# 2 questions in lenear algebra

1. Jan 31, 2008

### transgalactic

2. Jan 31, 2008

### HallsofIvy

Staff Emeritus
You complain "I got the same vector as before"- of course you did, you just went around in circles! Do exactly what I suggested before: apply the linear transformation to each of the basis vectors, and write the result in that basis. Of course, since your original matrix is the linear transformation in the $\{u_1, u_2, u_3\}$ basis, you have to apply the matrix to the new basis vectors, $\{u_1, u_1- u_2, u_1+ u_2+ u_3\}$ , written in that basis. The first "new" basis vector is just $u_1= 1(u_1)+ 0(u_2)+ 0(u_3)$ itself and
$$Tu_1= \left[\begin{array}{ccc} -2 & 1 & 3 \\ 0 & 1 & 2 \\ 0 & 2 & -2\end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right]= \left[\begin{array}{c} -2 \\ 0 \\ 0 \end{array}\right]$$
Since this involves only u1 which is the same in both bases, the first column of the matrix, in this new basis, is again [-2 0 0].

The second new basis vector is (1)u1+ (-1)u2+ 0u3:
$$\left[\begin{array}{ccc} -2 & 1 & 3 \\ 0 & 1 & 2 \\ 0 & 2 & -2\end{array}\right]\left[\begin{array}{c} 1 \\ -1 \\ 0 \end{array}\right]= \left[\begin{array}{c} -1 \\ 1 \\ 1 \end{array}\right]$$

That is, still, in the $\{u_1, u_2, u_3\}$ basis. We need to convert to the new $\{u_1, u_2, u_3\}$ basis by solving $-1(u_1)+ 1(u_2)+ 1(u_3)= a(u_1)+ b(u_1- u_2)+ c(u_1+ u_2+ u_3)$ for a, b, and c. We can "combine like terms" as $(-1-a- b- c)u_1+ (1+ b- c)u_3+ (1- c)u_3= 0$. Since a basis must be independent, each of those coefficients must be 0. 1- c= 0 so c= 1. 1+ b- c= 1+ b- 1= 0 so b= 0. Finally, -1-a-b-c= -1-a-0-1= 0 so a= 2. That vector can be written as $2u_1+ 0(u_1- u_2)+ 1(u_1+ u_2+ u_3)$ and the second column is [2 0 1].

The last new basis vector is $1u_1+ 1u_2+ 1u_3$ so
$$\left[\begin{array}{ccc} -2 & 1 & 3 \\ 0 & 1 & 2 \\ 0 & 2 & -2\end{array}\right]\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]= \left[\begin{array}{c} 2 \\ 3 \\ 0 \end{array}\right]$$
Again, that is in the $\{u_1, u_2, u_3\}$ basis. We have to solve $2u_1+ 3u_2+ 0u_3= a(u_1)+ b(u_1- u_2)+ c(u_1+u_2+u_3)$. That is the same as $(a+ b+ c)u_1+ (-b+ c)u_2+ cu_3= 2u_1+ 3u_2+ 0u_3$ and so c= 0, -b+ 0= 3 so b= -3, and a- 3+ 0= 2 so a= 5. The final column is [5 -3 0].

The matrix representation of T in this new basis is
$$\left[\begin{array}{ccc} -2 & 2 & 5 \\ 0 & 0 & -3 \\ 0 & 1 & 0 \end{array}\right]$$

3. Jan 31, 2008

### HallsofIvy

Staff Emeritus
Your other problem says "T: R3-> R3 is given by T(a, b, c)= (a- b, a+ 2b+ c, -2a+ b- c). Find T-1(0, 1, -1)."

And now you say "I don't know how to build a 3 by 3 matrix. The only thing I can build is a single vector by substituting (0, 1, -1) for (a, b, c)" . First you don't have to find a 3 by 3 matrix to answer this question but substituting (0, 1, -1) for (a, b, c) is going the wrong way. That would be finding T(0, 1, -1) and you want T-1(0, 1, -1). In other words you want to find (a, b, c) so that T(a, b, c)= (0, 1, -1). Since T(a, b, c)= (a- b, a+ 2b+ c, -2a+ b- c)= (0, 1, -1), you have to solve the equations a- b= 0, a+ 2b+ c= 1, -2a+ b- c= -1 for a, b, and c.

However, it is not at all difficult to find the matrix representing T (in a given basis) using the method I've told you about 2 or 3 times before including other threads.

Let's use the "standard" basis (1, 0, 0), (0, 1, 0), and (0, 0, 1). T(1, 0, 0)= (1- 0, 1+2(0)+ 0, -2(1)+ 0- 0)= (1, 1, -2). The first column of the matrix representation is [1 1 2].

T(0, 1, 0)= (0- 1, 0+ 2(1)+ 0, -2(0)+ 1- 0)= (-1, 2, 1). The second column is [-1 2 1].

T(0, 1, 1)= (0- 0, 0+ 2(0)+ 1, -2(0)+ 0- 1)= (0, 1, -1). The third column is [0 1 -1].

Once you have that matrix, you could find it's inverse and multipy by the column vector [0 1 -1]. That is, of course, the same as solving the three equations above.

4. Feb 1, 2008

### transgalactic

regarding the first question
nobody said that U1=(1,0,0)
u2=(0,1,0)
U3=(0,0,1)

i get the idea that if nobody tells us what does the u1 u2 u3 actually look like
we presume that its a standart basis
???

Last edited: Feb 1, 2008
5. Feb 1, 2008

### HallsofIvy

Staff Emeritus
u1, u2, u3 don't "look like" anything- it is just some abstract basis in some abstract vector space. They aren't necessarily even ordered sets of numbers and the is no "standard basis" in a general vector space. But saying it is a basis means that any vector can be written in the form v= au1+ bu2+ cu3 and we can identify it with the vector in R3 [a b c].

In particular, u1 written in the basis {u1, u2, u3} would be written 1u1+ 0u2+ 0u3 and represented as [1 0 0].

A general n-dimensional vector space can consist of any kind of "object" that we can add and multiply by numbers. But once we have selected a basis we can always write then as n-tuples of number, i.e. members of Rn. Once we've done that, the given basis is represented by the "standard basis in Rn".

6. Feb 1, 2008

### transgalactic

regarding the first question
i cant understand what is the meening of the first step
in your solultion(why dont we stop there)??
why is your resolt is not in the new basis
why its still in the old basis
i think that in the beginning it was in the old basis and by the
multiplication we transformed it into a matrix in the new basis
because after the operations we got a matrix the differs in appearence
it cannot be in the old basis after this step

http://img521.imageshack.us/my.php?image=img8240fo6.jpg

i found a solution to this problem in a different way, is it ok??

http://img175.imageshack.us/my.php?image=img8241wm9.jpg

regarding the second question i dont understand your interpretation of the question