If $\{v_1,\ldots,v_r\}$ span the row space of some $m\times n$ matrix $A$, and $\{v_{r+1},\ldots,v_n\}$ span the null space of $A$, I think while the union of the two subspaces is not necessarily a vector space, however, if I just take the union of the basis of the two subspaces, is it correct that I now have $n$ independent vectors and hence this set would span $\mathbb{R}^n$?
For an $m\times n$ matrix $A$, does the union of the basis of the row space and null space of $A$ span $\mathbb{R}^n$
linear algebravector-spaces
Related Solutions
This matrix reduces, through row reduction, to $\begin{pmatrix}1 & -1 & 0 & 0 \\ 0 & -1 & -4 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$. The row space has the three basis vectors, (1, -1, 0, 0), (0, -1, -4, 0), and (0, 0, 0, 1), so dimension 3. The column space has the three basis vectors (1, 0, 0, 0), (-1, -1, 0, 0), and (0, 0, 1, 0). (0, -4, 0, 0) is not independent because (0, -4, 0, 0)= 4(-1, -1, 0, 0)+ 4(1, 0, 0, 0). The column space has dimension 3. That's always true- the dimension of the row space of a matrix is equal to the dimension of the column space".
(x, y, z, t) is in the "null space" if and only if $\begin{pmatrix}11 & -2 & 36 & 2 \\ -2 & 1 & -4 & 0 \\ 3 & 0 & 12 & 1 \\ 1 & -1 & 0 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z\\ t\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0 \\ 0\end{pmatrix}$. That is equivalent to the four equations 11x- 2y+ 36z+ 2t= 0, -2x+ y- 4z= 0, 3x+ 12z+ t= 0, x- y= 0. From x- y= 0, of course, y= x so the other three equations can be written 9x+ 36z+ 2t= 0, -x- 4z= 0, and 3x+ 12z+ t= 0. From -x- 4z= 0, x= -4z so the other two equations can be written 2t= 0, 0= 0, and t= 0. Clearly t= 0 but we cannot solve for numerical values of x, y, and z. We can say that (x, y, z, t)= (-4z, -4z, z)= z(-4, -4, 1) where z can be any number. That is a basis for the null space is {(-4, -4, 1)} and the dimension of the null space (the "nullity") is 1.
Note that the dimension of the null space, 1, plus the dimension of the row space, 1+ 3= 4, the dimension of the whole space. That is always true. After finding a basis for the row space, by row reduction, so that its dimension was 3, we could have immediately said that the column space had the same dimension, 3, and that the dimension of the null space was 4- 3= 1 without any more computation.
To be precise, $$N(A)=\text{Span}\{v_1,...,v_n\},$$ indeed implies that $$Av_1+\cdot +Av_n=0,$$ or even that for all $\alpha _1,...,\alpha _n\in\mathbb R$ (if you work in a $\mathbb R-$vector space) that $$\alpha _1Av_1+...+\alpha _nAv_n=0.$$
But the converse doesn't hold. If $N(A)$ has $n$ dimension, that $v_1,...,v_n\in N(A)$ and that there are free, then indeed $N(A)=\text{Span}\{v_1,...,v_n\}$. But $Av_1+...+Av_n=0$ just allow you to conclude that $v_1+...+v_n\in N(A)$ and unfortunately nothing more.
Best Answer
The answer is yes. In particular, the row space and null space are orthogonal complements to each other. This is sometimes called (part of) the fundamental theorem of linear algebra.