Let $A\in\mathbb{M}_{m\times n}$. Prove:
(a) Let $\{\mathbf{v}_1,\ldots,\mathbf{v}_p\}\subseteq\mathbb{R}^n$. If $\{A\mathbf{v}_1,\ldots,A\mathbf{v}_p\}$ is linearly independent, then $r(A)\geq p$.
(b) Let $\{\mathbf{v}_1,\ldots,\mathbf{v}_p\}\subseteq\mathbb{R}^n$. If $\{A\mathbf{v}_1,\ldots,A\mathbf{v}_p\}$ is a spanning set for $\mathbb{R}^m$, then $r(A)=m$.
(c) Assume $r(A)=m$. If $\mathrm{Span}\{\mathbb{v}_1,\ldots,\mathbb{v}_p\}=\mathbb{R}^n$, then $\mathrm{Span}\{A\mathbb{v}_1,\ldots,A\mathbb{v}_p\}=\mathbb{R}^m$.
Hi, so I usually use this community to post questions / answers regarding calculus, but I'm really starting to struggle in my Linear Algebra course and I would really appreciate the help!
I just can't figure out how to connect all the dots in this question.. I know that in (a) the set is in the span of ColA, but how does it help me to know that it's linearly independent? And what's the relation to RankA?
in (b), if we know the set spans $R^m$, it means colA spans $R^m$, and in order for colA to be able to span $R^m$, it has to have m linearly independent vectors, and thus dimA = m = rank(A). Is this a right way to look at it?
in (c) I didn't even attempt it after not being able to solve the first two.. I don't feel confident enough with the material. :(
(a) It's important you realize that the span of the $Av_i$ is a subspace of the column space of $A$.
Proof: Take $x$ in the span of the $Av_i$, which means $x$ is a linear combination of the $Av_i$. Then $x=\sum c_i(Av_i)$ for some coefficients $c_i$. But $c_i(Av_i)=A(c_iv_i)$, so
$$x=\sum A(c_iv_i)=A\left(\sum c_iv_i\right)$$
But the rightmost vector is just a linear combination of the columns of $A$. Hence $x$ is in the column space of $A$. QED
Now you can also use the fact that if $U$ is a subspace of $V$, then
$$\dim U\leq \dim V$$
It follows from the fact we've proven above that the dimension of the column space of $A$, aka the rank of $A$, is at least the dimension of the span of the $Av_i$, which by the hypothesis of linear independence is $p$. In other words, $r(A)\geq p$.
(b) If the $Av_i$ give a spanning set for $\mathbb{R}^m$, then the span of the $Av_i$ is equal to $\mathbb{R}^m$. Using the fact we proved above that the span of the $Av_i$ is a subspace of the column space of $A$, we conclude $\mathbb{R}^m$ is a subspace of the column space of $A$. Thus $r(A)\geq m$. You also know $r(A)\leq m$ because the row rank (which is obviously at most $m$) equals the column rank (which in turn equals the plain old rank). Thus $r(A)=m$.
(c) Note that the span of the $Av_i$ is automatically a subspace of $\mathbb{R}^m$, so all we need to do is prove the other inclusion: $\mathbb{R}^m$ is a subspace of the span of the $Av_i$. Well, take a vector $x\in\mathbb{R}^m$. We want to show that $x$ is in the span of the $Av_i$.
Since $A$ has rank $m$, the column space has dimension $m$ and thus, being a subspace of $\mathbb{R}^m$, is in fact equal to $\mathbb{R}^m$. Therefore $x$ is a linear combination of the columns of $A$, say $x=Av$ for some $v\in\mathbb{R}^n$.
Now, using the hypothesis that the span of the $v_i$ is $\mathbb{R}^n$, we conclude $v$ can be written as a linear combination of the $v_i$, say $v=\sum c_iv_i$. Thus
$$x=Av=A\left(\sum c_iv_i\right)=\sum c_i(Av_i)$$
which is a linear combination of the $Av_i$. Thus $x$ is in the span of the $Av_i$.