Given that $u_1,u_2,v_1,v_2\in \mathbb{R}^n$ where $\mathbb{R}^n$ denotes row vector space over $\mathbb{R}$.
Also, given that $v_1,v_2$ are linearly independent and $A=u_1^tv_1+u_2^tv_2$.
Prove that $C(A)=\langle u_1^t,u_2^t \rangle$ where $C(A)$ denotes the column space of $A$.

I write $v_1=[v_{11}\; \dots \; v_{1n}]$ and $v_2=[v_{21}\; \dots \; v_{2n}]$.
Then I get $A=[v_{11}u_1^t+v_{21}u_2^t\; \dots \; v_{1n}u_1^t+v_{2n}u_2^t]$.

Let $y\in C(A)$. Then $y=Ax$ for some $x\in \mathbb{R}_n$.
Write $x=(x_1 \; \dots \; x_n)^t$.
Then $y=(x_1v_{11}+\dots+x_nv_{1n})u_1^t+\dots+(x_1v_{21}++x_nv_{2n})u_2^t$.
Hence $y\in \langle u_1^t,u_2^t \rangle$.

Now let $y\in \langle u_1^t,u_2^t \rangle$.
Then $y=c_1u_1^t+c_2u_2^t$ for some $c_1,c_2\in \mathbb{R}$.
Suppose that $Ax=y$ where $x\in \mathbb{R}_n$.
Write $x=[x_1 \; \dots \; x_n]^t$
Then $(x_1v_{11}+\dots+x_nv_{1n})u_1^t+(x_1v_{21}+\dots+x_nv_{2n})u_2^t=c_1u_1^t+c_2u_2^t$.
I want to show that there is a solution for $x$ so that I can conclude $y\in C(A)$. I think this should be related to the fact that $v_1,v_2$ are linearly independent.

  • $\begingroup$ Presumably, your vectors are row-vectors, as opposed to column vectors. Note that the more common convention is the opposite. That is, for most people, the elements of $\Bbb R^n$ are column-vectors. For future problems, specify that you're referring to row-vectors to avoid confusion. $\endgroup$ – Omnomnomnom Apr 6 '16 at 15:37
  • $\begingroup$ @Omnomnomnom Ok noted. Thanks. $\endgroup$ – Alan Wang Apr 6 '16 at 15:45

Here's the quick approach: we can write $$ u_1^Tv_1 + u_2^T v_2 = \underbrace{\pmatrix{u_1^T&u_2^T}}_{A} \underbrace{\pmatrix{v_1 \\ v_2}}_{B} = AB $$ Now, since $C(B)=\Bbb R^2$ (which is true since $v_1,v_2$ are independent), $C(AB) = C(A)$. The conclusion follows.

  • $\begingroup$ How to know that $B$ is a square matrix? Otherwise we cannot determine whether $B$ is invertible right? $\endgroup$ – Alan Wang Apr 7 '16 at 7:36
  • $\begingroup$ My mistake, see my edit. $B$ has full column rank. $\endgroup$ – Omnomnomnom Apr 7 '16 at 12:09
  • $\begingroup$ "How to know that B is a square matrix? Otherwise we cannot determine whether B is invertible right?" That's strangely worded. If B is not a square matrix, then we know immediately that is NOT invertible. $\endgroup$ – user247327 Apr 7 '16 at 12:12

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