Learning Linear Algebra on my own. Going through MIT Open Courseware lectures. There is a problem with solution and I understand the answer but would like to find a more systematic way to find it.


  1. a basis for the plane $x − 2y + 3z = 0$ in $R^3$
  2. a basis for the intersection of that plane with the $xy$ plane


Part a

This plane is the nullspace of the matrix $\begin{bmatrix}1&-2& 3\end{bmatrix}$ and also $A=\begin{bmatrix}1 & −2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$

The special solutions to $Ax = 0$ are

$v_1 = \begin{bmatrix}2 \\ 1 \\ 0 \end{bmatrix}$ and $v_2 = \begin{bmatrix} -3\\ 0 \\ 1\end{bmatrix}$

These form a basis for the nullspace of $A$ and thus for the plane.

Part b

The intersection of this plane with the $xy$ plane contains $v_1$ and does not contain $v_2$; the intersection must be a line. Since $v_1$ lies on this line it also provides a basis for it.

I understand how both parts were solved, but my question is about part b. Is there a more systematic way to find a solution? Say I can't see immediately that $v_1$ is in the plane. How would I solve the problem?

I did try solving the following system $\left \{\begin{array}{l}x-2y+3z=0\\x+y=0\end{array}\right.$

Wrote it in matrix form

$ \begin{bmatrix} 1 & -2 & 3 \\ 1 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} $

And then reduced it to the row reduced echelon form and got the following

$ \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix} $

Which led me to believe that the answer should be $ \begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix} $

But clearly I'm wrong. It seems to me I'm missing some important concept. I would really appreciate any help on this.


You correctly applied the definition of the "$xy$-plane" in your first approach, but not in the second. A point $(x,y,z)$ is on the $xy$-plane if and only if $z = 0$. As such, our system of equations should have been $$ \left \{\begin{array}{l}x-2y+3z=0\\z=0\end{array}\right. $$ Corresponding to the system of equations $Ax = 0$ with $$ A = \pmatrix{1 & -2 & 3\\0 & 0 & 1} $$ (by the way, I really don't understand your apparent preference for square matrices). From there, find the row-echelon form and confirm that $v_1 = (2,1,0)^T$ is our only special solution, so this vector forms a basis of the desired nullspace.

  • 1
    $\begingroup$ I would further remark that while it is good to ask the question of "how do I do this systematically", it is also good practice to embrace insights, such as that which led to your quicker answer. Having looked at the MIT course questions before and having taught out of Strang's text, I can say from experience that many questions are built to encourage that sort of alternative "non-algorithmic" perspective. $\endgroup$ – Omnomnomnom Aug 23 '17 at 0:57

Perhaps it is just because I learned Linear Algebra as being about vector spaces and learned matrices as a specific notation, but I have always disliked writing things in terms of matrices. Here is how I would do exercise (1): The equation x- 2y+ 3z= 0 can be written x= 2y- 3z so any point in that plane can be written (x, y, z)= (2y- 3z, y, z)= (2y, y, 0)+ (-3z, 0, z)= y(2, 1, 0)+ z(-3, 0, 1). So a basis for that plane is {(2, 1, 0), (-3, 0, 1)}.

Now, the intersection of that with x+ y= 0 or x= -y, allows us to write x- 2y+ 3z= 0 as -y- 2y+ 3z= 0 or -3y+ 3z= 0 or, finally, z= y. Then y(2, 1, 0)+ z(-3, 0, 1) becomes y(2, 1, 0)+ y(-3, 0, 1)= y(-1, 1, 1). That is, that one-dimensional space has {(-1, 1, 1)} as basis.


A basis for the plane $x − 2y + 3z = 0$ consists of two vectors perpendicular to $(1,-2,3).$ Thus we have

$$\{(0,3,2),(3,0,-1)\}, \:\: \{(0,3,2),(2,1,0)\}, \:\: \{(3,0,-1),(2,1,0)\}$$ as possible basis.

A basis for the intersection of planes $$\begin{cases}x-2y+3z&=0\\z&=0\end{cases}$$ consists of one vector satisfying both equations. Thus $$\{(2,1,0)\}$$ is a basis.


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