uniqueness of linear map

**James0502** · January 3rd 2009, 07:46 AM

I have a set of 4 vectors (a1,a2,a3,a4) in R3 and a set of 4 vectors (b1,b2,b3,b4)in R4. I need to show that there is precisely one linear map f: R3 --> R4 with f(ai) = bi

a1 = (1,0,0)
a2= (0,1,0)
a3 = (0,0,1)
a4 = (2,1,3)
b1 = (1,2,4,1)
b2 = (1,1,0,1)
b3=(-1,0,4,-1)
b4 = (0,5,20,0)

I have found the linear map (x+y-z,2x+y,4x+4z,x+y-z)
how would I show that this is unique? I then need to find the kernel and the image of f.. would the image just be (x+y-z,2x+y,4x+4z,x+y-z)

and the kernel would be (1,-2,-1)^T ?

many thanks

**JaneBennet** · January 3rd 2009, 09:13 AM

Let $g:\mathbb{R}^3\to\mathbb{R}^4$ be any linear transformation satisfying $g(a_i)=b_i$ for $i=1,2,3,4.$

Given any vector $\mathbf{u}=(x,y,z)\in\mathbb{R}^3,$ we have

$g(\mathbf{u})\ =\ g(xa_1+ya_2+za_3)$

___... $=\ xg(a_1)+yg(a_2)+zg(a_3)$

___... $=\ xf(a_1)+yf(a_2)+zf(a_3)$

___... $=\ f(xa_1+ya_2+za_3)$

___... $=\ f(\mathbf{u})$

Hence $g(\mathbf{u})=f(\mathbf{u})$ for all $u\in\mathbb{R}^3,$ proving the uniqueness of $f.$

and the kernel would be (1,-2,-1)^T ?

You need to write it down properly.

$\ker{f}\ =\ \{t(1,-2,-1)\in\mathbb{R}^3:t\in\mathbb{R}\}$

would the image just be (x+y-z,2x+y,4x+4z,x+y-z)

$\mathrm{image}\,{f}\ =\ \{(x+y-z,2x+y,4x+4z,x+y-z)\in\mathbb{R}^4:x,y,z\in\mathbb{R}\}$

Write $(x+y-z,2x+y,4x+4z,x+y-z)=(u,v,w,u)$ and express $v$ and $w$ in terms of $u$ .

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