# How to find the dimension of kernel $f$?

Consider a linear map $f∈L(Π2(\Bbb R),\Bbb R^2)$ whose matrix $f_{G,B}$ in the basis $\mathcal B$ of $\prod_2(\Bbb R)$ and

$G=\{(1,1) , (1,-1)\}$ of $\Bbb R^2$ is

$$f_{G,B} = \begin{pmatrix} -4 & -8 \\ 7 & 4\\ 5 & 0 \end{pmatrix} ^T$$ (This is suppose to be a $\Bbb R^{2x3}$ matrix)

From the information i need to find: dim ker $f$

Im a bit confused am i suppose to use the matrix of $f_{G.B}$? *From a previous part of the question i found the rank$f =2$ if that helps at all.

To find the kernel i multiplies the $fG,B$ matrix by $(x, y)^T$ and equated it to $(0,0)^T$ and solved this using the row reduced matrix . So I got $x1 + 1/2x3 = 0$ and $x2 + x3 = 0$ and let $x3= λ$. Therfore $x1 = -1/2λ$ , $x2 = -λ$ and $x3 = λ$

Im not sure if what i did so far is correct. I think i need to use span, but i dont know how exactly.

Would really appreciate your help, thank you so much guys!

You are done, a basis for the $\ker f$ is given by $(-1/2,-1,1)$, as you can directly check, thus the dimension of $\ker f$ is 1.

Note that to find the dimension of $\ker f$ (and thus also of Im(f)), it is sufficient to refer to the rank of the RREF matrix, we don't need to solve the system if we are not interested in finding a basis.

• So if i found the rank to = 2 how can i use this?
– Jess
Mar 22, 2018 at 8:22
• thanks for your help btw
– Jess
Mar 22, 2018 at 8:22
• refer to the rank-nullity theorem dim ker = n - r en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem
– user
Mar 22, 2018 at 8:23
• @Lily You are welcome!
– user
Mar 22, 2018 at 8:24
• @Lily Note also that in this case also without RREF, since we have a 2-by-3 matrix, $1\le rank \le 2$ and since the rows are not multiple we can immeditely conclude that r=2 and dim ker =1.
– user
Mar 22, 2018 at 8:26