# Parameter for row space of matrix

$M$ is a $2\times{4}$ matrix, and from its row reduces form, we have found the spanning vectors $$v_{1}= \begin{bmatrix}2\\1\\0\\0\end{bmatrix}\quad\text{and}\quad v_{2}= \begin{bmatrix}-1\\0\\2\\1\end{bmatrix}.$$ For what value of $T$ is $u= \begin{bmatrix}1&-2&3&T\end{bmatrix}$ in the row space of $M$?

• Should $v_1, v_2$ be row vectors instead? Are they the rows of $M$? – Arthur Jun 14 '16 at 0:18

If $u$ would be in the row space, then you could obtain it as a linear combination of $v_1$ and $v_2$ as a basis of row space. Now, assume that $u=av_1+bv_2$. From the second component, we have $a=-2$, and from the second one, we have $b=3/2$ while these two values do not satisfies the first component. So, there is no such a $T$.
• I know about this, but what if $v_{1}$ and $v_{2}$ are the vectors that span the column space? How can you be sure that these are the row spacing vectors? – kaboommath Jun 14 '16 at 0:27
• Your matrix is $2\times 4$. So, it has two rows which are $4-$tuple vectors and four columns which are $2-$tuple vectors. Now, how $v_1$ and $v_2$ can generate the column space which comprises $2-$tuple vectors? I think you may need to revise the question. – Majid Jun 14 '16 at 0:47