Consider the following real $4\times 5$ matrix

Question

$$A = \left( \begin{array} { c c c c c } { 1 } & { 4 } & { 15 } & { 1 } & { - 9 } \\ { 0 } & { 1 } & { 4 } & { 0 } & { - 3 } \\ { 2 } & { - 4 } & { - 18 } & { - 1 } & { 15 } \\ { 1 } & { 0 } & { - 1 } & { 0 } & { 2 } \end{array} \right)$$ Define a linear map $f _ { A } : \mathbb { R } ^ { 5 } \rightarrow \mathbb { R } ^ { 4 }$ by $f _ { A } ( \mathbf { v } ) = A v .$ Find a basis of the null space of $f _ { A }$ and a basis of the image of $f _ { A } ,$ respectively.

So whether the rank of null space and the image is 5? If it is, how can I find the basis of null space and basis? This is the first time I try to solve a linear map with no n*n matrix, I am confused.

Answer 1

The procedure is exactly the same as for square matrices. I'll do a simpler example for the nullspace, see if you can apply the idea to your case. Let's try to find the nullspace for $$ A=\begin{bmatrix}1&2&3&4\\1&1&1&4\\0&1&2&3\end{bmatrix}. $$ Do Gaussian Elimination to find the reduced row echelon form, $\operatorname{rref} A$. You should get $$\operatorname{rref}A=\begin{bmatrix}1&0&-1&-2\\0&1&2&3\\0&0&0&0\end{bmatrix}.$$ Now, put this in the form of equations. Say you have variables $w,x,y,z$, then $Ax=0$ (which is of course the system we want to consider since we're finding the nullspace of $A$) gives you the system $$\begin{cases}w=y+2z;\\x=-2y-3z.\end{cases}$$ Therefore, the vectors $x$ satisfying $Ax=0$ are exactly those of the form $$\begin{bmatrix}y+2z\\-2y-3z\\y\\z\end{bmatrix}=y\begin{bmatrix}1\\-2\\1\\0\end{bmatrix}+z\begin{bmatrix}2\\-3\\0\\1\end{bmatrix}.$$ But this tells you that the set $$\left\{\begin{bmatrix}1\\-2\\1\\0\end{bmatrix},\begin{bmatrix}2\\-3\\0\\1\end{bmatrix}\right\}$$ is a basis for the nullspace of $A$! We are done. Do the same steps with the different example in your case to get the desired solution.

Answer 2

The null space is the set of vectors $v$, such that $f_A(v) = 0$. This means that you need to solve $Av = [0, 0, 0, 0]^T$. This amounts to Gaussian elimination. In this case, the echelon form of A is $$ A=\left(\begin{array}{rrrrr} 1 & 0 & -1 & 0 & 2 \\ 0 & 1 & 4 & 0 & -3 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right) $$ Let $v = [v_1, v_2, v_3, v_4, v_5]^T$. Solving $Av = 0$ means that $$ \begin{cases} v_1 - v_3 + 2v_5 = 0\\ v_2 + 4v_3 -3v_5 = 0\\ v_4 + v_5 = 0 \end{cases} $$ If $Av = 0$, that means that $$ v = [v_3 - 2v_5, -4v_3 + 3v_5, v_3, -v_5, v_5]^T = v_3\cdot[1,-4,1,0,0]^T + v_5\cdot[-2,3,0,-1,1]^T $$ The two vectors are the basis of the null space.