Let $F$ be a field such that 1+1 is not equal to zero. Let $A$ element of $M_n(F)$ such that $A^T = -A$. a) Show that if $A$ is non-singular, then $n$ is even. b) Show that if $A$ is singular, then rank$A$ is even.
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We have $$\det(A)=\det(A^T)=\det(-A)=(-1)^n \det(A)$$ If $\det(A)\neq 0$, what can you deduce ? Why do you need that $2\neq 0$ ? The second question also has answers on MSE, e.g., see here.
Dietrich Burde
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Let $A \in M_{n}(F)$ and let $a_{ij}$, with $a_{ij}=-a_{ji}$ , be the entries of A.
Observe that $a_{ii}=0$
a) Now we construct A entry by entry. Let's start with $a_{ij}$ ($i\neq j$). If we add $a_{ij}$ we get rowrank 2.
Now we show that we add an element $a_{st}$ with i=s.
It's easy to see that the rowrank doesn't change. Because we can cancel $a_{st}$ and $a_{ts}$ out with one row-addition transformation.
Similar with $a_{st}$ with j=t.
Only if we add $a_{st}$, with $i\neq s$ and $ j\neq t$ for all i,j with $a_{ij}$ allready added, we get a rowrank change.
The change is +2 (one for $a_{st}$ and one for $a_{ts}$)
Now let's take a smart way to show a)
det(A)=det(A$^\top$)=det(-A)=(-1)$^{n}$det(A)
First and third equation by property of the determinant.
Second equation by assumption
Now we have det(A)=(-1)$^{n}$det(A) and thats only possible if n is even.
Next we have that a non-singular Matrix has complete rank.
Remark: If A is singular, then rank A is < n
The first prove of a) proves b) too.
Manuel G
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