Relationship between columns, rows, rank given a system of linear equations

Question

If $A\vec{x}\ = \left[\begin{array}{r}1\\0\\0 \end{array}\right]$ has no solution, where $A$ is an $m\times n$ matrix with rank $r$, apparently $r<m$ $(m=3)$. Why is this?

Also, if $A\vec{x}\ = \left[\begin{array}{r}0\\1\\0 \end{array}\right]$ has exactly one solution, why is $r=n<3$?

Answer 1

In each of the following statements, $A$ is a linear operator with domain $\Bbb R^n$ and codomain $\Bbb R^m$ with both $n$ and $m$ finite non-negative integers.

One can show that the following are all equivalent statements:

• $\text{rank}(A)=r$
• The number of linearly independent columns of (the matrix representation of) $A$ is $r$
• The number of linearly independent rows of (the matrix representation of) $A$ is $r$
• The dimension of the range (image) of $A$ is $r$

One has an immediate observation then that $0\leq\text{rank}(A)\leq \min(m,n)$

Furthermore, one has the "Rank-Nullity Theorem"

• $\text{rank}(A)+\text{nullity}(A)=n$

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Definition: $A$ is surjective (onto) if and only if for every $y$ in the codomain there exists at least one $x$ for which $Ax=y$. Equivalently worded, $A$ is surjective if and only if the image is equal to the codomain.

Theorem: $A$ is surjective if and only if $\text{rank}(A)=m$

Corollary: If $A$ is not surjective, then $\text{rank}(A)<m$

Proof of theorem:

$\Rightarrow)$ Suppose that $A$ is surjective. Then the image is all of $\Bbb R^m$ which is of dimension $m$. Thus $\text{rank}(A)=m$.

$\Leftarrow)$ Suppose that $\text{rank}(A)=m$. Then the dimension of the image is $m$. As the image is a subspace of the codomain and the only $m$-dimensional subspace of $\Bbb R^m$ is $\Bbb R^m$ itself, the image is in fact equal to $\Bbb R^m$. Thus $A$ is surjective.

The corollary follows by contraposition of the statement and the fact that $\text{rank}(A)\leq \min(m,n)\leq m$

What this means for your problem: Since $Ax=\begin{bmatrix}1\\0\\0\end{bmatrix}$ has no solution, either $m\neq 3$ or $A$ is in fact not surjective. If we are told that $m=3$, then that implies that $A$ is not surjective, and then by the corollary the rank is strictly less than $3$.

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Definition: $A$ is injective (one-to-one) if and only if for every $x_1$ and $x_2$ where $x_1\neq x_2$ in the domain, one has $Ax_1\neq Ax_2$. Equivalently worded, if $Ax_1=Ax_2$ then $x_1=x_2$. Equivalently worded, if $Ax=y$ has a solution, it has exactly one solution.

Lemma: $A$ is injective if and only if $\ker(A):=\{x~:~Ax=0\}=\{0\}$

Proof of Lemma:

$\Leftarrow)$ Suppose that $A$ is not injective: Then there exist $x_1$ and $x_2$ with $x_1\neq x_2$ such that $Ax_1=Ax_2$. Then $Ax_1-Ax_2=A(x_1-x_2)=0$ and so $(x_1-x_2)\in\ker(A)$

$\Rightarrow)$ Suppose that $A$ is injective. Then $Ax\neq A0=0$ for all $x\neq 0$, thus $\ker(A)=\{0\}$

Theorem: $A$ is injective if and only if $\text{rank(A)}=n$

Proof of theorem:

$\Rightarrow)$ Suppose $A$ is injective. Then by the lemma $\ker(A)=\{0\}$ and so $\text{nullity}(A)=\text{dimension}(\ker(A))=0$. Then by the rank-nullity theorem, $\text{rank}(A)+\text{nullity}(A)=n=\text{rank}(A)+0=\text{rank}(A)$

$\Leftarrow)$ Suppose $A$ is not injective. Then by the lemma $\ker(A)\neq \{0\}$ and therefore $\text{nullity}(A)>0$ as the kernel is a subspace of the domain. Then by the rank-nullity theorem, $\text{rank(A)}<\text{rank}(A)+\text{nullity}(A)=n$ and so $\text{rank}(A)<n$

What this means for your problem: Since $Ax=\begin{bmatrix}0\\1\\0\end{bmatrix}$ has exactly one solution, by the theorem above this implies that $\text{rank}(A)=n$. If this is the same matrix as in the first part, where we had learned that $\text{rank}(A)<3$, then this implies that $\text{rank}(A)=n<3$.

Answer 2

Firstly, for $Ax=b$ to have no solution, there must be a row in the reduced row echelon form of $A$ that is the zero vector (in $\Bbb K^n$ assuming $A\in\Bbb K^{m\times n}$). That would mean that there are not $m$ nonzero rows in the RREF form of $A$, which means that the rank of $A$ has to be less than $m$. Note that this is necessary but not sufficient to say $Ax=b$ has no solution.

Secondly, if $Ax=b$ has one solution, that means that the rank must equal the number of columns (i.e. $r=n$), which can be determined from the rank-nullity theorem. Moreover, $r$ must be less than or equal to $m$ (in this case $3$) because the rank of an $m\times n$ matrix is less than the minimum of $n$ and $m$ (i.e. $r\leq \text{min}(m,n)$). Although, in the question it says strictly less than, which is incorrect (that is, $r$ would equal $3$ if $n=3$).