Question

Find the matrix B of T with respect to the basis B = {~u1, ~u2}, where u1 is a unit vector that points along y = x toward increasing x and y and ~u2 is a unit vector that is perpendicular to ~u1 and 90 degrees counterclockwise from ~u1.

Answer 1

Answer:

The matrix B is given as

$B=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\end{array}\right]$

Step-by-step explanation:

As the vectors are given by u1 and u2 where u1 is given as

As y=x thus $\theta=tan^{-1}(\frac{y}{x})=tan^{-1}{1}=\pi/4$

So the value of x and y is given as

Thus u_1 is given as

$u_1=\left[\begin{array}{c}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{array}\right]$

Now the vector u_2 is given as perpendicular to u_1 i.e.

$u_1.u_2=0\\\left[\begin{array}{c}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{array}\right].\left[\begin{array}{c}u_{2x}\\u_{2y}}\end{array}\right]=0\\u_{2x}=-\frac{1}{\sqrt{2}}\\u_{2y}=-\frac{1}{\sqrt{2}}\\\\u_2=\left[\begin{array}{c}-\frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}}\end{array}\right]$

So the B matrix is given as

$B=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\end{array}\right]$

So the matrix B is as above.