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Let $v_1=(1,1)$ and $v_2=(-1,1)$ vectors in $\mathbb{R}^2$. They are clearly linearly independent since each is not an scalar multiple of the other. The following information about a linear transformation $f: \mathbb{R}^2 \to \mathbb{R}^2$ is given: $$f(v_1)=10 \cdot v_1 \text{ and } f(v_2)=4 \cdot v_2$$

  1. Give the transformation matrix $_vF_v$ with respect to ordered basis $\mathcal{B}=(v_1,v_2)$

  2. Give the transformation matrix $_eF_e$ with respect to the ordered standard basis $e=(e_1,e_2)$ of $\mathbb{R}^2$

Recall that $$ \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix} $$ We need a matrix $_eF_e$ such that: $$_eF_e\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} 10 & -4 \\ 10 & 4 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 10 & 0 \\ 0 & 4 \end{bmatrix}$$ then $$_eF_e=\begin{bmatrix} 10 & -4 \\ 10 & 4 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}^{-1} =\begin{bmatrix} 10 & -4 \\ 10 & 4 \end{bmatrix}\begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix}=\begin{bmatrix} 7 & 3 \\ 3 & 7 \end{bmatrix}$$ Okay so I'm pretty sure that $$_eF_e=_eF_v \cdot _vF_v \cdot _vF_e$$ And i figured I could find $_eF_e=\begin{bmatrix} ? & ? \\ ? & ? \end{bmatrix}$ in the following equation $$\begin{bmatrix} ? & ? \\ ? & ? \end{bmatrix} \text{ } \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}= \begin{bmatrix} 10 & -4 \\ 10 & 4 \end{bmatrix} \\ \Rightarrow \begin{bmatrix} 10 & -4 \\ 10 & 4 \end{bmatrix} \text{ } \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix}= \begin{bmatrix} 7 & 3 \\ 3 & 7 \end{bmatrix} \\ \Rightarrow {}_eF_e=\begin{bmatrix} 7 & 3 \\ 3 & 7 \end{bmatrix} $$

Now, how can I find ${}_v{F}_v$? I got a feeling that I'm making it more difficult than necessary

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    $\begingroup$ From your conditions $f(v_1)=10 \cdot v_1$ $f(v_2)=4 \cdot v_2$ you can see directly that $\ {}_v{F}_v=\left[\begin{array}{cc}10&0\\0&4\end{array}\right]$, because $v_1$ and $v_2$ are eigenvectors of eigenvalues $10$ and $4$ respectively. $\endgroup$ – Hector Blandin Dec 7 '17 at 14:19
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Recall that $v_1$ and $v_2$ are linearly independent eigenvectors of $f$ since: $$f(v_1)=10\cdot v_1$$ $$f(v_2)=4\cdot v_2$$

The vectors $v_1$ and $v_2$ form a basis of $\mathbb{R}^2$ consisting of eigenvectors of $f$. So $_vF_v$ is similar to a diagonal matrix whose elements in the diagonal are the eigenvalues $\lambda_1=10$ and $\lambda_2=4$ of $f$.

So your matrix $_vF_v$ that represents $f$ w.r.t. the ordered basis $\mathcal{B}=(v_1,v_2)$ should be the diagonal form bellow: $$_vF_v=\left[\begin{array}{cc} 10&0\\ 0&4\end{array}\right]$$ Also, if you let $P$ to be the matrix whose columns are $v_1$ and $v_2$: $$P=\left[\begin{array}{c} \\ \ v_1\\ \\ \end{array} \middle\vert \begin{array}{c} \\ \ v_2\\ \\ \end{array}\right]=\left[\begin{array}{cc} 1&-1\\ &\\1& \ 1\end{array}\right]$$ then $$_{e}F_e = P \cdot{}_vF_v\cdot P^{-1}$$ or you can do this: $$(x,y)=\alpha\cdot v_{1}+\beta\cdot v_{2} =\alpha\cdot(1,1)+\beta\cdot(-1,1)=(\alpha-\beta,\alpha+\beta)$$ so $$\alpha=\frac{x+y}{2}$$ $$\beta=\frac{y-x}{2}$$ then: $$ (x,y)= \frac{(x+y)}{2}\cdot v_1 + \frac{(y-x)}{2}\cdot v_2$$ Apply $f$ in both sides to get: $$f(x,y) = \frac{(x+y)}{2}\cdot f(v_1) + \frac{(y-x)}{2}\cdot f(v_2) $$ $$f(x,y) = \frac{(x+y)}{2}\cdot 10\cdot v_1 + \frac{(y-x)}{2}\cdot 4\cdot v_2$$ $$ = 5(x+y)\cdot v_1+2(y-x)\cdot v_2$$ $$ = (5x+5y,5x+5y)+(2x-2y,2y-2x)=(7x+3y,7y+3x)=(7x+3y,3x+7y)$$ So $$f(x,y)=(7x+3y,3x+7y)$$ then $$f(1,0)=(7,3)$$ $$f(0,1)=(3,7)$$ this implies that: $$_{e}F_e = \left[\begin{array}{cc} 7&3\\ &\\3& 7\end{array}\right] $$

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  • $\begingroup$ Thanks Hector. This makes really good sense. $\endgroup$ – Alex5207 Dec 7 '17 at 14:54
  • $\begingroup$ Your welcome :) $\endgroup$ – Hector Blandin Dec 7 '17 at 14:55
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Let be $\vec v \in \mathbb{R^2}$ a vector and consider two different basis:

$$B_v:{\vec v_1,\vec v_2}$$ and $$B_w:{\vec w_1,\vec w_2}$$

Vector $\vec v $ can be represented in two ways:

$$\vec v =a_1 \cdot \vec v_1+a_2 \cdot \vec v_2 =x_1 \cdot \vec w_1+x_2 \cdot \vec w_2$$

or in matrix form:

$$\vec v =V \cdot a=W\cdot x$$

NOTE matrices V and W have the corresponding vectors of the basis as columns

Now suppose we know the components of $\vec v$ with respect to basis $B_v$ and we want to find the $x$ components of $\vec v$ with respect to basis $B_w$, thus:

$$V \cdot a=W\cdot x \implies x=W^{-1}Va=Ma$$

matrix $M=W^{-1}V$ is the matrix of change of basis from $B_v$ to $B_w$.

NOTE

This concept can be extended to vectors $\vec v \in \mathbb{R^n}$.

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