What are all the possible cases among three planes.

Question

Can someone help me to understand geometrically the system of equations in $3$ equations and $3$ unknowns in terms of all possible combinations of $\operatorname{Rank}(A)$ and $\operatorname{Rank}(A|B)$ each explained both analytically and geometrically?Most of the books only mention geometrically the cases but do not categorize them in terms of matrix rank.

Answer 1

In summary, either

• $\operatorname{Rank}(A|b) = \operatorname{Rank}(A)$, in which case $Ax = b$ has solutions, or
• $\operatorname{Rank}(A|b) = \operatorname{Rank}(A) + 1$, in which case $Ax = b$ has no solution.

If $Ax = b$ has solutions,

• If $\operatorname{Rank}(A) = 0$, then $A = 0, b= 0$, and all vectors $x$ are solutions.
• If $\operatorname{Rank}(A) = 1$, then the set of all solutions is an affine plane (a plane that need not pass through the origin).
• if $\operatorname{Rank}(A) = 2$, then the set of all solutions is an affine line.
• if $\operatorname{Rank}(A) = 3$, then the solution is unique (and $\operatorname{Rank}(A|b)$ is necessarily $3$ as well, so it doesn't need checked).

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Let $A$ be an $m\times n$ matrix and consider $$A = \begin{bmatrix}A_1 & A_2 & \ldots & A_n\end{bmatrix}$$ where each $A_i$ is the $i$-th column of $A$, an $m$-dimensional column vector. If $$x = \begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}$$ then $$Ax = A_1x_1 + A_2x_2 + ... + A_nx_n$$ which is a linear combination of the columns $A_1, ..., A_n$. The column space of $A$ is the set of all such linear combinations - which is therefore the space $\{Ax\mid x \in \Bbb R^n\}$. The Rank of $A$ is just the dimension of that space.

Therefore, when $m=n=3$,

• If $\operatorname{Rank}(A) = 0$, then $A$ maps all $\Bbb R^3$ to $0$. I.e., $A = 0$. By the Rank-Nullity theorem, $\operatorname{Null}(A) = 3$.
• If $\operatorname{Rank}(A) = 1$, then $A$ maps all $\Bbb R^3$ onto a line passing through $0$. At least one of the $A_i \ne 0$, and the other $A_j = c_jA_i$ for some constant $c_j$. $\operatorname{Null}(A) = 2$.
• If $\operatorname{Rank}(A) = 2$, then $A$ maps all $\Bbb R^3$ onto a plane passing through $0$. At least one of the $A_i$ can be expressed as a linear combination of the other two. The other vectors are not multiples of each other. $\operatorname{Null}(A) = 1$.
• If $\operatorname{Rank}(A) = 3$, then $A$ maps $\Bbb R^3$ onto all of itself. The three columns $A_i$ are linearly independent. $\operatorname{Null}(0) = 0$.

Note that if $x$ and $x'$ both satisfy $Ax = b$ and $Ax' = b$, then $$A(x - x') = Ax - Ax' = b - b = 0$$ Conversely, if $Ax = b$ and $A(x - x') = 0$, then $Ax' = b$. Therefore, if you know one solution to the equation $Ax = b$, all the other solutions differ from it by a vector from the null space of $A$ (the space of all vectors $v$ with $Av = 0$).

Now look at the augmented matrix $$A|b = \begin{bmatrix}A_1 & A_2 & \ldots & A_n& b\end{bmatrix}$$ As noted above, the rank of this matrix is the dimension of the column space of all linear combinations $$c_1A_1 + c_2A_2 + ... + c_nA_n + c_0b$$ Adding in $b$ does not change whether any set of the $A_i$ are linearly dependent or not. So the rank of $A|b$ cannot be any less than the rank of $A$ itself. So there are two cases:

• $b$ is not linearly independent of the $A_i$. Then $\operatorname{Rank}(A|b) = \operatorname{Rank}(A)$. And there exist scalars $x_i$ such that $$b = x_1A_1 + x_2A_2 + ... +x_nA_n$$. But that means $Ax = b$, where $x = \begin{bmatrix}x_1 & x_2 & ... & x_n\end{bmatrix}^T$. So $Ax = b$ has a solution.
• $b$ is linearly independent of the $A_i$, Then $\operatorname{Rank}(A|b) = \operatorname{Rank}(A) + 1$ If $Ax = b$ for some $x$, then $$A_1x_1 + A_2x_2 + ... + A_nx_n - b = 0$$ which would mean $b$ is not linearly dependent.

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Addendum:

To relate the planes of each equation to the rank of $A$, we need to look at the row space instead of the column space. The matrix $A$ can also be expressed as $$A = \begin{bmatrix} {A^1}^T\\{A^2}^T\\{A^3}^T\end{bmatrix}$$ where each $A^i$ is a transposed row of $A$ (where I am indexing by superscripts to distinguish them from the column vectors above). If $b = \begin{bmatrix} b_1 & b_2 & b_3\end{bmatrix}^T$, then the three equations are $${A^1}^Tx = b_1\\{A^2}^Tx = b_2\\{A^3}^Tx = b_3$$ Now for any non-zero column vector $V$, the equation $V^Tx = c$ is the equation of the unique plane normal to $V$ and passing though the point $c\frac V{\|V\|^2}$

The row space of a $A$ is the space of all possible linear combinations of the $A^i$. The dimension of the row space is always the same as the column space - i.e., it is the rank of $A$. So,

• If $\operatorname{Rank}(A) = 0$, then $A = 0$ and there is a solution if and only if $b = 0$, in which case the solution is all of space. (There are no planes in this case, because all three row vectors are $0$ as well.)
• If $\operatorname{Rank}(A) = 1$, then the three row vectors span a 1-dimensional space, so they must all be multiples of each other. Since they all point in the same direction, their normal planes are parallel. A solution can only exist if all three planes all pass through the same point, and being parallel, this requires them to be the same plane. This requires that the ratio of $b_i$ to $b_j$ be the same as $A^i$ to $A^j$ for all $i,j$. Since the planes are all the same, that common plane is the set of all solutions.
• If $\operatorname{Rank}(A) = 2$, then the normal vectors $A^1, A^2, A^3$ all lie in some plane through the origin. If $N$ is a normal vector to that plane, then ${A^i}^TN = 0$ for all $i$. The three planes ${A^i}^Tx = b_i$ all extend in the direction $N$ (if $x$ is a point on the plane, then so will be $x + tN$ for any $t$). Either any two planes are parallel, or they intersect in a line parallel to $N$. These lines are all parallel, and so never intersect the third plane. This is the prism case you mentioned. Only when the three lines of intersection all turn out to be the same line is there a point that lies on all three planes. I.e., a solution to the equation. In this case, all points on the common line of intersection are solutions.
• If $\operatorname{Rank}(A) = 3$, then the three normal vectors are linearly independent. No two planes are parallel, and the line of intersection of those two planes will not be parallel to the third plane, either. Therefore that line of intersection will intersect the third plane in some point $x$. That point $x$ is the unique solution to $Ax = b$.