# How basis of kernel relates to basis of range

I have the following question that I am trying to figure out:

Suppose that $A$ is a $m \times n$ matrix of rank $r$ and $\{\vec{v}_{1}, \vec{v}_{2},...,\vec{v}_{n}\}$ is an orthogonal basis for $\mathbb{R}^{2}$. Show that if $\{ \vec{v}_{r+1}, ... , \vec{v}_{n}\}$ is a basis for ker($A$), then $\{A\vec{v}_{1}, ... ,A\vec{v}_{r}\}$ is a basis for the range.

My first thought is to use the Fundamental Theorem of Linear Algebra $$\dim(\ker(A)) = n - r \qquad \dim(\text{rng}(A)) = \text{rank}(A) = r$$ However, I'm not really sure how this theorem applies to the basis of both the range and kernel of the matrix.

• Why don't you just show that $Av_1,...,Av_r$ are linearly independent and span the range of $A$? – copper.hat May 5 '15 at 14:55
• But how do I show linear independence of $Av_{1}, ...Av_{r}$? The only ways I know how are by taking the determinant and by putting a matrix into row-echelon form and seeing if there are no free variables. – user238081 May 5 '15 at 14:57
• Just use the definitions. – copper.hat May 5 '15 at 14:57
• I think you meant "basis for $\mathbb{R}^n$". Also, $r$ vectors spanning the range of $A$ form a basis of the range since the latter has dimension $r$. – Thibaut Dumont May 5 '15 at 14:59

Suppose $w$ belongs to the range of $A$; then $w=Av$ for some $v$.

Therefore $v=a_1v_1+\dots+a_rv_r+a_{r+1}v_{r+1}+\dots+a_nv_n$ and $$w=Av=\sum_{i=1}^r a_iAv_i+\sum_{i=r+1}^n a_iAv_i=\sum_{i=1}^r a_iAv_i$$ because $Av_{r+1}=\dots=Av_n=0$.

Therefore $\{Av_1,\dots,Av_r\}$ is a spanning set of the range. Since, by the rank-nullity theorem, the range has dimension $n-\dim\ker A=n-(n-r)=r$, the set is a basis.

Since $A(\sum_{k=1}^n \alpha_k v_k) = A(\sum_{k=1}^r \alpha_k v_k)$, it is clear that $Av_1,...,A v_r$ span the range of $A$.

To see that they are linearly independent, suppose $\sum_{k=1}^r \alpha_k A v_k=A(\sum_{k=1}^r \alpha_k v_k)=0$, then $\sum_{k=1}^r \alpha_k v_k \in \ker A$, which means $\sum_{k=1}^r \alpha_k v_k = \sum_{k=r+1}^n \beta_k v_k$ for some $\beta_k$, (take the right hand side to be zero if $r=n$) which implies that $\alpha_k = 0$ for all $k$, since the $v_k$ are linearly independent.

The fact that the $v_k$ are orthonormal is irrelevant here.

• One question. Then, the problem is wrong because if n=r (something that you are using in your answer) The part of v(r+1) is wrong. May you fix the question? I guess that R2 is also wrong. – Beginner May 5 '15 at 15:44
• @Beginner: I am not relying on $n=r$, why do you think that? If $n=r$, then the kernel of $A$ is trivial which means it is invertible. It follows that $Av_k$ are linearly independent, hence a basis for the range. – copper.hat May 5 '15 at 15:58
• May you explain your first step? – Beginner May 5 '15 at 17:03
• @Beginner: $A$ is linear and $Av_k = 0$ for $k>r$. – copper.hat May 5 '15 at 17:08