I have a problem that asks the following:

Let $V$ be a vector space, and $B_1 = \{x_1,x_2,x_3\}$ be a basis of $V.$ Let $y_1 = x_1 + x_3, y_2 = x_2 + x_3$, and $y_3 = x_3$

a) Prove that $B_2 = \{y_1, y_2, y_3\}$ is also a basis of $V.$

b) If $x \in V,$ and the coordinates of $x$ in basis $B_1$ are $[v]_{B_1} =\left[ \begin{array}{cc} 2\\ 5\\ -1\\ \end{array} \right]$, find coordinates of $v$ in basis $B_2$.

What I know:

A set of vectors forms a basis if the vectors are linearly independent.

So essentially I need to prove that $B_2$ meets this criteria right?

We have $$\alpha_1 y_1 + \alpha_2 y_2 +\alpha_3 y_3 = 0$$ which is the same as saying $$\alpha_1[y_1]_B + \alpha_2[y_2]_B + \alpha_3[y_3]_B = 0$$ which is $$\alpha_1[x_1+x_3]_B + \alpha_2[x_2+x_3]_B + \alpha_3[x_3]_B = 0.$$ Correct me if I'm wrong, but I believe this suggests that the vectors are linearly independent. Therefore, $B_2$ forms a basis of $V.$

As for the given coordinate, I'm not sure how I would use it to find the coordinates in basis $B_2$. What should I be doing here?

  • $\begingroup$ Linearly independent means that those coefficients $\alpha_i$ must all be zero for the right hand side to be zero. You must show that the only linear combination of $y_1,y_2,y_3$ that yields zero is the one with zero-coefficients. $\endgroup$
    – bames
    Feb 26, 2018 at 1:50
  • $\begingroup$ Note proper MathJax usage as in my edits to the question. $\endgroup$ Feb 26, 2018 at 2:10

3 Answers 3


To prove that $\{y_1,y_2,y_3\}$ is a basis, it suffices to show that it is independent. To this end, suppose $c_1y_1+c_2y_2+c_3y_3=0$, for some $ c_1,c_2,c_3 $ in the underlying field. Then we have $c_1 (x_1+x_3)+c_2 (x_2+x_3)+c_3y_3=0$ which implies $c_1x_1+c_2x_2+(c_1+c_2+c_3)x_3=0$ that leads to $c_1=c_2=c_3=0$,

i.e $\{y_1,y_2,y_3\} $ is independent and consequently is a basis. For the second part, first notice that $x_3=y_3,x_2=y_2-y_3, x_1=y_1-y_3$. The latter fact with $v=2x_1+5x_2-x_3$ implies $v=2y_1+5y_2-9y_3$.


For the first half: Not quite, but it's pretty close. You still need to distribute the scalars and recombine everything: $$ \alpha_1[x_1+x_3]_B + \alpha_2[x_2+x_3]_B + \alpha_3[x_3]_B = 0 \\ [\alpha_1x_1+\alpha_1x_3]_B + [\alpha_2x_2+\alpha_2x_3]_B + [\alpha_3x_3]_B = 0 \\ [\alpha_1x_1+\alpha_1x_3 + \alpha_2x_2+\alpha_2x_3 + \alpha_3x_3]_B = 0 \\ [\alpha_1x_1 + \alpha_2x_2 + (\alpha_1 +\alpha_2 + \alpha_3)x_3]_B = 0 \\ \alpha_1[x_1]_B + \alpha_2[x_2]_B + (\alpha_1 +\alpha_2 + \alpha_3)[x_3]_B = 0 \\ $$

And now, because the $x_i$ form a basis of $V$, we may conclude that: $$ \alpha_1 = \alpha_2 = \alpha_1+\alpha_2+\alpha_3 = 0 \\ 0 + 0 + \alpha_3 = 0 \\ \alpha_3 = 0 $$

Which shows that yes, the $y_i$ are linearly independent and therefore a basis of $V$.

As for the given coordinate, I'm not sure how I would use it to find the coordinates in basis $B_2$.

Remember what the question actually means. We're being asked to find a triplet of scalars, $\beta_i$, where: $$ 2x_1 + 5x_2 - 1x_3 = \beta_1(x_1 + x_3) + \beta_2(x_2 + x_3) + \beta_3x_3. $$ By inspection, $\beta_1$ must be 2, and $\beta_2$ must be 5, because those are the only terms which contain $x_1$ and $x_2$ respectively. For concreteness, you might imagine $[x_1, x_2, x_3] = [\hat{i}, \hat{j}, \hat{k}]$, so that it is "visually obvious" why this needs to be so (because the $\beta_1$ term is the only one with an $x$ component, and the $\beta_2$ term is the only one with a $y$ component, so if you want the components on the left to equal the corresponding components on the right, their respective coefficients must match).

Now that we have those picked out, the remaining algebra may serve as an exercise for the reader (Hint: Substitute $\beta_1$ and $\beta_2$ back into the equation, expand, and cancel like terms).


Check your teacher's/instructor's definitions but for $B_2$ to be a base for $V$ you should need every vector in $V$ to be decomposable into a linear combination of the vectors in $B2$. If that was not the case, $B2$ could still be a base for some sub-space of $V$. There is a different theorem to state that if 3 vectors are linearly independent and non-zero then they form a basis for a 3-dimensional vector space, but don't confuse theorems with definitions.

Having said that, I believe you are on the right track, but your tried thinking a bit backwards. Let's do it forward:

If $B_1$ is a base for $V$, then for every $v\in V$ there exists constants $\alpha_1,\alpha_2$, and $\alpha_3$ such that:

$$ v=\alpha_1 x_1+ \alpha_2 x_2 +\alpha_3 x_3 $$

And it should be possible to show the same for $B_2$ with constants $\beta_1,\beta_2$, and $\beta_3$. Note that taking the previous equation is possible to rearrange some terms: $$ v=\alpha_1 (x_1+x_3)+ \alpha_2 (x_2+x_3) +(\alpha_3-\alpha_1-\alpha_2) x_3 $$

Then by setting: $$ \alpha_1= \beta_1 $$

$$ \alpha_2= \beta_2 $$

$$ \alpha_3-\alpha_1-\alpha_2= \beta_3 $$

It is shown that for every vector $v\in V$: $$ v=\beta_1 y_1+ \beta_2 y_2 +\beta_3 y_3 $$

As for the second part, the coordinates you were given are the $\alpha$ values, so plugging them in the equations above yields the coordinates in $B_2$, which will be $[v]_{B_2}=[2,5,-8]^T$.

You'll probably learn better methods soon enough though.


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