible. Can trace of $A$ be non-zero number ?

Any suggestions how to go about these proofs. I understand that the trace of $A$ must be $0$ to satisfy skew symmetry but I am unsure how to go about the proofs for $A - I$ and $A + I$. Any help is appreciated!

Answer 1

If $A- \lambda I$ were not invertible, then there would exist some $v \ne 0$ such that $(A-\lambda I)v=0$ i.e. $Av=\lambda v$. Then $$\lambda v^\top v = v^\top A v = (A^\top v)^\top v = - (Av)^\top v = - \lambda v^\top v,$$ which implies $\lambda = 0$. [All quantities above are assumed to be real.]

Answer 2

If $A$ is not a real skew-symmetric matrix, $A+E$ coule be non-invertible. Ex. Given that $$A= \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix} $$ It is easy to find that $det(A+E)=0$.

Answer 3

an amusing way to do this is to prove both at the same time:

I assume scalars in $\mathbb R$, with real constant $c\neq 0$

$\det\Big(A+cI\Big)^2$
$\det\Big(\big(A+cI\big)^2\Big)$
$ = \det\big(A+cI\big)\det\big(A+cI\big) $
$= \det\big(A+cI\big)^T\det\big(A+cI\big) $
$= \det\big(A^T+cI\big)\det\big(A+cI\big)$
$= \det\Big(\big(A^T+cI\big)\big(A+cI\big)\Big)$
$= \det\Big(A^TA + cA^T + cA + c^2I\Big)$
$= \det\Big(A^TA - cA + cA + c^2I\Big)$
$= \det\Big(A^TA + c^2I\Big)$
$\neq 0$

where the final line makes use the the fact that a positive semidefinite matrix ($A^TA$) plus a positive definite matrix ($c^2I$) gives a positive definite matrix which is necessarily nonsingular (and in fact has a positive determinant).

post script:
I take some issue with the idea "they want us to prove this using basic properties of matrices" because in some sense that is false. We are using some basic properties that we get with scalars in $\mathbb Q$ and $\mathbb R$ (and with a little care, an obvious generalization in $\mathbb C$), but these don't automatically hold in other fields. The proof above explicitly called on positive definiteness to finish.

Bad behavior comes with scalars in $\mathbb F_2$ where e.g.
$A = \begin{bmatrix} 0&1&1\\ -1&0&1\\ -1&-1&0\\ \end{bmatrix}= \begin{bmatrix} 0&1&1\\ 1&0&1\\ 1&1&0\\ \end{bmatrix}$
is skew-symetric but $A+I = A-I$ and both have rank one, hence are singular.

Slightly more pecuiliarly, in $\mathbb F_2$
$A=\begin{bmatrix} 1&1&1\\ 1&1&1\\ 1&1&1\\ \end{bmatrix}$
is also skew symmetric, not traceless, and $A+I = A-I$ which has rank 2, hence is still singular.

Answer 4

HINT: $A$ real and skew-symmetric implies $i A$ is hermitian, and so all the eigenvalues of $i A$ are real. But that means that all the eigenvalues of $A$ are in $i \mathbb{R}$ ( and are $0$ or in conjugate pairs). Therefore, the eigenvalues of $I \pm A$ are $1$ or conjugate pairs $1 \pm i \lambda$. Thus $\det (I\pm A)\ge 1$, and so $\ne 0$.

Answer 5

Since $A+I=-(A-I)^T$, we only need to consider $A-I$. In general:

If $\operatorname{char}(F)\ne2$ and $v^Tv\ne0$ for every nonzero vector $v\in F^n$, then $A-I$ is nonsingular for every skew-symmetric matrix $A\in M_n(F)$.

The proof of the above statement can be found in angryavian's answer. Essentially, suppose $A$ is skew-symmetric and $(A-I)v=0$ for some vector $v$. Then $Av=v$ and $v^TAv=v^Tv$. Taking transposes on both sides and using $A^T=-A$, we also have $-v^TAv=v^Tv$. Therefore $v^Tv\,(=v^TAv)=-v^Tv$. Since $\operatorname{char}(F)\ne2$, we must have $v^Tv=0$. Hence $v=0$ and $A-I$ is nonsingular.

Note that this proof does not involve any inner product. It is valid, for instance, when $n=2$ and $F=GF(3)$.

The converse of the above statement is (basically) also true:

Unless $n=1$, if $\operatorname{char}(F)=2$ or $v^Tv=0$ for some nonzero vector $v\in F^n$, there always exists some skew-symmetric matrix $A$ such that $A-I$ is singular:

• If $n\ge2$ and $\operatorname{char}(F)=2$, consider $A=\pmatrix{0&1\\ 1&0}\oplus0$.
• If $n\ge2$ and $v^Tv=0$ for some $v\ne0$, pick any vector $u$ such that $u^Tv=1$. Let $A=vu^T-uv^T$. Then $Av=vu^Tv-uv^Tv=v$ and hence $A-I$ is singular.