Row picture of the system of linear equations contains planes corresponding to each equation. Column picture of the system of linear equations contains vectors corresponding to each unknown.

While learning linear algebra from Gilbert Strang's book, I came across following statement:

If $n$ planes have no point in common (i.e. no solution), or infinitely many points (i.e. infinite solutions), then $n$ columns (vectors representing those columns having coefficients of unknowns as their entries) lie in the same plane.

For example, consider we have following system of three equations:


If planes representing these equations intersect in zero or infinite points, then the columns (or vectors represented by these columns): $[a_1,b_1,c_1],[a_2,b_2,c_2],[a_3,b_3,c_3]$ lie in the same plane. (I should have arranged them vertically instead of horizontally.)

Q1. Though I can plot (in matlab or octave) and see that in no solution and infinite solution cases, the vectors do indeed lie in the same plane, I am not able to grasp visually / geometrically / intutively, why this happens?

Q2. Also in the row picture system of linear equations, the solution to the system is

  • visually: the intersection of the planes.
  • algebraically: the set of values for unknowns/variables satisfying equations of all given planes.

However what does the solution in the column picture takes form visually and algebraically?


1 Answer 1


This eventually boils down to the condition for a matrix $\mathbf{A}$ so the equation $\mathbf{Ax=b}$ has no solution, one solution or infinitely many solutions. You can read that here.

For your question, consider the matrix equation which represents solutions for system of equations for $n$ variables $\mathbf{Ax}=\mathbf{b}$ where $$A=\begin{pmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{n,1} & a_{n,2} & \cdots & a_{n,n} \end{pmatrix}, \mathbf{x}= \begin{pmatrix} x_1\\ \vdots \\ x_n \end{pmatrix}, \mathbf{b}= \begin{pmatrix} b_1\\ \vdots \\ b_n \end{pmatrix}$$ We can rewrite this as a linear combination of column vectors: $$x_1 \begin{pmatrix} a_{1,1}\\ \vdots \\ a_{n,1} \end{pmatrix} +x_2 \begin{pmatrix} a_{1,2}\\ \vdots \\ a_{n,2} \end{pmatrix}+ \ldots+ x_n \begin{pmatrix} a_{1,n}\\ \vdots \\ a_{n,n} \end{pmatrix}= \begin{pmatrix} b_1\\ \vdots \\ b_n \end{pmatrix}. \tag{2}$$ This system has either no solution or infinitely many solutions if and only if $\text{rank}A<n$, or in other words, the set of $n$ column vectors $\left \{ \begin{pmatrix} a_{1,i}\\ \vdots \\ a_{n,i} \end{pmatrix}| 1\le i \le n \right \}$ are linearly dependent. This means there exists one column vector that can be written as a linear combination of other column vectors. WLOG, we can assume that $$ \begin{pmatrix} a_{1,n}\\ \vdots \\ a_{n,n} \end{pmatrix}=\sum_{i=1}^{n-1} \alpha_i \begin{pmatrix} a_{1,i}\\ \vdots \\ a_{n,i} \end{pmatrix}, \tag{1}$$ for $\alpha_i$ not all zero.

Consider $(n-1)\times n$ matrix $B$ (which is eventually transpose of $A$ but without last column vector in $A$). $$B=\begin{pmatrix} a_{1,1} & a_{2,1} & \cdots & a_{n,1} \\ a_{1,2} & a_{2,2} & \cdots & a_{n,2} \\ \vdots & \vdots & \vdots & \vdots \\ a_{1,n-1} & a_{2,n-1} & \cdots & a_{n,n-1} \end{pmatrix}$$ According to the size of $B$ ($n-1<n$), the system $\mathbf{Bx}=0$ will have infinitely many solutions. Let's say one of such is $\mathbf{x}=\begin{pmatrix} \beta_1\\ \vdots \\ \beta_n \end{pmatrix}$ ($\mathbf{x}\ne 0$). Therefore, from $\mathbf{Bx}=0$ we find $n-1$ column vectors $\begin{pmatrix} a_{1,i}\\ \vdots \\ a_{n,i} \end{pmatrix} \; (1 \le i \le n-1)$ belong to the hyperplane $$P= \left \{ \begin{pmatrix} x_1\\ \vdots \\ x_n \end{pmatrix}: \beta_1x_1+\beta_2x_2+ \ldots+ \beta_nx_n=0 \right \}.$$ And since the last column vector is a linear combination of such $n-1$ column vectors (from $(1)$) so it is also in the hyperplane $P$. In conclusion, all $n$ column vectors lie in the hyperplane $P$.

For your second question, if you see system of equations as linear combination of column vectors (as in $(2)$) then I would say that the column vectors are points in $n$-dimensional plane which may or may not lie on a hyperplane (depending on number of solutions for $\mathbf{Ax=b}$ as mentioned above). Note that rank of matrix $A$ is eventually refer to the dimension of the span of $n$ column vectors. If $\text{rank}A=n$ then the system has one unique solution.


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