Find the transformation matrix that has kernel the span of two vectors

Question

Find a linear map $T : R^4\to R^3$ whose kernel is generated by $v=(1, 2, 3, 4)$, $w=(0, 1, 0, 1)$

This question has been answered but I don't understand the workings. In particular I don't understand why they multiply the matrix with respect to the basis $B$ with the inverse of the matrix $M$

Here is the link: http://www.math.mcgill.ca/goren/MATH251.2006/Sol3.pdf

Excericse 2 part (b)

If anyone could explain! Thanks!

Answer 1

There are lots of ways to do this. Here's one way.

Let \begin{align*} \vec v_1 &= (1,2,3,4) & \vec v_2&=(0,1,0,1) \end{align*} Our strategy will be to first extend $\{\vec v_1,\vec v_2\}$ to a basis $\{\vec v_1,\vec v_2,\vec v_3,\vec v_4\}$ of $\Bbb R^4$. Once we have done this, we can define $T:\Bbb R^4\to\Bbb R^3$ by $$ T(\vec v_k) = \begin{cases} \vec 0 & k=1,2 \\ \vec u & k=3 \\ \vec v & k=4 \end{cases} $$ where $\vec u,\vec v\in\Bbb R^3$ are arbitrary nonzero vectors. Of course by "define $T$" we mean to define $T$ on the basis $\{\vec v_1,\vec v_2,\vec v_3,\vec v_4\}$ and extend linearly.

One of the advantages to using this strategy is we avoid matrices and get practice in working with linear maps.

Is this enough to get you started or is this confusing?

Answer 2

The author of the given answer constructed a basis $B=(v,w,e_2,e_3)$ where $e_2,e_3$ are the standard vectors of the standard basis of $\Bbb R^4$. He then constructed a linear transformation $T$ that fulfills the desired condition. Notice that the matrix of $T$ denoted $[T]_{B\to St}$ where $St$ is the standard basis of $\Bbb R^3$. Now to write the matrix relative to the standard basis of $\Bbb R^4$ we write the change matrix

$$M_{B\to St}=(v \;w \;e_2\; e_3)$$ from the standard basis $St$ of $\Bbb R^4$ to the basis $B$ and finally the desired matrix is

$$[T]_{St\to St}= [T]_{B\to St} (M_{B\to St})^{-1}$$

Remark The two standard basis of $\Bbb R^4$ and of $\Bbb R^3$ are both denoted by $St$ and by the context we can understand the meaning of the notation.

Answer 3

here is one way to do this. you are looking for a $3 \times 4$ matrix whose null space is the span of $\left\{\pmatrix{1&2&3&4}^T, \pmatrix{0&1&0&1}^T \right\}$ each row $\pmatrix{x&y&z&w}$is orthogonal to each of the vectors in the basis. you have two equations for $4$ variables. you can set $z$ and $w$ as free variables and solve for $x$ and $y.$ two linearly independent solutions are $\pmatrix{-3&0&1&0}^T, \pmatrix{-2&-1&0&1}^T$ these are two rows you need and the third row can be any linear combination of these two. there are several matrices that can do the job. but here is one: $$\pmatrix{-3&0&1&0\\-2&-1&0&1\\-2&-1&0&1} $$