Find the basis and dimension of vector space over $\mathbb R$:
a) vector space generated by $\{u+v+w,v+w+z,w+z+u,z+u+v\}$, where $u,v,w,z$ are linearly independent vectors of some vector space
b) $R(A) \cap S(A) \cap Ker(A)$ for matrix $$
A= \begin{pmatrix}
1 & 2 \\
3 & 5 \\
\end{pmatrix}$$
$R(A)$ is row matrix space, $S(A)$ is column matrix space and $Ker(A)$ is kernel of the matrix.
How to reduce this vector in a) to get the basis? I'm not sure how to calculate $R(A) \cap S(A) \cap Ker(A)$. Will $R(A)$ and $S(A)$ be the same for matrix 2×2? Could someone help me please?
Best Answer
For a), start by checking whether the generators of the vector space are linearly independent. You know that $(u,v,w,z)$ are linearly independent so if $au + bv + cw + dz = 0$ you must have $a = b = c = d = 0$. Write a linear combination of the generators, collect terms and check what you get.
For b), note that $A$ has rank $2$ so $R(A) = S(A) = \mathbb{R}^2$ so you only need to find $\ker(A)$ which amounts to solving a linear system of homogeneous equations. Since $\dim \ker(A) = 2 - \mathrm{rank} A = 0$, you actually don't need to solve a thing.