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I want to show a few equivalences for the quadratic matrix $A \in K^{n\times n}$:

  1. The equation $Ax=b$ has no solution for at least one $b \in K^n$.
  2. The equation $Ax=b$ has multiple solutions for at least one $b \in K^n$.
  3. The equation $Ax=b$ has exactly one solution for no $b \in K^n$.

I already thought about using the Rank–nullity theorem: $$\operatorname{dim}(\operatorname{im} (A)) + \operatorname{dim} (\ker (A)) = \operatorname{dim} (A)$$ I know the dimension of $A$, which is $n$. Do I know the dimension of $b$? Is it $n$ as well? I think if I know the dimension of the kernel I can say something about the amount of solutions.

I'm not allowed to use the term "determinant".

Thanks in advance!

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    Yes, you have written $b\in K^n$, so the dimension of $b$ is $n$. – 2017-01-04
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    right, does this tell me something about $dim(im(A))$? – 2017-01-04
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    No, nothing. $b$ must be of dimension $n$, otherwise multiplication with $A$ would not be possible. $A$ is of dimension $n\times n$, so $b$ must be of dimension $n\times 1$. – 2017-01-04
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    $\dim \mathrm{Im}(\mathbf{A})$ can be any integer from $0$ to $n$. If $\mathbf{b} \notin \mathrm{Im}(\mathbf{A})$, what do you think of the linear equation $\mathbf{A}\mathbf{x} = \mathbf{b}$. Similarly, if $\mathbf{b} \in \mathrm{Im}(\mathbf{A})$, what can you say about the equation $\mathbf{A}\mathbf{x} = \mathbf{b}$ ? – 2017-01-04
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    In the first case there exists no solution for $Ax=b$, in the second case there is one or multiple solutions for $x$. – 2017-01-04

1 Answers 1

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First recall that $Ax=0$ has always at least one solution $x=0$. Here $0$ denotes the $n$-dimensional vector $(0,0,\dots,0)\in K^n$. So

  • $(1.)\implies (2.)$ If there exists such a $b$, then $b\notin \operatorname{im}(A).$ Hence $\dim(\operatorname{im}(A))\le n-1$, hence $\dim(\ker (A))\ge 1$ and hence $Ax=0$ has multiple solutions.
  • $(2.)\implies (3.)$ Let $Ax=b_1$ have multiple solutions, i.e. let $b_1$ be such that $Ax_1=b_1$ and $Ax_2=b_1$ where $x_1\neq x_2$. Assume that there exists $b_0$ be such that $Ax=b_0$ has a unique solution $x=x_0$. But then $$A\left(x_0+\frac{x_1-x_2}{2}\right)=b_0$$ So $x_0+\frac{x_1-x_2}{2}\neq x_0$ is a second solution for $Ax=b_0$ which is a contradiction.
  • $(3.)\implies (2.)$ This immediate for $b=0$.
  • $(2.)\implies (1.)$ Again dimension formula.

Of course you can go directly from $(3.)$ to $(1.)$ if you take $b=0$.

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    If $b\notin im(A)$, then $A$ is not surjective. Therefore $dim(im(A)) < n$. So $A$ has not the full rang. Is this right? – 2017-01-04
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    Yes, this is right (and probably better phrased than in my answer). – 2017-01-04
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    k, cool. Why does $Ax=b$ have multiple solutions if $dim(ker(A)) > 0$? What does $dim(ker(A))$ tell me about the set of solutions? – 2017-01-04
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    $Ax=0$ has multiple solutions if $\dim \ker (A)>0$. I do not understand what you mean with $Ax=b$ – 2017-01-04
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    So if $dim(ker(A))>0$, we choose $b = 0$ so we know that $Ax=0$ has multiple solutions. But why does it have multiple? What's the relationship of $dim(ker(A))$ and the amount of solutions? – 2017-01-04
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    I think I got it: If $dim(ker(A))>0$ then the $ker(A)$, which is the set of solutions of $Ax=0$ can be represented for example by a line and therefore $Ax=0$ has multiple solutions. – 2017-01-04
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    Yes, almost, just write that $\ker(A)$ is the set of solutions to $Ax=0$ instead of $b$. Otherwise correct. – 2017-01-04
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    Already edited that one :P Why is it that $x_0 + \frac{x_1−x_2}{2}$ is a second solution for $Ax=b_0$? – 2017-01-04
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    Ok, sorry, you are too fast :)! – 2017-01-04
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    $x_1 = b_1 * A^{-1}$ and $x_2 = b_1 * A^{-1}$. Therefore $x_1 - x_2 = 0$. Why do you divide by 2? – 2017-01-04
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    No, you do not know that $A^{-1}$ exists. You only know that $Ax_1=b_1$ and $Ax_2=b_1$ hence $A(\frac12x_1+\frac12x_2)=\frac12b_1-\frac12b_1=0$. – 2017-01-04
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    I think you mean $A(\frac{1}{2}x_1-\frac{1}{2}x_2)$, right? Why is it if $A(\frac{1}{2}x_1-\frac{1}{2}x_2) = 0$ that $x_0 + \frac{x_1−x_2}{2} \neq x_0$? – 2017-01-04
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    Yes, I mean $-$ instead of $+$. $x_1$ and $x_2$ are different by assumption, so $x_0+\frac{x_1-x_2}{2}$. Ok, have a nice day! – 2017-01-04
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    Got now everything, except how you can go directly from (3.) to (1.). Because $Ax=0$ has at least one, and not no solution. Thank you very much! – 2017-01-04
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    (3.) implies that $Ax=0$ has more than one solutions. (it has at least one, by (3.) it cannot have exactly one, so it has more than one). Then you know that $\dim\ker (A)\ge 1...$ – 2017-01-04