Question

Given vectors v=[51], b1=[11] and b2=[11] all written in the standard basis, what is v in the basis defined by b1 and b2? You are given that b1 and b2 are orthogonal to each other.

Answer 1

Answer:

&Recall what the matrix [\, T\,]_B is:

If you write a vector x\in\Bbb R^2 in terms of the basis B=$\{\,b_1,b_2\,\}$ x=$\alpha_1 b_1+\alpha_2 b_2,$ then if you multiply [\, T\,]_B by the coordinate vector x_B=$\bigl[{\alpha_1\atop\alpha_2}\bigr]$ you get the coordinate vector of T(x) with respect to B. That is $\tag{1} [T(x)]_B$ = $[\,T\,]_B x_B$

Now the columns of the matrix [\,T\,]_E where E=$\{e_1,e_2\}$ are the vectors T(e_1) and T(e_2). To find these vectors, we can use the matrix [\, T\,]_B. There are three steps involved here. Considering the vector e_2, we have to

Find the coordinates of e_2 with respect to the basis B.

Find the coordinates of T(e_2) with respect to the basis B.

Find T(e_2) expressed in the standard basis.

Step 1: For e_2=(0,1), we first find the coordinates of e_2 in terms of the basis B. Towards this end, we have to solve the system $\Bigl[{0\atop1}\Bigr]$= $\alpha_1 \Bigl[{-1\atop-3}\Bigr] +\alpha_2\Bigl[{-3\atop-10}\Bigr]$. Doing so gives: $\alpha_1=3,\quad \alpha_2$=-1 The coordinate vector of e_2 with respect to B is $\bigl[{3\atop-1}\bigr]$.

Note we could have done this differently: the coordinate vector $\bigl[{\alpha_1\atop\alpha_2}\bigr]$ of x with respect to B satisfies $[b_1 \ b_2] \bigl[{\alpha_1\atop\alpha_2}\bigr]$=x; so $\bigl[{\alpha_1\atop\alpha_2} \bigr]$=$[b_1 \ b_2]^{-1} x$ Thus we could have found $[b_1\ b_2]^{-1}$ and just multiplied this by e_2. This is actually preferable, since we can use the inverse when considering e_1 later.

Step 2: Using (1) now, the coordinate vector of T(e_2) with respect to B is $[\,T\,]_B (e_2)_B = \Bigl[ \matrix{4&4\cr 4&5 }\Bigr]\Bigl[{3\atop -1} \Bigr] = \Bigl[{8\atop 7} \Bigr]$.

Step 3: But note that T(e_2) is not the vector $\bigl[{8\atop 7} \bigr]$; this vector gives the coordinates of T(e_2) with respect to the basis B. In general, if x_B is the coordinate vector of x with respect to B, then x=$[b_1\ b_2] x_B$; so T(e_2)=$[b_1\ b_2]\Bigl[{8\atop 7} \Bigr] =8\, b_1+7\, b_2$= $8\Bigl[{-1\atop -3} \Bigr] +7 \Bigl[{-3\atop -10} \Bigr]$ =$\Bigl[{-29\atop -94} \Bigr]$

Thus, the second column of [\,T\,]_E is $\bigl[{-29\atop -94} \bigr]$

To find the first column of [\,T\,]_E, apply the same procedure to the vector e_1. The first step here would be to write e_1 in terms of the basis B. To do that, you need to solve the system $\Bigl[{1\atop0}\Bigr]$= $\alpha_1 \Bigl[{-1\atop-3}\Bigr] +\alpha_2\Bigl[{-3\atop-10}\Bigr]$. or compute: $\Bigl[{\alpha_1\atop\alpha_2} \Bigr]$=$[b_1 \ b_2]^{-1} e_1$