Consider the function $T:\mathbb R^2 \to \mathbb R^2$ given by $$T \begin{pmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \end{pmatrix} = \begin{bmatrix} 2x+y \\ -3x \end{bmatrix}$$ Prove that $T$ is linear, one-to-one, and onto. Here is what I 've got so far:

Let $u,v \in \mathbb R^2$ such that $u = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix}$ and $v = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}$ then, $$T(u + v) = T\begin{pmatrix} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} + \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} \end{pmatrix} =\ldots \text{ (skipping tedious steps)} \ldots = T(u) + T(v) \to $$ preserves addition and,

$$T(ru) = T\begin{pmatrix} r \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} \end{pmatrix}= \ldots = rT(u) \to$$ preserves scalar multiplication $\therefore T$ is linear.

To prove it is one-to-one I believe I need to show that it is impossible for $u=v$ right? on this i am not positive. Also to prove that it is onto I am pretty sure I need to show that $sp(T) = sp(\mathbb R^2)$ again not really sure so I don't know where to start.

Any help will be greatly appreciated.

Edit 1:

please correct me if im wrong here:

Let $T(v)=0$, then $0=T\begin{pmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} \end{pmatrix} = \begin{bmatrix} 2v_1 + v_2 \\ -3v_2 \end{bmatrix}$, so $2v_1 + v_2=0$ and $-3v_1=0$

$\therefore v=\begin{bmatrix} 0 \\ 0 \end{bmatrix} = 0$ and $T$ is one-to-one

  • $\begingroup$ It is enough to show the kernel is trivial: $T(x)=0\Rightarrow x=0$ $\endgroup$ – Ranc Dec 9 '15 at 10:11
  • $\begingroup$ @Ranc that proves one to one correct? $\endgroup$ – justin shores Dec 9 '15 at 10:13
  • $\begingroup$ Yes, in the finite dimensional case; If $T\colon E\rightarrow E$ is a linear map, then if $T$ is $1-1$ then it must be onto. This is due to the relation: $\dim E = \dim \mathrm {Im} T +\dim \ker T$ $\endgroup$ – Ranc Dec 9 '15 at 12:39

The matrix of $T$ is $$A = \begin{bmatrix} &2 &1 \\ &-3 &0 \end{bmatrix}$$ The kernel of $T$ is $\bf0$, because the only solution to $A * [x,y]^T = \bf 0$ is $x=y=0$. Because $$ \operatorname{dim}\,\operatorname{ker}\,A + \operatorname{dim}\,\operatorname{rank}\,A = \text{# of columns of $A$}, $$ it follows that $\operatorname{dim}\,\operatorname{rank}\,A = 2$. Thus $A$ is invertible, so $T$ has an inverse — that is, $T$ is an injection and a bijection.


Solve $2x+y = w, -3x=z$ for $x,y$ to get $x=-{1 \over 3} z$ and $y={2 \over 3} z+w$.

Hence for any $(w,z)$ there is a unique $(x,y)$ such that $T(x,y) = (w,z)$.

It follows that $T$ is injective and surjective.



  1. One-to-one: Let $u, v$ as you have them. Show that $T(u)=T(v) \implies u=v$. You will need to solve a two-by-two system of equations.
  2. Onto: Show that $\dim \Im(T)=2$. How? By finding two vectors in the image of $T$ that are linearly independent. For example: what is the image of the vector $e_2=\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ under $T$? Can you find another one, say $z$ so that $T(e_2)$ and $T(z)$ are linearly independent (and therefore span $\mathbb R^2$)?

Of course as @kevinzakka comments, it suffices to show that $T$ is one-to-one. This is equivalent to $\dim Ker (T)=0$ and hence by the Rank - nullity Theorem you can conclude that $\dim \Im (T)=2$.

  • 1
    $\begingroup$ Since the image and the co-image are the same, all you have to show is one or the other. Usually proving it is one-to-one is easier. $\endgroup$ – Kevin Zakka Dec 9 '15 at 15:14

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