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Say you have $k$ linear algebraic equations in $n$ variables; in matrix form we write $AX=Y$. Give a proof or a counterexample for each of the following:

a) If $n=k$ there is always at most one solution.

b) If $n > k$ you can always solve $AX=Y$ .

c) If $n > k$ the nullspace of A has dimension greater than zero.

d) If $n < k$ then for some Y there is no solution of $AX=Y$ .

e) If $n < k$ the only solution of $AX=0$ is $X=0$ .

I solved c), answer:

Rank Theorem: $\dim N(A) = n − \dim I(A) \geq \dim I(A) \geq n-k \geq 1 $.

For a) I think the zero matrix is a valid counterexample, but I don't know how to argue it.

For b) when $A$ is the zero matrix. Is it mean that zeros elements matrix for A? What about $X$ and $Y$?

e) Let $A$ be the zero matrix. Can anyone give a counterexample for that please?

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1 Answer 1

a)

Consider the system of equations

$$ 2x-y=4\\ -2x+y=-4 $$ This is two equations in two variables, so $n=k$, and yet there's more than one solution - $x=2,y=0$, and $x=1,y=-2$, and $x=3,y=2$, etc.

b)

Consider the system of equations

$$ 3x+2y-5z=2\\ 5z-2y-3x=0 $$ Two equations in three unknowns, so $n>k$, but there are no solutions.

d) demands a proof, as it's true. Think of the fact that, with more equations than variables, there must be at least one equation that is a linear combination of other equations, and therefore changing the constant for such an equation will change it from solvable to unsolvable. I'll leave the proving itself to you.

e)

Consider the system of equations

$$ 2x-y=0\\ y-2x=0\\ 4x-2y=0 $$ Three equations in two unknowns, so $n<k$, and $Y=0$, but $x=1,y=2$ is a solution.

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