00:04
For chapter 12, question 23, provided f equals zero, then it's true.
00:27
Because, hold on, there's another truth we need to look at.
00:36
It is true.
00:37
And if dim v equals 1, then f, k1u and k2u equals k -substrip 1, k -substcript 1, k -substcript 2, as a function of you and u multiply equals zero so f equals zero then then v is greater than one and f equals zero because it does not equal zero because there will now exist u1 u2 as an element of v so f of u1 u2 doesn't equal 0.
02:43
Now, multiplying u1 by an appropriate factor, we can suggest that f, u1, u2, equals 1.
03:23
So f of u2 and u1 equals negative 1.
03:36
U1 and u2 are linearly independent.
03:51
Because if u2 equals ku1, then a function of u1 and u2 equals a function of u1, k u1, equals a function of u1, k u1, equals 0 times a function of u1 u1 equals 0.
04:29
Now, if u equals a span of u1, u2, then a, the matrix representation of the restriction of f to u and the basis, u1, u2, is 0, 1, 1 ,000, is 0 ,1, 1 ,000, negative 1, 0, b, if u is an element of u, u equals a times u -subscript 1 plus b times u -substcript 2, then the function of you and u -subscript 1 can equal f, a, u -times 1 plus b, u -times -subscript 2, u -subscript 1 equals negative b, and f -u -u -substrip 2 equals f -a times u -substcript 1, plus b -u -substcript 2, u -subscript 2 equals a.
06:36
Now, if w has those vectors, w is an element of v, you can say such that, the function of w, u -subscript 1, equals 0...