A linear transformation $T\colon\mathbb{R}^3\to\mathbb{R}^2$ whose matrix is $$\left(\begin{array}{ccc} 1 & 3 & 3\\ 2 & 6 & -3.5+k \end{array}\right)$$ is onto if and only if $k\neq$__________

I'm a little confused by the notation here, so is the matrix given here supposed to be the matrix $A$ such that $XA \Rightarrow Y$? And what does the $\mathbb{R}^3$ and $\mathbb{R}^2$ notation mean? Does that mean a $3\times 3$ matrix and a $2\times 2$ matrix respectively?

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    $\begingroup$ $\mathbb{R}^n$ denotes $n$-dimensional Euclidean space. The notation $T : \mathbb{R}^3 \to \mathbb{R}^2$ means that $T$ send a $3$-dimensional vector to a $2$-dimensional vector. A map $T : X \to Y$ is onto if every element $y \in Y$ can be realized by a point $x \in X$ (I.e., for every element $y$ in $Y$, there is an element $x$ such that $T(x) = y$). The question wants you to find the value(s) of $k$ such that the transformation $T : \mathbb{R}^3 \to \mathbb{R}^2$ is onto. $\endgroup$ – JavaMan Mar 5 '12 at 16:45

$T$ takes in a column vector $(a_1, a_2, a_3)^T$, i.e. an element of $\mathbf R^3$, and sends it to \[ \begin{pmatrix} 1 & 3 & 3 \\ 2 & 6 & -3.5 + k \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}. \] Convince yourself that this results in a $2 \times 1$ matrix, i.e. an element of $\mathbf R^2$.

For the surjectivity: the image of the transformation is the span of the columns of the matrix (why?). You also know that $\mathbf R^2$ is spanned by any two non-zero vectors that are not parallel. What is the span of the first two columns of the given matrix?

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  • $\begingroup$ Let me know if I should say more. I have to hop on a train for a bit, though. $\endgroup$ – Dylan Moreland Mar 5 '12 at 16:53
  • $\begingroup$ How do I find the span of the first two columns? $\endgroup$ – Kyle V. Mar 5 '12 at 18:13
  • $\begingroup$ @StickFigs The span $W$ of the first column is the space of scalar multiples of $(1, 2)$. So, vectors of the form $(x, 2x)$ for all $x \in \mathbf R$. Now, is the second column an element of $W$, or not? $\endgroup$ – Dylan Moreland Mar 5 '12 at 18:57
  • $\begingroup$ Yes, it is. It's x=3 for (x,2x). $\endgroup$ – Kyle V. Mar 5 '12 at 21:38
  • $\begingroup$ @StickFigs Exactly. So it's already contained in the span of the first column. For the "column space" to be all of $\mathbf R^2$, we need the third column to not be contained in $W$ (is it clear why?). What restriction does this give? $\endgroup$ – Dylan Moreland Mar 5 '12 at 22:12

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