# Linear operator

1. Oct 14, 2012

### Clandry

Hi. I attached the problem and my work.
I'm not sure if I did part a) right. In the past problems I've done, they usually provide you with 3 vectors that are linearly independent, thus giving you unique values for C1, C2, C3. The matrix for this one forms:
1 1 1
0 1 3
0 0 0
Which is obviously linearly dependent.

In my work I solved it and C3 ended up canceling out. Does the free variable always cancel out when solving for T(vector) if the matrix above is linearly dependent?

For part b)
I said no, because if the 3rd element (element in 3rd row) in T(v) is nonzero, then the system is inconsistent.

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2. Oct 15, 2012

### jbunniii

As you noted, the three vectors (1,0,0), (1,1,0), and (1,3,0) are linearly dependent. You can express (2,-5,0) as a linear combination of any two of them. For example,
$$(2,-5,0) = 7(1,0,0) -5(1,1,0)$$
Note that these are unique coefficients, because now you are working with two linearly independent vectors, (1,0,0) and (1,1,0).

If you use all three vectors, then there are many ways to write (2,-5,0) as a linear combination of the three. For example, you could make $C_3=0$:
$$(2,-5,0) = 7(1,0,0) -5(1,1,0) + 0(1,3,0)$$
(This is the solution you found.)

Or you could make $C_2=0$:
$$(2,-5,0) = (11/3)(1,0,0) + 0(1,1,0) -(5/3)(1,3,0)$$
Or you could make $C_1=0$:
$$(2,-5,0) = 0(1,0,0) + (11/2)(1,1,0) - (7/2)(1,3,0)$$
There are also infinitely many solutions where none of the coefficients are zero, for example:
$$(2,-5,0) = 9(1,0,0) -8(1,1,0) + 1(1,3,0)$$
Note that you will get the same answer for $T(2,-5,0)$ no matter which of the above solutions you use.