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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 don't know what it wants me to do beyond this point however. Could someone please explain what a transition matrix is?

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  • $\begingroup$ Do you perhaps know what change of basis matrix is? $\endgroup$ – user160738 Mar 4 '16 at 13:19
  • $\begingroup$ $A_1$ looks incorrect to me. $\endgroup$ – amd Mar 4 '16 at 20:00
  • $\begingroup$ @amd I just rechecked it and can't find a mistake. $R_2 \to R_2-3R_1$ then $R_3 \to R_3 +R_2$ gives the reduced matrix. $\endgroup$ – GGG Mar 6 '16 at 3:02
  • $\begingroup$ The reduced matrix isn’t the issue. The original matrix is. Its first and third columns don’t match any of the given vectors, so $\mathbf A_1$ certainly isn’t “the vectors of the subset $B1$ as the columns of a matrix.” $\endgroup$ – amd Mar 6 '16 at 3:16
  • $\begingroup$ @amd, Thanks for catching that! I should have said $$\mathbf{A}_1: = \begin{pmatrix} 1&2&4 \\ 2&3&6 \\ 1&0 &1 \end{pmatrix} \to \begin{pmatrix} 1&2&4 \\ 0&-1&-2 \\ 0&0 &1 \end{pmatrix} \implies \det(\mathbf{A}_1) =(1)(-1)(1) = -1$$ $\endgroup$ – GGG Mar 6 '16 at 3:40
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The given vector $x$ $$ x = 4 e_1+ 4 e_3 $$ can be represented according the first or the second basis too. \begin{align} x &= x_1^{(0)} b_1^{(0)} + x_2^{(0)} b_2^{(0)} + x_3^{(0)} b_3^{(0)} = (x_1^{(0)},x_2^{(0)},x_3^{(0)})^T = [x]_0 \\ &= x_1^{(1)} b_1^{(1)} + x_2^{(1)} b_2^{(1)} + x_3^{(1)} b_3^{(1)} = (x_1^{(1)},x_2^{(1)},x_3^{(1)})^T = [x]_1 \end{align} This gives a set $[x]_i$ of coordinates, a coordinate vector, for each basis. You are asked to give the matrix $T$, which would transform the first set into the second set of coordinates. $$ [x]_1 = T [x]_0 $$

Solution:

From $B_0$ to standard basis $e_i$ we get via the matrix $$ A_0 = (u_0, u_2, u_3) = \begin{pmatrix} 1 & 2 & 4 \\ 2 & 3 & 6 \\ 0 & 0 & 1 \end{pmatrix} $$ From $B_1$ to standard basis we get via the matrix $$ A_1 = (u_1, u_2, u_3) = \begin{pmatrix} 1 & 2 & 4 \\ 2 & 3 & 6 \\ 1 & 0 & 1 \end{pmatrix} $$ From $B_0$ to $B_1$ we get by going from $B_0$ to standard basis via $A_0$ and from standard basis to $B_1$ via $A_1^{-1}$: \begin{align} T &= A_1^{-1} A_0 \\ &= \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ -1 & 0 & 1 \end{pmatrix} \end{align} We have \begin{align} [x]_0 &= A_0^{-1} [x] \\ &= \left( \begin{array}{rrr} -3 & 2 & 0 \\ 2 & -1 & -2 \\ 0 & 0 & 1 \end{array} \right) \begin{pmatrix} 4 \\ 0 \\ 4 \end{pmatrix} \\ &= \left( \begin{array}{rrr} -12 \\ 0 \\ 4 \end{array} \right) \end{align} and \begin{align} [x]_1 &= A_1^{-1} [x] \\ &= \left( \begin{array}{rrr} -3 & 2 & 0 \\ -4 & 3 & -2 \\ 3 & -2 & 1 \end{array} \right) \left( \begin{array}{rrr} 4 \\ 0 \\ 4 \end{array} \right) \\ &= \left( \begin{array}{rrr} -12 \\ -24 \\ 16 \end{array} \right) \end{align}

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  • $\begingroup$ Thank you! I've found two transition matrices: $$T = \begin{pmatrix} 3&1&6 \\ -1&0&-4 \\ 0&0 &1 \end{pmatrix}$$ So that $[x]_1 = T [x]_0$ And $$T' = \begin{pmatrix} 0&-1&-4 \\ 1&3&6 \\ 0&0 &1 \end{pmatrix}$$ So that $[x]_0 = T' [x]_1$ However, they map the coordinates from one non-standard basis to next, so I'm never in a position to use the given fact $x = 4 e_1+ 4 e_3$. It's possible that I completely misunderstand what's going on, however. $\endgroup$ – GGG Mar 4 '16 at 15:39
  • $\begingroup$ The matrix has to work for arbitrary $x$. The special $x$ was given for the last part of the task, calculating its coordinates regarding the $B_i$. $\endgroup$ – mvw Mar 4 '16 at 16:54
  • $\begingroup$ Thank you very much! I've been struggling with this question for hours and hours! I'll study your post, and put it in my notes! I can't thank you enough! $\endgroup$ – GGG Mar 6 '16 at 3:07
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    $\begingroup$ @GGG Notice that, since the two bases differ only in their first vectors, you can immediately write down two of the columns of the transition matrix. $\endgroup$ – amd Mar 6 '16 at 3:23

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