Thread: Find the basis of the nullspace

1. Find the basis of the nullspace

Let $\displaystyle V$ and $\displaystyle W$ be vector-spaces over $\displaystyle \mathbb{R}$. Let $\displaystyle v_1, v_2, v_3, v_4$ be a Basis $\displaystyle B$ of $\displaystyle V$ and $\displaystyle w_1, w_2, w_3$ be a basis $\displaystyle B'$ of $\displaystyle W$. Let $\displaystyle f: V \rightarrow W$ be a linear map such that:

$\displaystyle _{B'}M_B(f) = \left[\begin{matrix} 2 & -1 & 3 & 4 \\ -1 & 6 & 4 & 9 \\ 5 & -12 & -2 & -9 \end{matrix}\right]$

Let $\displaystyle \left[\begin{matrix} 3 \\ 2 \\ 1 \\ 1 \end{matrix}\right]$, $\displaystyle \left[\begin{matrix} 1 \\ 0 \\ -2 \\ -3 \end{matrix}\right]$ be the coordinate vectors of vectors $\displaystyle x, y \in V$ in relation to the basis $\displaystyle B$.

1. Find the basis of the nullspace of $\displaystyle f$.
2. What are the coordinate vectors of $\displaystyle f(x)$ and $\displaystyle f(y)$ relative to the basis $\displaystyle B'$?

2. Originally Posted by bmp05
Let $\displaystyle V$ and $\displaystyle W$ be vector-spaces over $\displaystyle \mathbb{R}$. Let $\displaystyle v_1, v_2, v_3, v_4$ be a Basis $\displaystyle B$ of $\displaystyle V$ and $\displaystyle w_1, w_2, w_3$ be a basis $\displaystyle B'$ of $\displaystyle W$. Let $\displaystyle f: V \rightarrow W$ be a linear map such that:

$\displaystyle _{B'}M_B(f) = \left[\begin{matrix} 2 & -1 & 3 & 4 \\ -1 & 6 & 4 & 9 \\ 5 & -12 & -2 & -9 \end{matrix}\right]$
1) $\displaystyle \left[\begin{matrix} 2 & -1 & 3 & 4 \\ -1 & 6 & 4 & 9 \\ 5 & -12 & -2 & -9 \end{matrix}\right]\left[\begin{matrix}a \\ b \\ c \\ d\\\end{matrix}\right]= \left[\begin{matrix}2a- b+ 3c + 4d \\ -a+ 6b+ 4c+ 9d \\ 5a- 12b- 2c- 9d\end{matrix}\right]$

Let $\displaystyle \left[\begin{matrix} 3 \\ 2 \\ 1 \\ 1 \end{matrix}\right]$, $\displaystyle \left[\begin{matrix} 1 \\ 0 \\ -2 \\ -3 \end{matrix}\right]$ be the coordinate vectors of vectors $\displaystyle x, y \in V$ in relation to the basis $\displaystyle B$.

1. Find the basis of the nullspace of $\displaystyle f$.
If $\displaystyle \left[\begin{matrix} a \\ b \\ c \\ d \end{matrix}\right]$ is in the null space of A then we must have $\displaystyle \left[\begin{matrix}2a- b+ 3c + 4d \\ -a+ 6b+ 4c+ 9d \\ 5a- 12b- 2c- 9d\end{matrix}\right]= \left[\begin{matrix}0 \\ 0 \\ 0 \\ 0 \end{matrix}\right]$.

We must have 2a- b+ 3c+ 4d= 0, -a+ 6b+ 4c+ 9d= 0, 5a- 12b- c- 9d= 0. That is three equations in four unknowns. You should be able to solve for three of them, say, a, b, c in terms of d. In that case, replacing a, b, c, by those you can write the vector in terms of d only. Factor d out of that and you have d times a vector which is the basis vector.

It might happen that you the equations above are not independent. In that case you might be only able to solve for two of the unknowns in terms of the other two or only one in terms of the other three. In that case you can write the vector above as a sum of one variable times a vector plus the other times a vector. The basis vectors are those two vectors and similarly for three vectors if you can solve for only one of a, b, c, d in terms of the other three.

2. What are the coordinate vectors of $\displaystyle f(x)$ and $\displaystyle f(y)$ relative to the basis $\displaystyle B'$?
Multiply that matrix by each of the given vectors x and y. Yes, it really is that easy!