Give an example of a linear transformation whose kernel is the line spanned by vector: $\begin{bmatrix} -1 & 1 & 2\end{bmatrix}^T$

Question

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?"

Answer 1

Guide:

• $$2(-1) + 0(1) +1(2)=0$$

• $$0(01) + (-2)(1)+1(2)=0$$

• 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 $\{ a_i^Tx = 0 : i \in \{ 1, \ldots, m\}\}$.

Answer 2

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.

Answer 3

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.