# Linear Algebra - matrix problem

idn1984

## Homework Statement

Consider the following.
T:R3 → R3, T(x, y, z) = (2x − y + z, x + y + 2z, -x − 2y − z),
B ' = {(2, 1, 6), (1, 1, 6), (1, 1, 12)}

## Homework Equations

Find the matrix A ' for T relative to the basis of B '.

## The Attempt at a Solution

i tried to create the relevant matrix according to the known basis but without luck.
i answer should be as 3x3 matrix

Gold Member
What method are you using to find the matrix?

idn1984
i tried to set the coordinates of the basis B in T(x,y,z) and then to get the standard matrix.

any idea will be great.

Gold Member
i tried to set the coordinates of the basis B in T(x,y,z) and then to get the standard matrix.

any idea will be great.

What do you mean by set the coordinates?

Did you find the matrix for T in the usual R3 basis? Did you find the change of basis matrix from B' to the usual R3 basis?

idn1984
no, i don't know how to do that...

Gold Member
Ah, I see. Change of basis etc. is a fairly large topic. Are you aware that there are two matricies, which are inverses of each other, that map between the basis?
I.e. if you have two basis, B1 and B2, you will have two matrices $$A_{12}$$ and $$A_{21}$$ such that
$$A_{12}B_1^T=B_2$$
and
$$A_{21}B_2^T=B_1$$

And that if you have a point P1 described in B1, then the same point in terms of B2, P2, is given by
$$A_{21}P_1^T=P_2$$

Have you covered this kind of thing?

Now if the above is the case, then finding B' in terms of {(1,0,0),(0,1,0)(0,0,1)} is fairly trivial. It's quite clear to see that the columns of the matrix will be the transposes of the elements of B'.

Does this make sense?

Now you could fairly easily find the mapping for T in terms of the bases {(1,0,0),(0,1,0),(0,0,1)}, then you will have to think of some way of putting that in terms of B'.

Do you know how to find the matrix of T in terms of {(1,0,0),(0,1,0)(0,0,1)}?

idn1984
not really , can you be more specifically?

thanks

Gold Member
Oh dear. Well, I will try to explain it as well as I can, but (see my signature) I'm not the worlds best writer. I assume you have notes on the topic, you should give them a good and thorough read through, checking out the examples of what basis are, change of basis matrix, etc. building up to this question.

Primarily, do you understand what a basis is?
A basis is a spanning set that is linearly independent. If every element of R3 can be described as a (not neccasarily unique) linear combination of the elements of the set, then it spans R3. Linear independence means that no element of the basis can be formed by a linear combination of the other elements of the set. Putting these together on one set means that each point in R3 can be described by a unique linear combination of the elements of the basis.

E.G., let the basis be our B', and let b1=(2, 1, 6), b2=(1, 1, 6), b3=(1, 1, 12), then a point can be described as 3b1+4b2-2b3. This is the point would call (6+4-2,3+4-2,18+24-24)=(8,5,18), or in B we would call 8(1,0,0)+5(0,1,0)+18(0,0,1).
Does this make sense?

A change of basis matrix is a basic operator that simply transforms the elements of one basis to another.

For instance, if we take a wee looksie at our example with B being the usual basis {(1,0,0),(0,1,0)(0,0,1)}, and B' being the basis {(2, 1, 6), (1, 1, 6), (1, 1, 12)}, then a change of basis matrix A will be the matrix that when multiplied to each of the elements of B in turn will give the elements of B'.
For instance (with ^T denoting the transpose): A(1,0,0)^T=(2,1,6)

Now, for B this matrix is trivial if you are very familiar with matrix multiplication. If A(1,0,0)^T=(2,1,6), then the left hand column of A has to be 2 1 6, correct? Then if you let A(0,1,0)^T=(1, 1, 6), the middle column must be 1,1,6. Have a wee look at this and see if you can understand why. Then find what the third column must be.

Now it is fairly trivial that the matrix that takes the elements of the basis B' to the elements of B will be the inverse of A. I'm assuming you can work out why.

Now here is where it's a little confusing: if you have a point in terms of B', say 2B1+5B2+7B3, then you can use the matrices to find out what it is in terms of B. You do this by simply by multiplying (2,5,7) by the matrix that maps the desired basis to the basis the point is in terms of - in this case you have a point in terms of B', so you will have to use the matrix that maps B to B'! This is the matrix that maps the basis in the opposite direction!
If A maps B to B', and A' maps B' to B, then to find 2B1+5B2+7B3 in terms of B, you must use the matrix A, ie A(2,5,7)^T=(...), where ... is in terms of B.

Does this make sense so far? Hope it's helping.