Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
What are $[1, 0; 0, 0]$ and $v$? One vector matrix one scalar? Two Matrices? It isn't clear enough. –  moray95 May 2 '13 at 18:34
    
f(v) is defined by the multiplication of the 2*2 matrix [1, 0; 0, 0] by the column matrix in R^2 v. –  JesseP_613 May 2 '13 at 18:34
    
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$1$. –  moray95 May 2 '13 at 18:40
    
How about I say the image of f is precisely $\mathbb R\times\{0\}$ ? It's a line of dimension 1 –  G.T.R May 24 '13 at 18:07

3 Answers 3

$\dim(f)=\dim([1,0;0,0])$ because matrix of $f$ in standard basis is exactly $[1,0;0,0]$

share|improve this answer

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 $(1,0)$.

share|improve this answer

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

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

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

share|improve this answer
    
Thank you, this was a great explanation! –  JesseP_613 May 2 '13 at 18:44
    
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. =) –  Patrick Da Silva May 25 '13 at 20:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.