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I need to find the dimension of the image of the linear transformation $f(v)$, where $f\colon \Bbb R^2 \to \Bbb R^2$ is defined by

$$f(v) = \begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}v$$

I have already found that the linear transformation is neither injective nor surjective by finding contradictory examples for both, but I'm not sure where to proceed. I know that the $\text{Im} (f) = \{f(v) \} v$ elements of vector space $V$}, but I'm not sure how to derive the dimension.

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  • $\begingroup$ What are $[1, 0; 0, 0]$ and $v$? One vector matrix one scalar? Two Matrices? It isn't clear enough. $\endgroup$
    – moray95
    May 2 '13 at 18:34
  • $\begingroup$ f(v) is defined by the multiplication of the 2*2 matrix [1, 0; 0, 0] by the column matrix in R^2 v. $\endgroup$
    – JesseP_613
    May 2 '13 at 18:34
  • $\begingroup$ Then it's just a simple matrix multiplication rule : a matrix of dimension $a$x$b$ multiplied by a matrix $b$x$c$, gives a matrix $a$x$c$... So the dimension of the image will be $2$x$1$. $\endgroup$
    – moray95
    May 2 '13 at 18:40
  • $\begingroup$ How about I say the image of f is precisely $\mathbb R\times\{0\}$ ? It's a line of dimension 1 $\endgroup$ May 24 '13 at 18:07
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$\dim(f)=\dim([1,0;0,0])$ because matrix of $f$ in standard basis is exactly $[1,0;0,0]$

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Another perspective: the image of the linear transformation that sends $\vec{x}$ to $A \vec{x}$ for some matrix $A$ is the column space of $A$, the subspace spanned by the columns of $A$.

In your example the columns of $A$ are $(1,0)$ and $(0,0)$. Given a finite set $S = \{\vec{v}_1,\dots,\vec{v}_k$ of vectors that span a vector space $V$, you can delete vectors from $S$ to obtain a basis in the following manner. Drop a vector $\vec{v}_i$ from the list if it is a linear combination of $\vec{v}_1,\dots, \vec{v}_{i-1}$, and keep it otherwise. The vectors that you keep will be a basis for $V$. (It is a good exercise to prove this statement.)

In your example, you would keep $\vec{v}_1 = (1,0)$. (In the algorithm, you always keep the first vector $\vec{v}_1$ for your basis.) Your second vector $\vec{v}_2 = (0,0)$ is a linear combination of $\vec{v}_1$, namely $\vec{v}_2 = 0 \vec{v}_1$, so you drop it. Your basis consists of just the single vector $\vec{v}_1 = (1,0)$, so you conclude that the image of your transformation is one-dimensional and in fact the image is simply the span of the vector $(1,0)$.

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$$f(\left( \begin{array}{c} v_1\\ v_2 \end{array}\right))=\left( \begin{array}{cc}1&0\\ 0&0\end{array}\right)\cdot\left( \begin{array}{c} v_1\\ v_2 \end{array}\right)=\left( \begin{array}{c} v_1\\ 0 \end{array}\right)$$

Hence the null-space of $f$ is $$N(f)= \left\langle \left( \begin{array}{c} 0\\ 1 \end{array}\right) \right\rangle$$ then use

$$\dim(N(f))+\dim(R(f))=\dim(\mathbb{R}^2)$$

so that

$$\dim(R(f))=\dim(\mathbb{R}^2)-\dim(N(f))=2-1=1$$.

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  • $\begingroup$ Thank you, this was a great explanation! $\endgroup$
    – JesseP_613
    May 2 '13 at 18:44
  • $\begingroup$ I saw the $<$ and $>$ comment by Zev Chonoles on your answer about non-generators and Frattini subgroups so I looked up your other answers for fun. You should use the \begin{pmatrix} or \begin{bmatrix} command instead of always typing \left( \begin{array}{c} ... all the time. It'll make your code a lot easier to read and save you some time. =) $\endgroup$ May 25 '13 at 20:59

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