Let $W$ be a vector space with $dim(W ) = 3$. Find all vectors $x ∈ W$ such that there exists some basis α for which $[x]_α$ = $\begin{bmatrix} 1 \\ 0 \\0 \end{bmatrix}$

I'm stuck on this question here, not completely sure how to go about it. Here are the thoughts I've had so far:

$Ax = \begin{bmatrix} 1 \\ 0 \\0 \end{bmatrix}$. So, I'm solving for some vector $x$ where all $A$ matrices give that matrix? Could $x$ be $\begin{bmatrix} x_1 \\ 0 \\0 \end{bmatrix}$? Do I need to somehow find $a$, the basis to solve this?

Would greatly appreciate any help on this, thanks!


1 Answer 1


The answer is: every vector other than the null vector.

If $x=0_W$, then the problem has no solution, since every coordinate of nuell vector with respect to any basis is $0$.

And if $x\ne0_W$, just take two vectors $y$ and $z$ such that $B=\{x,y,z\}$ is linearly independent. Then $B$ is a basis of $W$ and$$x=1.x+0.y+0.z.$$

  • $\begingroup$ is this around the same idea as what I wrote for $x$? $\endgroup$
    – eddie
    Dec 4, 2021 at 19:31
  • $\begingroup$ I don't see any reason to use matrices here. $\endgroup$ Dec 4, 2021 at 19:36

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