1
$\begingroup$

I'm trying to solve this problem. V is a vector space with bases:

$$B = \lbrace v_1, v_2, v_3\rbrace\quad \text{and}\quad B' = \lbrace w_1, w_2, w_3\rbrace.$$

Assume:

\begin{align*} v_1 &= w_1 + w_2 + 3w_3 \\ v_2 &= 2w_1 + 3w_2 + 2w_3\\ v_3 &= 3w_1 + 4w_2 + 4w_3 \end{align*}

Intuitively, I think that this represents the matrix equation:

$$ \begin{bmatrix} 1&1&3\\2&3&2 \\ 3&4&4\\ \end{bmatrix} \begin{bmatrix} w_1\\w_2\\w_3\\ \end{bmatrix} = \begin{bmatrix} v_1\\v_2\\v_3\\ \end{bmatrix} $$

And therefore the matrix $$ \begin{bmatrix} 1&1&3\\2&3&2 \\ 3&4&4\\ \end{bmatrix}$$ is the transformation matrix from $B' \to B$. However by the definition of a transformation matrix (columns being the basis vectors), the matrix should be the transpose $$ \begin{bmatrix} 1&2&3\\1&3&4 \\ 3&2&4\\ \end{bmatrix}.$$

Which of these is right? And why?

Thank you!

$\endgroup$
  • $\begingroup$ What do you mean by $\begin{bmatrix}w_1&w_2&w_3\end{bmatrix}^T$ and $\begin{bmatrix}v_1&v_2&v_3\end{bmatrix}^T$, whose elements are themselves vectors? $\endgroup$ – amd May 25 '17 at 18:43
  • $\begingroup$ Instead of writing those vectors as columns, you would write them as rows. $\endgroup$ – Abteen May 26 '17 at 3:41
0
$\begingroup$

Let's revisit the theorem relating to change of basis once again.We will assume the field to be $\mathbb{R}$

Suppose $P$ is an $n\times n$ invertible matrix over $\mathbb{R}.$ Let $V$ be an $n$-dimensional vector space over $\mathbb{R},$ and let $\scr C$ be an ordered basis of $V.$ Then there is a unique ordered basis $\scr D$ of $V$ such that

\begin{align} \tag{1}[\alpha]_{\scr C} = P[\alpha]_{\scr D} \end{align} for every vector $\alpha \in V.$

$\Big($ $[\alpha]_{\scr B}$ is the $\color{red}{\text{coordinate matrix}}$ of $\alpha$ relative to the ordered basis $\scr B.$ $\Big)$

Here the columns of $P$ are the $\color{red}{\text{coordinates}}$ of the basis vectors of $\scr D$ with respect to the ordered basis $\scr C.$

So note that $\color{red}{\text{coordinate matrices}}$ in your case will always be $3\times 1$ matrices with entries from $\mathbb{R}.$

Suppose in your case I consider the ordered basis $B.$ A vector $\alpha \in V$ with coordinate matrix $\begin{bmatrix} 2\\5\\3\\ \end{bmatrix}$ means $\alpha =2v_1+5v_2+3v_3. $

The matrix equation you wrote has NO coordinate matrices.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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