0
$\begingroup$

Consider the 2-dimensional vector subspace $V = \{(x, y, z) : x + y + z = 0\}$ of $\mathbb{R}^3$ , and two bases: $\alpha_1 = (1,−1,0)$, $\alpha_2 = (1,0,−1)$, and $\beta_1 = (0,1,−1)$, $\beta_2 = (1,1,−2)$.

Find the change-of-basis matrix $A$.

I (foolishly) tried the standard way from linear algebra to find it and I found that $A$ is a 2*2 matrix $\begin{bmatrix} -1 &1 \\ 2&-1 \end{bmatrix}$, which doesn't make sense since the vectors in $V$ are of $\mathbb{R}^3$ any help?

$\endgroup$

marked as duplicate by user99914, Claude Leibovici calculus Dec 16 '17 at 6:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

0
$\begingroup$

No, it does make sense.

Because the transformation matrix is not used to be multiplied by the basis vectors itselves, it is used for multiplication of coordinates of the point in the space with respect to the specified basis.

So in your example, say you have a point $P=(x,y,z)$ which satisfies $x+y+z=0$, then the coordinates of P wrt first and second basis are say, $(a,b)$ and $(c,d)$ respectively, that is, $P = a*\alpha_1 + b*\alpha_2 = c*\beta_1 + d*\beta_2$

then you have

$\begin{bmatrix} -1 &1 \\ 2&-1 \end{bmatrix}$ $\begin{bmatrix} a \\ b \end{bmatrix}$ = $\begin{bmatrix} c \\ d \end{bmatrix}$

$\endgroup$
0
$\begingroup$

finding change of coordinate from $\beta$ to $\alpha$ $[I_V]^\beta_\alpha$ $$\begin{aligned} \alpha_1&=b_1\beta_1+b_2\beta_2 =\begin{pmatrix}1 \\ -1 \\ 0 \end{pmatrix} = b_1 *\begin{pmatrix}0 \\ 1 \\ -1 \end{pmatrix} +b_2*\begin{pmatrix}1 \\ 1 \\ -2 \end{pmatrix} \end{aligned}$$ So, $$ \begin{bmatrix}1 \\-1\\0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 1 \\-1 & -2 \end{bmatrix} \begin{bmatrix} b_1 \\b_2\end{bmatrix}$$ we can append $\alpha_2$ we need the row reduced echolon form so
$$rref\left ( \begin{bmatrix} 0 & 1&|& 1 &1\\ 1 & 1 &|&-1 &0\\-1 & -2 &|&0&-1 \end{bmatrix} \right ) = \begin{bmatrix} 1 & 0&|& -2 &-1 \\ 0 & 1 &|&1 &1 \\0 & 0 &|&0&0 \end{bmatrix}$$ Making $$[I_V]^\beta _\alpha = \begin{pmatrix} -2& 1 &0 \\-1& 1 &0 \end{pmatrix}$$ $$rref(\alpha_1,\alpha_2:\beta_1,\beta2) =\begin{bmatrix} 1 &0 & | &-1 & -1 \\0& 1 &| &1 &2 \\ 0 &0 &| &0 &0 \end{bmatrix}$$ making $$[I_V]^\alpha_\beta= \begin{pmatrix} -1 &1 &0 \\ -1& 2& 0\end{pmatrix}$$ knwowing that can also help use other theorems and learn things about the problem

$\endgroup$
  • 2
    $\begingroup$ How can a non-square matrix be the correct answer? $\endgroup$ – amd Mar 8 '16 at 19:04
  • $\begingroup$ @amd does change of coordinate matrix have to be square? I'll rather get something wrong here than in an exam. $\endgroup$ – Tiger Blood Mar 9 '16 at 18:25
  • $\begingroup$ It’s an endomorphism, i.e., a mapping from a vector space to itself, so its matrix is going to be square. What’s not entirely clear to me from the problem as it is posed (and all of the clones of it that regularly appear here) is whether they’re looking for a $2\times 2$ matrix or a $3\times 3$. See corbah’s answer for why a $2\times 2$ matrix makes sense here even though $\mathbb R^3$ is three-dimensional. $\endgroup$ – amd Mar 10 '16 at 2:11

Not the answer you're looking for? Browse other questions tagged or ask your own question.