I know that a linear transformation could be a projection onto the plane with normal vector $\begin{bmatrix} -1 & 1 & 2\end{bmatrix}^T$, but finding the projection would be too difficult.

I could easily think up a matrix where multiplied by $\begin{bmatrix} -1 & 1 & 2\end{bmatrix}^T = 0$, but I'm not sure on how to choose a matrix where $\begin{bmatrix} -1 & 1 & 2\end{bmatrix}^T$ is the only element of the kernel.

Also, can you please explain this Hint: "to describe a subset as a kernel means to describe it as an intersection of planes?"

  • $\begingroup$ it's not too hard to cook up a matrix with rank 2 that kills this vector $\endgroup$ – qbert Jan 31 '18 at 4:53
  • $\begingroup$ @qbert Yeah but how can I ensure nothing else send the matrix to 0? $\endgroup$ – Goldname Jan 31 '18 at 4:59
  • $\begingroup$ "cook up" a $3\times 3$ matrix, making sure that (-1,1,2) is in the kernel and then show that the rank of this matrix is 2. And you know that rank + nulity = 3, so $(-1,1,2)$ must span the kernel. $\endgroup$ – Doug M Jan 31 '18 at 5:00
  • $\begingroup$ if it has rank 2, it's kernel is rank 1. If it kills your given vector, then that spans the kernel $\endgroup$ – qbert Jan 31 '18 at 5:00
  • $\begingroup$ ah I see, I forgot about that theorem, thanks $\endgroup$ – Goldname Jan 31 '18 at 5:18


  • $$2(-1) + 0(1) +1(2)=0$$
  • $$0(01) + (-2)(1)+1(2)=0$$

  • Verify that $\begin{bmatrix} 2 & 0 & 1 \end{bmatrix}$ and $\begin{bmatrix} 0 & -2 & 1 \end{bmatrix}$ are linearly independent.

  • Now think of someway to form your matrix and prove that span of $\{\begin{bmatrix} -1 & 1 & 2 \end{bmatrix}^T\}$ is a basis to the kernel.

As for the explanation of the hint: To describe $x$ which satisfy $Ax=0$ where $a_i^T$ are the $i$-th row means $x$ is in the intersection of $\{ a_i^Tx = 0 : i \in \{ 1, \ldots, m\}\}$.

  • $\begingroup$ I don't understand what your equations are supposed to mean $\endgroup$ – Goldname Jan 31 '18 at 4:59
  • $\begingroup$ I can easily think of a matrix, but I don't understand what is required so that [-1 1 2] is the only element in the kernel $\endgroup$ – Goldname Jan 31 '18 at 4:59
  • $\begingroup$ $[-1, 1, 2]$ is not the only element, any multiple of $[-1, 1, 2]$ is in the kernel. $\endgroup$ – Siong Thye Goh Jan 31 '18 at 5:01
  • $\begingroup$ yeah that's what I meant $\endgroup$ – Goldname Jan 31 '18 at 5:03
  • $\begingroup$ rank-nullity theorem is useful. geometrically, two non-parallel 3D plane intersect at a line. $\endgroup$ – Siong Thye Goh Jan 31 '18 at 5:03

Expanding upon the comment: Choosing the following matrix works $$ \begin{bmatrix} 1&-1&1\\ 2&0&1\\ 1&1&0 \end{bmatrix} $$ It is clear that $(-1,1,2)$ is in the kernel. The dimension of the image is $2$, so the dimension of the kernel is just $1$, by rank nullity. So the only vectors that map to $0$ under the matrix are your vector and multiples of it.


Pick any two vectors that together with $[-1,1,2]^T$ form an ordered basis of $\mathbb R^3$. Relative to this basis, the matrix $\operatorname{diag}(1,1,0)$ represents a transformation with the required properties. Apply a change of basis to it to get a matrix relative to the standard basis. For that matter, since a linear transformation is completely determined by its action on a basis, the description $T(v_1)=v_1$, $T(v_2)=v_2$, $T([-1,1,2]^T)=0$, where $v_1$ and $v_2$ are the two vectors you chose, is a complete description of the transformation, but that’s likely not the answer that whoever gave you this problem is looking for.

As far as the hint goes, the kernel is a line through the origin, which can be described as the intersection of a pair of planes. The normals to those planes span the orthogonal complement of the line, so you can use them to construct the above basis.

  • $\begingroup$ How did you conclude that "the matrix $diag(1,1,0)$ represents a transformation with the required properties"? Thanks $\endgroup$ – gbox Jan 31 '18 at 9:53
  • 1
    $\begingroup$ @gbox The columns of a transformation matrix are the images of the basis vectors. In fact, for any linear transformation one can find bases for the domain and codomain such that the matrix has the form $\tiny{\left[\begin{array}{c|c}I&0\\\hline0&0\end{array}\right]}$. $\endgroup$ – amd Jan 31 '18 at 19:39
  • $\begingroup$ @gbox BTW, $\operatorname{diag}(1,1,0)$ gives you a projection. If you instead choose some other pair of linearly independent vectors for the first two columns, you’ll get a different type of rank-2 transformation. $\endgroup$ – amd Jan 31 '18 at 19:47

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