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Check if it's linear mapping: $F:\Bbb{R}^3 \to \Bbb{R}^2, \, F(x,y,z)=(2x+y,z+1)$

I don't know if I'm doing it correct: $v=(x_1,y_1,z_1)$ and $w=(x_2,y_2,z_2)$.

$F(av+bw)?=aF(v)+bF(w)$ if it's true it's a linear mapping so

Left-hand side:$$F(ax_1+bx_2,ay_1+by_2,az_1+bz_2)=\big( 2(ax_1+bx_2)+ay_1+by_2,az_1+bz_2+1\big) $$ Right-hand side:$$a(2x_1+y_1,z_1+1)+b(2x_2+y_2,z_2+1)=(2ax_1+ay_1+2bx_2+by_2,az_1+a+bz_2+b)$$

than $a+b=1$ so it's not a linear mapping?

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    $\begingroup$ You have the right idea, but on the RHS calculation you changed from a vector to a scalar. The addition should be vector addition. $\endgroup$ – Dave Jan 30 '17 at 15:25
  • $\begingroup$ RHS=a(2x1+y1,z1+1)+b(2x2+y2,z2+1)=2ax1+ay1+2bx2+by2,az1+a+bz2+b like this? $\endgroup$ – sswwqqaa Jan 30 '17 at 15:31
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    $\begingroup$ You forget the brackets, but yeah. $\endgroup$ – Mathematician 42 Jan 30 '17 at 15:31
  • $\begingroup$ thanks for checking my solution. $\endgroup$ – sswwqqaa Jan 30 '17 at 15:32
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    $\begingroup$ Looks good (remember to put brackets around the vector). $\endgroup$ – Dave Jan 30 '17 at 15:33
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You should know that if $F$ is a linear map, then $F(0)=0$. In this case $F(0,0,0)=(0,1)$, hence it's not linear.

Let me prove that fact. If $F$ is linear, we have that $F(0)=F(0+0)=F(0)+F(0)=2F(0)$, subtract $F(0)$ from both sides to obtain $0=F(0)$. Of course, all zeroes here are zero vectors of the proper dimension.

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  • $\begingroup$ sorry, I didn't write that I have to use vector calculation method. $\endgroup$ – sswwqqaa Jan 30 '17 at 15:30

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