Find linear transformation $F$ in canonical bases given

$ F: \Bbb R^4 \to \Bbb R^3 $

$ \ker F=\operatorname{span}\left\{\begin{bmatrix}1\\2\\3\\4\end{bmatrix}, \begin{bmatrix}0\\1\\1\\1\end{bmatrix} \right\} $

I tried expanding these two vectors to a basis of $\Bbb R^4$, I tried adding two linear dependent vectors, but nothing seems to work. I would be glad for a little hint.

  • 1
    $\begingroup$ Here's basically the same question. $\endgroup$ Feb 9, 2015 at 15:02
  • $\begingroup$ Can anyone tell me how to proceed if instead of the given kernel the range instead is given. For example : $T:V_3 \rightarrow V_3$ such that the equation of $R(T)$ is $4x_1-3x_2+x_3 = 0$. Should I post a new question for this ? $\endgroup$
    – roni
    Jul 29, 2015 at 4:51

1 Answer 1


Let $u=(1,2,3,4)$ and $v=(0,1,1,1)$. Clearly, $(u,v)$ is an independent family of $\mathbb{R}^4$, and can be completed to a basis of $\mathbb{R}^4$ with two vectors. It's very easy to see that adding $e_1=(1,0,0,0)$ and $e_2=(0,1,0,0)$ to this family yields a basis of $\mathbb{R}^4$ (compute the rank of $(u,v,e_1,e_2)$).

Now a linear mapping is uniquely determined by the image of a basis of its domain. We already know the image of $u$ and $v$ (the nil vector of $\mathbb{R}^3$), hence we only need to determine the image of $e_1$ and $e_2$ by $F$. A bit later on, we'll need the coordinates of a vector of $\mathbb{R}^4$ in the basis $(u,v,e_1,e_2)$, so we might as well do this right now. A straightforward system solving yields: $$\forall (x,y,z,t)\in\mathbb{R}^4,\quad (x,y,z,t)=(-z+t)u+(4z-3t)v+(x+z-t)e_1+(y-2z+t)e_2.$$

There's a caveat: we can't take just any vectors $a$ and $b$ of $\mathbb{R}^3$ for the images of $e_1$ and $e_2$, since we could create some other vectors in the kernel of $F$. The restriction is that the family $(a,b)$ should be independent (think of the Rank–Nullity Theorem).

We then have the general solution of your problem: a linear mapping $F:\mathbb{R}^4\longrightarrow\mathbb{R}^3$ satisfies all your requirements if and only if there exists an independent family $(a,b)$ of vectors of $\mathbb{R}^3$ such that: $$\forall(x,y,z,t)\in\mathbb{R}^4,\quad F(x,y,z,t)=(x+z-t)a+(y-2z+t)b.$$

For example, the linear mapping $F:\mathbb{R}^4\longrightarrow\mathbb{R}^3$ defined by: $$\forall(x,y,z,t)\in\mathbb{R}^4,\quad F(x,y,z,t)=(x+z-t,y-2z+t,0)$$ fulfills all your requirements. The matrix of this $F$ in the standard bases of $\mathbb{R}^4$ and $\mathbb{R}^3$ is: $$[F]_{\text{std}(\mathbb{R}^4),\text{std}(\mathbb{R^3})}=\begin{pmatrix}1&0&1&-1\\0&1&-2&1\\0&0&0&0\end{pmatrix}.$$

But there are lots more (I actually gave you all of them!).

The example given by YourAdHere corresponds to the case $a=(3,3,0)$ and $b=(1,2,0)$ (which is, in some respect, not the simplest).

Now, given a $3\times4$ matrix, how can you determine whether its the matrix (in the standard bases of $\mathbb{R}^4$ and $\mathbb{R}^3$) of a linear mapping satisfying your requirements? Easy:

  • Check that the two first columns are not proportional,
  • Check that $C_3=C_1-2C_2$ and $C_4=-C_1+C_2$.

This fact is easily seen, e.g., from the form of the matrix I gave above.

  • 2
    $\begingroup$ Please clarify: how did you get to: $$\forall (x,y,z,t)\in\mathbb{R}^4,\quad (x,y,z,t)=(-z+t)u+(4z-3t)v+(x+z-t)e_1+(y-2z+t)e_2.$$ ? $\endgroup$
    – Dor
    Jan 3, 2015 at 19:57
  • 3
    $\begingroup$ @Dor: solve the linear system:$$\begin{cases}\alpha+\gamma=x\\2\alpha+\beta+\delta=y\\3\alpha+\beta=z\\4 \alpha +\beta=t\end{cases}$$The coefficients $\alpha,\beta,\gamma,\delta$ you obtain are the coefficients in front of $u,v,e_1,e_2$ repectively. $\endgroup$ Jan 3, 2015 at 20:41

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