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Let A be a real $3\times3$ matrix which of the following conditions does not imply that A is invertible?

(D) The set of all vectors of the form $Av$, where $v \in \mathbb R^3$, is $\mathbb R^3$.

(E) There exist 3 linearly independent vectors $v_1$, $v_2$, $v_3 \in \mathbb R^3$ such that $Av_i\neq 0$ for each $i$.

The answer is E. I see why E is the answer by coming up with a counter example. But why does D imply that A is invertible? Also, how to solve this problem in a more efficient way (instead of coming up with a counter example)?

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$A$ is invertible iff the rank is 3. The rank is the dimension of the image. So D) says that the rank is $3$. For E) there may be three vectors with nonzero image, but they may all be sent to the same vector so it doesnt guarantee that the image has dimension $3$.

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  • $\begingroup$ Thanks. It may be really basic, but just wanna clarify: so when we say a vector is $R^n$, we are saying the rank is $n$; and it's different from saying a vector is in $R^n$? $\endgroup$
    – Chu Ma
    Sep 11, 2016 at 1:21
  • $\begingroup$ The first doesnt make any sense. A SET of vectors can SPAN $n$ dimensional space, in which case the rank of the set will be $n$. $\endgroup$
    – Rene Schipperus
    Sep 11, 2016 at 1:25
  • $\begingroup$ Thanks a lot! I overlooked the "set" part in the problem. $\endgroup$
    – Chu Ma
    Sep 11, 2016 at 1:35
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Let $e_1 = (1,0,0)^T, e_2 = (0,1,0)^T, e_3 = (0,0,1)^T$, D implies the existence of $v_1, v_2, v_3$ such that $Av_1 = e_2, Av_2 = e_2, Av_3=e_3.$ If $V = \left[ \begin{matrix} v_1 & v_2 & v_3 \end{matrix} \right]$ then $AV = I$ which means $A$ is invertible and $V$ is the inverse.

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