0

Let $W$ be the vector space over $\mathbb R$ defined by

$$ W= \begin{Bmatrix} \left( \begin{matrix} x-y&2x+y+3z\\ -14x-7y-21z&-3x+3y \end{matrix} \right) | x,y,z\in \mathbb R \end{Bmatrix} $$

Find a basis for this subspace of $\mathbb R^{2x2}$ from matrices whose rank is only 1

My way:

$$x \left( \begin{matrix} 1&2\\ -14&-3 \end{matrix} \right) +y \left( \begin{matrix} -1&1\\ -7&3 \end{matrix} \right) +z \left( \begin{matrix} 0&3\\ -21&0 \end{matrix} \right) $$

Now: $$ \left( \begin{matrix} 1&2&-14&-3\\ -1&1&-7&3\\ 0&3&-21&0\\ \end{matrix} \right) \to \left( \begin{matrix} 1&2&-14&-3\\ 0&3&-21&0\\ 0&0&0&0\\ \end{matrix} \right) $$

So the basis I have found is:

$$\begin{Bmatrix} \left( \begin{matrix} 1&2\\ -14&-3 \end{matrix} \right) , \left( \begin{matrix} 0&3\\ -21&0 \end{matrix} \right)\end{Bmatrix} $$

But their rank is 2 and not 1 and I don't know how to continue from here.

Ro168
  • 517
  • 1
    Subtle hint: rank-1 matrix could be written as $\begin{pmatrix} a_1b_1 & a_2b_1 \ a_1b_2 & a_2b_2 \end{pmatrix}$ (see https://math.stackexchange.com/q/1545118/113703). – arseniiv Dec 19 '17 at 13:31
  • OK but which 2 vectors I need to multiply? – Ro168 Dec 19 '17 at 16:45
  • 1
    Then not so subtle, try substituting $u = x - y, v = 2x + y + 3z$ in a matrix from $W$, and… – arseniiv Dec 19 '17 at 17:13
  • Then it's: $$ \begin{Bmatrix} u \left( \begin{matrix} 1&0\ 0&3 \end{matrix} \right) + v \left( \begin{matrix} 0&1\ -7&0 \end{matrix} \right) \end{Bmatrix} $$ The rank is still 2, What do I miss? – Ro168 Dec 19 '17 at 17:57
  • 1
    To early to expand yet—now find precisely when its rows (or columns) are linearly dependent, in terms of $u$ and $v$. – arseniiv Dec 19 '17 at 18:09

0 Answers0