# Finding a change of basis matrix when the basis vectors of one basis are in terms of the other

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!

• 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? – amd May 25 '17 at 18:43
• Instead of writing those vectors as columns, you would write them as rows. – Abteen May 26 '17 at 3:41

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.