# Change of Basis question .

Given standard basis $B$ of $\Bbb{R}^4$ and the basis $B'=((1,2)^T,(1,1)^T)$ of $\Bbb{R}^2$ and the linear tranformation $f:\Bbb{R}^4 \to\Bbb{R}^2,(x_1,x_2,x_3,x_4)^T\to(4x_1-x_4,2x_1+2x_2+2x_3-3x_4)^T$

Find a the matrix that represents this linear transformation. So my textbook does the following

$f((1,0,0,0)^T)=(4,2)^T=-2(1,2)^T+(6)(1,1)^T$

$f((0,1,0,0)^T)=(0,2)^T=2(1,2)^T+(-2)(1,1)^T$

$f((0,0,1,0)^T)=(0,2)^T=2(1,2)^T+(-2)(1,1)^T$

$f((0,0,0,1)^T)=(-1,-3)^T=-2(1,2)^T+(1)(1,1)^T$

Thus the matrix A representing f is $$\begin{pmatrix} -2 & 2 & 2 & -2\\ 6 & -2 & -2 & 1 \\ \end{pmatrix}$$

I was wondering how this works in terms of matrix multiplication,

Intuitevely $$\begin{pmatrix} 4 & 0 & 0 & -1\\ 2 & 2 & 2 & 3 \\ \end{pmatrix}$$ sends the standard basis vectors of $\Bbb{R}^4$ to standard basis of $\Bbb{R}^2$

So now to get the represantation of this matrix in terms of Basis B' we multiply this matrix with

$$\begin{pmatrix} 1 & 1\\ 2 & 1 \\ \end{pmatrix}$$

But the results arent the same..., It turns out that multiplying with the inverse of this matrix gives the same result, Can you please explain why this happens and given an intuition ???

• You've got a typo. The matrix $\begin{pmatrix} -2 & 2 & 2 & -2\\ 6 & -2 & -2 & 1 \\ \end{pmatrix}$ should be $\begin{pmatrix} -2 & 2 & 2 & -2\\ 6 & -2 & 6 & 1 \\ \end{pmatrix}$. – j0equ1nn Mar 16 '17 at 9:41
• fixed the mistake thanks – asddf Mar 16 '17 at 9:42
• I think you mean something else when you say "sends the standard basis vectors of $\mathbb{R}^4$ to standard basis of $\mathbb{R}^2$". A matrix like that would be, for instance, $\begin{pmatrix}1&0&1&0\\0&1&0&1\end{pmatrix}$. – j0equ1nn Mar 16 '17 at 9:46
• but with respect to the linear transformation f ? – asddf Mar 16 '17 at 9:47
• That phrase doesn't really make sense in this context. Can you say exactly what you mean? – j0equ1nn Mar 16 '17 at 9:49

Because $\pmatrix{1 & 1 \\ 2 & 1}$ lets you to send a vector in basis $B'$ to canonical (standard) basis. Example: $\pmatrix{1 & 1 \\ 2 & 1} \pmatrix{1 \\ 0} = \pmatrix{1 \\ 2}$, representation of first basis vector of $B'$ in canonical basis. You need just the opposite, hence the inversion.
• but you are sending $(1,0)^T$ to $(1,2)^T$ doesnt this mean you convert it to B' basis ? – asddf Mar 16 '17 at 9:43
• No. Let's take space $\mathbb{R}^1[x]$ (linear polynomials) to clearly distinguish between vector and its representation as a ($n \times 1$) matrix. We have canonical basis $\{e_1 = 1, e_2 = x\}$; we have basis $B' = \{b_1 = 1+2x, b_2 = 1+x\}$. We have vector $v = 1+2x$ (which is the same no matter what the basis). In basis $B'$, $v = 1b_1+0b_2 \sim (1,0)^T$. In canonical basis, $v = 1 e_1 + 2 e_2 \sim (1,2)^T$. – Abstraction Mar 16 '17 at 9:54
• @asddf For the same example, if you take vector $u$ corresponding to $(1,0)^T$ in canonical basis, it would be $u = 1e_1+0e_2 = 1$ and its representation in $B'$ would be $u = 2b_2 - b_1 \sim (-1, 2)^T$. Note that $1 \times n$ matrix itself doesn't describe a vector, you need to keep track what basis you wrote that matrix representation in. – Abstraction Mar 16 '17 at 10:11