00:01
Let a be an invertible 3 by 3 matrix.
00:06
Prove that the set b, composed by the vectors uv and w, is a basis for r3 if and only if the set c composed by the vector a times u, a times b, and a times w, is a basis for r3.
00:31
So we are going to use the fact that any basis in r3 has three vectors.
00:46
That is because the dimension of r3 is 3.
01:04
And we say that because this means that if we want to prove that a set is base for r3, is sufficient to prove that is linearly independent.
01:25
So to prove that a set of three vectors is a base for the space r3, it is sufficient to prove that the set is linearly independent.
02:18
And that's what are we going to do here.
02:20
We have three elements, three vectors in each set.
02:25
So to prove that one of the sets is a basis for r3, we only have to prove that the set is linearly independent.
02:33
That's very important to notice because in other situations, we will need the two requirements to prove that the set is a basis of certain linear spaces.
02:47
Is they are that the set is not empty, that there is a zero vector.
02:56
Sorry, the set is not empty.
02:58
We have three or more elements or two elements dependent on the problem.
03:03
The set is linearly independent and the set span the linear space.
03:08
In this case, we are sure that if the set has three elements and is linearly independent, it's going to be basis for the linear.
03:18
Space that is going to span the space also.
03:22
Okay, so we're going to prove this way first.
03:31
That is, we're going to prove that if the set b composed by the elements u -b -w is a basis for r3, then the set c composed by the vector a -u -a -w is a basis for r3.
03:53
So let alpha, beta, and gamma, b, any real numbers such that alpha times a u or a times u plus beta times a times v plus gamma times a times w is zero that is we start with a linear combination of the three vectors in set c equal to zero we have to prove that these three coefficients, alpha, beta and gamma, are all zero to prove that the set c is linearly independent.
04:55
Our hypothesis in this case is that the set b is a basis that is the elements, u, v, and w are linearly independent.
05:06
Okay, so we start with it.
05:08
And a's matrix, alpha, beta, and gamma are constants, that is coefficients or scalers.
05:15
So this equation is the same as a times alpha u plus a times beta b plus gamma times a, sorry, plus a times gamma w equals zero.
05:42
And this is the same as a times the vector alpha u, beta b plus comma w because we know if we multiply a and the matrix a by this vector we distribute this way it's a linear operation so this is the same but now a is invariable the matrix is invariable means there exists another matrix we call a to negative one and has a property that a inverse times a equal a times a is a inverse equal a times a identity metrics.
06:35
Then we can multiply both sides of this equation by the inverse of a.
06:42
We get this...