MHB How do you change the basis of vectors in $\mathbb{R}^3$?

[Guest2]

Consider the following set of vectors in $\mathbb{R}^3:$ $u_0 = (1,2,0),~ u_1 = (1,2,1), ~u_2 = (2,3,0), ~u_3 = (4,6,1)$ Explain why each of the two subsets $B_0 = \left\{u_0, u_2,u_3\right\}$ and $B_1 = \left\{u_1, u_2, u_3\right\}$ forms a basis of $\mathbb{R}^3$. If we write $[\mathbf{x}]_0$ and $[\mathbf{x}]_1$ for the coordinates of the vector $\mathbf{x}$ in terms of these two basis, find the precise transition matrix which inter-relates these two sets of coordinates. If $\mathbf{x} = 4\mathbf{e}_1+4\mathbf{e}_3$, what are $\mathbf [\mathbf{x}]_0$ and $[\mathbf{x}]_1$?

Writing the vectors of the subset $B_0$ as the columns of a matrix we have:

$\mathbf{A}_0: = \begin{pmatrix}
1&2&4 \\
2&3&6 \\
0&0 &1
\end{pmatrix} \to \begin{pmatrix}
1&2&4 \\
0&-1&-4 \\
0&0 &1
\end{pmatrix} \implies \det(\mathbf{A}_0) =(1)(-1)(1) = -1 $

As the columns of a $3 \times 3$ matrix whose determinant is non-zero, these vectors form a basis for $\mathbb{R}^3.$

Similarly, writing the vectors of the subset $B_1$ as the columns of a matrix we have:

$\mathbf{A}_1: = \begin{pmatrix}
1&1&2 \\
3&2&6 \\
0&1 &1
\end{pmatrix} \to \begin{pmatrix}
1&1&2 \\
0&-1&0 \\
0&0 &1
\end{pmatrix} \implies \det(\mathbf{A}_1) =(1)(-1)(1) = -1 $

As the columns of a $3 \times 3$ matrix whose determinant is non-zero, these vectors form a basis for $\mathbb{R}^3.$

I'm stuck changing the basis. I think one transition matrix is:

$\mathbf{M} = \begin{pmatrix} 3&1&6 \\ -1&0&-4 \\ 0&0 &1 \end{pmatrix}$

This maps $[x]_1 \mapsto \mathbf{M} [x]_0$; and the one below maps $[x]_0 \mapsto \mathbf{M}' [x]_1$

$\mathbf{M}' = \begin{pmatrix} 0&-1&-4 \\ 1&3&6 \\ 0&0 &1 \end{pmatrix}$

I know I'm supposed to multiply by $[4,0,4]^{T}$ at some point, but when can I do that if I'm always in non-standard basis?

 

[Deveno]

If: $[\mathbf{x}]_0 = a_1u_0 + a_2u_2 + a_3u_3$, and:

$[\mathbf{x}]_1 = b_1u_1 + b_2u_2 + b_3u_3$

then the transition matrix from the basis $\{u_0,u_2,u_3\}$ to the basis $\{u_1,u_2,u_3\}$ is the matrix that maps the coordinate triple $(a_1,b_2,a_3) \mapsto (b_1,b_2,b_3)$.

The trick is to pick suitable values for $a_1,a_2,a_3$ so that the matrix is easy to calculate. We know that:

$u_2 = 0u_0 + 1u_2 + 0u_3$, so our matrix maps $(0,1,0)$ to $(0,1,0)$, since the coordinates of $u_2$ in the second basis are: $(0,1,0)$ which represents $0u_1 + 1u_2 + 0u_3$.

This tells us the second column of our transition matrix $M$ is $\begin{bmatrix}0\\1\\0\end{bmatrix}$.

Similarly, the third column must be: $\begin{bmatrix}0\\0\\1\end{bmatrix}$.

So, we really only need to work to find the first column.

To do that, we need to know what $u_0$ (which is $(1,0,0)$ in the first basis) is in the second basis.

So we want to find $b_1,b_2,b_3$ such that:

$(1,2,0) = b_1(1,2,1) + b_2(2,3,0) + b_3(4,6,1)$.

This is equivalent to solving the system of equations:

$b_1 + 2b_2 + 4b_3 = 1$
$2b_1 + 3b_2 + 6b_3 = 2$
$b_1 + b_3 = 0$.

We could form a matrix, and row-reduce it, but the third equation gives us:

$b_3 = -b_1$. Subbing this back into the first two equations gives:

$2b_2 - 3b_1 = 1$
$3b_2 - 4b_1 = 2$

Multiplying the first equation by 3, and the second by -2, and adding them together gives:

$b_1 = 1$, which leads to $b_2 = 2$.

So, if we've done the arithmetic properly, we should have:

$(1,2,0) = (1,2,1) + 2(2,3,0) - (4,6,1)$, which we do.

This, then tells us our transition matrix is:

$M = \begin{bmatrix}1&0&0\\2&1&0\\-1&0&1\end{bmatrix}$

Now let's take a random vector $a_1u_0 + a_2u_2 + a_3u_3$, and make sure this works.

So I'll pick $-2u_0 + u_2 + 3u_3$, which is (in the standard coordinates):

$-2(1,2,0) + (2,3,0) + 3(4,6,1) = (-2,-4,0) + (2,3,0) + (12,18,3) = (12,17,3)$

To find the $B_1$-coordinates of this, we compute:

$\begin{bmatrix}1&0&0\\2&1&0\\-1&0&1\end{bmatrix}\begin{bmatrix}-2\\1\\3\end{bmatrix} = \begin{bmatrix}-2\\-3\\5\end{bmatrix}$

And we check that $-2u_1 - 3u_2 + 5u_3 = -2(1,2,1) - 3(2,3,0) + 5(4,6,1) = (-2,-4,-2) + (-6,-9,0) + (20,30,5) = (12,17,3)$-woah, magic!

This, unfortunately, does not address the question of how to transition from the standard basis to either $B_0$ or $B_1$, for which you will need at least one more transition matrix (I'd try the one that takes $(4,0,4) = 4e_1 + 4_3$ to its $B_0$ coordinates. The easiest way I can think of is to find out what $e_1,e_2$ and $e_3$ are in the $B_0$ basis).

 

[Guest2]

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Thank you very much. I really do appreciate your help here and elsewhere.