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I've got the linear map: f: $\mathbb{R^4} \rightarrow \mathbb{R^3}$ with $ \left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array} \right) \mapsto \left( \begin{array}{c} x_1+2x_2+x_3 \\ 2x_1+x_3+x_4 \\ 3x_1+2x_2+3x_3+x_4 \end{array} \right) $ .

Until now I've also calculated the image(f) and the kernel(f), but I don't know how to start with calculating the dimensions of the image and the kernel of f. I hope you can help me. Thanks already!

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    $\begingroup$ If you have calculated those linear space (image and kernel) then you should know their dimensions! They are equal the the cardinalities of bases. $\endgroup$ – Thomas Jun 2 '15 at 17:23
  • $\begingroup$ But can I calculate that in any way? I just can't simply write down their dimensions are 3... $\endgroup$ – marlenee161 Jun 2 '15 at 17:25
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The matrix of the application is $$\left(\begin{matrix}1&2&1&0\\2&0&1&1\\3&2&3&1\end{matrix}\right)$$

Use Gauss method to compute its rank. This will be the dimension of the image of the linear application.

Then apply that if $f:\Bbb R^m\to\Bbb R^n$ is linear, then $\dim\ker f+\dim \mathrm{Im} f=m$.

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The "kernel" of a linear transformation, f, is set of all v such that f(v)= 0.

Here, we must have $x_1+ 2x_2+ x_3= 0$, $2x_1+ x_3+ x_4= 0$, and $3x_1+ 2x_2+ 3x_3+ x_4= 0$. That is three equations in four unknowns so we would expect that there will be "one degree of freedom" or dimension 4- 3= 1. Specifically, we can solve the first equation for $x_3$: $x_3= -x_1- x_2$. The second equation then becomes $2x_1+ (-x_1- x_2)+ x_4= x_1- x_2+ x_4= 0$ and the third equation becomes $3x_1+ 2x_2+ 3(-x_1- x_2)+ x_4= -x_2+ x_4= 0$. We can then write $x_4= x_2$ so the $x_1- x_2+ x)2= x_1+ 2x_2= 0$. Then $x_1= -2x_2$, $x_4= x_2$, and $x_3= -x_1- x_2= 2x_2- x_2= x_2$. $(x_1, x_2, x_3, x_4)= (-2x_2, x_2, x_2, x_2)= x_2(-2, 1, 1, 1). Yes, the kernel is one-dimensional, spanned by the single vector (-2, 1, 1, 1).

Once we know that the kernel has dimension 1, we know that the image has dimension 4- 1= 3.

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