# Calculate the coordinates with respect to the basis B given the coordenates of that vector with respect to another basis

first time I post. I can't solve an exercise that I know it's easy, It's so frustrating...

First of all: Consider the basis $B_1 =$ {$u_1, u_2, u_3$} in $\mathbb R^3$.

1) Prove that the set $B_2 =$ {$v_1, v_2, v_3$} given by $v_1 = u_1 - u_2; v_2 = u_2 + u_3; v_3 = u_1 - u_3$ is another basis.

2) (Post title) If the vector $w_1$ has coordinates (2, 1, 1) with respect to the basis $B_1$, calculate the coordinates of $w_1$ with respect to the basis $B_2$.

Well, the answer of the question 1) I think is related with the definition of "basis of $R^3$, thus with the definition of linear independence. But I can't figure up the answer smoothly.

The part 3) of the exercise is the inverse of 2) and I can solve it without any difficulty. (Or so I think)

Can you help me? Thanks!

## 1 Answer

1) If $B_1=\{\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3\}$ is a basis for $\mathbb{R}^3$, then that means the equation $\mathbf{0}=c_1\mathbf{u}_1+c_2\mathbf{u}_2+c_3\mathbf{u}_3$ has only the solution $c_1=c_2=c_3=0$.

Now, to show that $B_2=\{\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\}$ is a basis for $\mathbb{R}^3$, we can show that the equation $\mathbf{0}=k_1\mathbf{v}_1+k_2\mathbf{v}_2+k_3\mathbf{u}_3$ has only the trivial solution.

We have

$\mathbf{0}=k_1\mathbf{v}_1+k_2\mathbf{v}_2+k_3\mathbf{k}_3=k_1(\mathbf{u}_1-\mathbf{u}_2)+k_2(\mathbf{u}_2+\mathbf{u}_3)+k_3(\mathbf{u}_1-\mathbf{u}_3)$

$=(k_1+k_3)\mathbf{u}_1+(k_2-k_1)\mathbf{u}_2+(k_2-k_3)\mathbf{u}_3$

Since we know $(k_1+k_3)\mathbf{u}_1+(k_2-k_1)\mathbf{u}_2+(k_2-k_3)\mathbf{u}_3=\mathbf{0}$ has only the trivial solution, we know that $k_1+k_3=k_2-k_1=k_2-k_3=0$ must hold.

Thus, $k_1=-k_3$, $k_2=k_1$, and $k_2=k_3$ must hold.

Since $k_2=k_1$ and $k_2=k_3$, then $k_1=k_3$. But we also have that $k_1=-k_3$. We know that $k_1=k_3=-k_3$ will only hold if $k_3=0$. Then we have $k_1=k_2=k_3$, thus $k_1=k_2=k_3=0$.

Since, this is the only solution that works, that tells us that $B_2=\{\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\}$ is a linearly independent subset of $\mathbb{R}^3$ containing $3$ elements (which implies $\dim B_2 = 3$). Thus, $B_2$ is also a basis for $\mathbb{R}^3$.

2) If $\mathbf{w}_1$ has coordinates $(2,1,1)$ with respect to the basis $B_1$, that means that we can write $\mathbf{w}_1$ as a linear combination of the elements of $B_1$ in exactly one way:

$\mathbf{w}_1=2\mathbf{u}_1+\mathbf{u}_2+\mathbf{u}_3$

One can see by inspection that $u_1=\frac{1}{2}(\mathbf{v}_1+\mathbf{v}_2+\mathbf{v}_3)$

Then from our original equations:

$\mathbf{v}_1=\mathbf{u}_1-\mathbf{u}_2$

$\mathbf{v}_2=\mathbf{u}_2+\mathbf{u}_3$

$\mathbf{v}_3=\mathbf{u}_1-\mathbf{u}_3$

We have $\mathbf{u}_3=\mathbf{u}_1-\mathbf{v}_3=\frac{1}{2}(\mathbf{v}_1+\mathbf{v}_2-\mathbf{v}_3)$ and $\mathbf{u}_2=\mathbf{v}_2-\mathbf{u}_3=\mathbf{v}_2-\frac{1}{2}(\mathbf{v}_1+\mathbf{v}_2-\mathbf{v}_3)=\frac{1}{2}(-\mathbf{v}_1+\mathbf{v}_2+\mathbf{v}_3)$

Then we can substitute:

$\mathbf{w}_1=2(\frac{1}{2}(\mathbf{v}_1+\mathbf{v}_2+\mathbf{v}_3))+(\frac{1}{2}(-\mathbf{v}_1+\mathbf{v}_2+\mathbf{v}_3))+(\frac{1}{2}(\mathbf{v}_1+\mathbf{v}_2-\mathbf{v}_3))=\mathbf{v}_1+2\mathbf{v}_2+\mathbf{v}_3$ Thus, the coordinates with respect to $B_2$ are $(1,2,1)$ (There are various other ways to do this).