# vector spaces – Find the linear map given image and kernel and a parameter

Let $$W_{k}$$ be a subspace and $$k$$ a real parameter.

$$W_{k} = <(1,2,k),(1,k,1),(2,4,3)>$$

I have to determine for what values of $$k$$ exist a linear map $$F_{k}:mathbb{R^3} to mathbb{R^3}$$ such that

$$Im(F_{k})=W_k\ker(F_k)=<(1,2,2)>$$

I found that for $$k neq 2$$ and $$k neq dfrac{3}{2}$$ the vectors $$(1,2,k),(1,k,1),(2,4,3)$$ form a basis of $$W_k$$ and $$W_k$$ is basically $$mathbb R^3$$.

The associated matrix should have 3 rows and 3 columns.

Since $$dim(ker(F_k)) = 1$$, $$dim(Im(F_k))$$ must be $$2$$ since the domain is $$mathbb R^3$$.

The image must be equal to $$W_k$$, so I should take values of $$k$$ so that $$dim(W_k)=2$$.

For example I could take $$k=2$$ and the image would be generated by $$(1,2,2)$$ and $$(1,2,1)$$ that would be the first two columns of the matrix and to find the third column I need to assure that $$F(1,2,2) = (0,0,0)$$. Is that correct?