# Null spaces and invertible matrix

If $$A$$ and $$B$$ are $$n×n$$ matrices, show that they have the same null space if and only if $$A = UB$$ for some invertible matrix $$U$$.

I started the question by saying $$Ax = 0$$ for some vector $$x$$ in $$\text {null}(A)$$. Now I'm lost. Could someone please help me out with this question? Thank you very much.

• Are you familiar with the orthogonality relations between the fundamental subspaces? Mar 14, 2019 at 0:30
• Hi! I learned that the row space is orthogonal to null space but thats about it Mar 14, 2019 at 0:31

First, note that $$A=UB$$ for an invertible $$U$$ means that $$A$$ and $$B$$ are row equivalent. This means that $$A$$ can be obtained from $$B$$ with elementary row operations.

Next, recall that the orthogonal complement of the null space $$\operatorname{Null}(M)$$ of any matrix $$M$$ is the row space $$\operatorname{Row}(M)$$. Succinctly, this relation is written as $$\operatorname{Row}(M)=\operatorname{Null}(M)^\perp$$.

Now, in our situation, we have two same-sized matrices $$A$$ and $$B$$ satisfying $$\operatorname{Null}(A)=\operatorname{Null}(B)$$. Taking orthogonal complements gives $$\operatorname{Null}(A)^\perp=\operatorname{Null}(B)^\perp$$ which reduces to $$\operatorname{Row}(A)=\operatorname{Row}(B)$$.

Finally, the equation $$\operatorname{Row}(A)=\operatorname{Row}(B)$$ tells us that $$\operatorname{rref}(A)=\operatorname{rref}(B)$$. This means that there are elementary matrices $$\{E_1,\dotsc,E_r\}$$ and $$\{F_1,\dotsc,F_s\}$$ satisfying the equations $$E_r\dotsb E_1A = F_s\dotsb F_1 B = \operatorname{rref}(A)$$ Inverting each elementary matrix $$E_i$$ and solving for $$A$$ gives $$A=E_1^{-1}\dotsb E_r^{-1}F_s\dotsb F_1B$$ Putting $$U=E_1^{-1}\dotsb E_r^{-1}F_s\dotsb F_1$$ gives our desired equation $$A=UB$$.

I will give a proof not based on properties of matrix reduction (such as that every matrix has a unique row reduced echelon form) as I consider these rather hard to prove formally. I will however assume known that every family of linearly independent vectors can be extended to a basis of the whole space. Also given any basis $$[b_1,\ldots,b_n]$$ of $$F^n$$ (where $$F$$ is the field of scalars one is working with), and any family $$[w_1,\ldots,w_n]$$ of vectors, there is a linear map that sends each $$b_i$$ to $$w_i$$, namely the one sending an arbitrary vector written as $$\lambda_1b_1+\cdots+\lambda_nb_n$$ to $$\lambda_1w_1+\cdots+\lambda_nw_n$$.

Take a basis $$[v_1,\ldots,v_k]$$ of $$\def\null{\operatorname{null}}\null(A)$$ and extend it to a basis $$[v_1,\ldots,v_n]$$ of $$F^n$$. Then the vectors $$Av_{k+1},\ldots,Av_n$$ are linearly independent, since a relation $$0=\lambda_{k+1}Av_{k+1}+\cdots+\lambda_nAv_n$$ implies via $$A(\lambda_{k+1}v_{k+1}+\cdots+\lambda_nv_n)=0$$ that $$\lambda_{k+1}v_{k+1}+\cdots+\lambda_nv_n\in\null(A)=\operatorname{span}(v_1,\ldots,v_k)$$, which forces $$\lambda_{k+1}=\cdots=\lambda_n=0$$ since $$[v_1,\ldots,v_n]$$ is a basis. Similarly $$Bv_{k+1},\ldots,Bv_n$$ are linearly independent, since $$\null(B)=\null(A)$$. Now extend each of these linearly independent families to a basis of $$F^n$$, calling them respectively $$\def\B{\mathcal B}\B_A$$ and $$\B_B$$. Now let $$f:F^n\to F^n$$ be a linear map sending each vector of the basis $$\B_B$$ to the corresponding vector of the basis $$\B_A$$, and let $$U$$ be the matrix of$$~f$$ (with respect to the standard basis). We have in particular $$(UB)v_i=U(Bv_i)=f(Bv_i)=Av_i$$ for $$i=k+1,\ldots,n$$. But one also has $$(UB)v_i=U(Bv_i)=Av_i$$ for $$i=1,\ldots,k$$ since both side are zero. Then matrices $$UB$$ and $$A$$, giving the same results when applied to all vectors of the basis $$[v_1,\ldots,v_n]$$, must be equal matrices.

It remains to show that the matrix $$U$$ is invertible; this is so because the map $$f$$ is, its inverse being the linear map that sends each vector of the basis $$\B_A$$ to the corresponding vector of the basis $$\B_B$$.

The opposite implication is clear, since $$UBx=0\iff Bx=U^{-1}UBx=0$$.