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Notation: (A|b) denotes the equation system $Ax=b$

True or false?

Let $K$ be a field, $n\in \mathbb{N}$. Furthermore let $A \in M(n \times n, K)$ a $n \times n$ matrix over $K$ and $b \in K^n$.

a) If the equation system $(A | b)$ has exactly one solution, then for all $c \in K^n$ the equation system $(A | c)$ has exactly one solution.

b) If the equation system $(A | b)$ has no solution, then there exists a $c \in K^n$ such that the equation system $(A | c)$ has more than one solution.

c) If the equation system $(A | b)$ has more than one solution, then the set of all $ c \in K^n$ such that $(A | c)$ has more than one solution forms a true subspace of $K^n$.

d) Let $K$ be a finite field with more than two elements. If the equation system $(A | b)$ has no solution, then the cardinality of the set $M=\{c ∈ K^n |$ the equation system $(A|c)$ has no solution $\}$ is bigger than the cardinality of its complement, $K^n \setminus M$.


My solutions:

a) True. If $(A | b)$ has exactly one solution, then $(A | 0)$ only has got the trivial solution $x=0$. Thus the nullspace is trivial, $A$ is injective and surjective. Thus for every $c \in K^n$ there exists exactly one solution for $(A | c)$.

b) True. If $(A | b)$ has no solution, there exist linearly dependant rows in $A$, thus the rank of $A$ is not full and there exists a non-trivial solution for $(A | 0)$. Given a particular solution for $(A | c)$, the sum of the particular solution and the non-trivial solution for $(A | 0)$ also are a solution for $(A | c)$.

c) True. If $(A | b)$ has more than one solution, then $(A | 0)$ has got the trivial solution $x=0$ and another non-trivial one.

If $(A | b)$ and $(A | b')$ both have more than one solution, then $(A | b+b')$ has got more than one solution, namely the sum of the particular solutions and the trivial solutions.

If $(A | b)$ has got more than one solution, then $(A | \alpha b)$ for $\alpha \in K$ has got more than one solution, namely multiples of $\alpha$ and the particular and the trivial solutions$.

I am not sure about those solutions, especially about b). I also do not know how to tackle d). Any corrections and hints are welcome.

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1 Answer 1

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Part (d), let $|K|=q>2$.

The set $S=K^n\backslash M$ of all $c\in K^n$ such that $(A\mid c)$ has a solution forms a subspace of $K^n$. Then $|S|=q^k$ for some $k< n$ since $b\notin S$. Thus, $|M|=q^n - q^k$. Since $q^n > 2q^{n-1} \geq 2q^k$, we have $$ |M|=q^n - q^k > q^k = |S|=|K^n\backslash M|. $$

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