Consider F: R^2 -> R^2 where F(1,1) = (3,-2) and F(3,-2) = (4,8). Find the matrix representation of F in the standard basis. Assuming B{(1,1),(1,-1)} is a basis, find the matrix representation of F from B to the standard basis.
Consider F: R^2 -> R^2 where F(1,1) = (3,-2) and F(3,-2) = (4,8). Find the matrix representation of F in the standard basis. Assuming B{(1,1),(1,-1)} is a basis, find the matrix representation of F from B to the standard basis.
Hey lisa92.
Hint: Remember that your matrix is 2x2 [a b; c d] and F(1,1) = (3,-2) means <a*1 + b*1, c*1 + d*1> = <3,-2> so a + b = 3 and c + d = -2.
Now you have another relationship which will give you two more equations which means four equations in four unknowns which is enough to solve for these four unknowns if the matrix is non-singular (i.e. has non-zero determinant).
suppose the matrix for F (in the standard basis) is T, where:
$\displaystyle T = \begin{bmatrix}a&b\\c&d \end{bmatrix}$.
we are given that:
$\displaystyle \begin{bmatrix}a&b\\c&d \end{bmatrix} \begin{bmatrix}1\\1 \end{bmatrix} = \begin{bmatrix}3\\-2 \end{bmatrix}$
and:
$\displaystyle \begin{bmatrix}a&b\\c&d \end{bmatrix} \begin{bmatrix}3\\-2 \end{bmatrix} = \begin{bmatrix}4\\8 \end{bmatrix}$
this gives us:
a + b = 3
3a - 2b = 4
and
c + d = -2
3c - 2d = 8
these 2 systems of two equations in two unknowns are simple enough to solve by elimination and substitution (the "high-school method"), mutliply the first equation in each system by 3 and subtract the second from the first:
3a + 3b = 9
3a - 2b = 4
3c + 3d = -6
3c - 2d = 8
so
5b = 5 --> b = 1, so a = 2
5d = -14 --> d = -14/5, so c = 4/5
thus
$\displaystyle T = \begin{bmatrix}2&1\\ \frac{4}{5}&-\frac{14}{5}\end{bmatrix}$
now finding the representation of T from the basis B = {(1,1),(1,-1)} to the standard basis is a bit of a chore. what we need is a "change of basis" matrix, P.
the idea is: P turns "B-coordinates" into "standard coordinates", and one we have THOSE, we can apply T to them.
now in "B-coordinates" (1,1) is [1,0]_{B}, which is short-hand for: 1(1,1) + 0(1,-1). so the matrix P should take (1,0) to (1,1) = 1(1,0) + 1(0,1)
((1,0) and (0,1) are the "standard basis vectors"). that is: the first COLUMN of P should be (1,1)^{T}.
similarly, P should take [0,1]_{B} to (1,-1) so the second column of P should be (1,-1)^{T}, that is:
$\displaystyle P = \begin{bmatrix}1&1\\1&-1 \end{bmatrix}$.
so the representation of F from B to the standard basis, should be:
$\displaystyle T' = TP = \begin{bmatrix}2&1\\ \frac{4}{5}&-\frac{14}{5}\end{bmatrix}\begin{bmatrix}1&1\\1&-1 \end{bmatrix} = \begin{bmatrix}3&1\\-2&\frac{18}{5} \end{bmatrix}$
let's verify that T' does what we want it to.
first, we need to express (1,1) and (3,-2) in "B-coordinates". well, (1,1) is easy: (1,1) = [1,0]_{B}.
to express (3,-2) in "B-coordinates", we COULD find P^{-1} (which is a bit of a pain), or just solve:
u(1,1) + v(1,-1) = (3,-2) for u and v (which i think is easier). this gives:
u + v = 3
u - v = -2
so 2u = 1, thus u = 1/2 and so v = 5/2. so in "B-coordinates" we have (3,-2) = [1/2,5/2]_{B}.
so we want to show that:
T'([1,0]_{B}) = (3,-2), which is obvious because that's the first column of T'.
and:
$\displaystyle \begin{bmatrix}3&1\\-2&\frac{18}{5} \end{bmatrix} \begin{bmatrix}\frac{1}{2}\\ \frac{5}{2} \end{bmatrix} = \begin{bmatrix}4\\8 \end{bmatrix}$
which means T' maps the B-coordinate versions of (1,1) and (-3,2) to what they should be (as determined by F): (3,-2) and (4,8).
to get the matrix of F from B TO B, we need to do the following:
first we use the change of basis matrix P (which turns B-coordinates into "standard" coordinates (what we're used to)).
in the standard basis, we know the matrix that works is T. so we take T of P([v]_{B}).
this is TP applied to [v]_{B.}
now that answer is in "standard coordinates", and we want out answer in B-coordinates, so we need to "undo what P does".
this means we need to use the matrix P^{-1}, and apply that to TP([v]_{B}).
so the matrix we're looking for is:
P^{-1}TP.
the change from T to P^{-1}TP is sometimes called a "similarity transform", it turns "standard basis matrix representations" into "B-coordinate matrix representations".
this rule actually holds for any 2 bases B and C, and any linear transformation T, with matrix representations [T]_{B},[T]_{C}:
if P([v]_{B}) = [v]_{C},
[T]_{B} = P^{-1}[T]_{C}P.
often, to make things "easier on the mind", textbook exercises use the standard basis for B or C. the thing about bases is: vector spaces don't care, any basis is as good as any other.
we, on the other hand, with our desire for "easy calculation" often have a preference. but that preference is a CHOICE, vector spaces don't always come with "a tailor-made basis".