00:01
Suppose that t from r2 to r2 is a linear transformation such that t at vector 1, negative 1 is vector negative 5, 9 and t at vector negative 3, negative 2 is vector 5, negative 7.
00:18
We want to write t as a matrix transformation, that is, for any vector v in r2, we want to write the linear transformation t as t at v equal a times v, where a is a matrix of two rows and two columns.
00:36
So, the key thing here is to notice that vectors 1, negative 1 and negative 3, negative 2 form a basis of r2, and that is because they are two vectors and they are linearly independent.
00:53
It's easy to verify that they are linearly independent.
00:57
So, we know that taking any vector in r2, we can write that vector as a linear combination of these two vectors 1, negative 1 and negative 3, negative 2.
01:07
Let's see that.
01:07
So, let's take a vector v in r2.
01:15
Let's say it has coordinates x, y, and we want to find scalars alpha and beta, real numbers, such that alpha times the first vector 1, negative 1, plus beta times vector negative 3, negative 2, be equal to vector v, that is, x, y.
01:43
And we are going to verify that we can solve always this equation in terms of x and y, that is, put alpha and beta in terms of x and y in a unique way, and that comes from the fact that vectors 1, negative 1 and negative 3, negative 2 form a basis of r2.
02:07
So, this equality here means a linear system.
02:11
We have alpha minus 3 and beta, that's the first component of the linear combination on the left, must be equal to the first component of v, that is, x, and the second component of the linear combination on the left is negative alpha minus 2, beta, equal, or must be equal to, the second component of v, that is, y.
02:36
Now we can add up these two equations to get negative 5 beta, because alpha and negative alpha cancel out, equal x plus y, and from here we can write beta in terms of x and y as negative x plus y, divided by 5.
03:02
Now we can find coefficient alpha from the first equation or the second equation.
03:12
From the first equation, alpha minus 3 beta equal x, we get alpha equal 3 beta plus x, and that is 3 times, beta can be replaced with this expression here, negative x plus y divided by 5, and that plus x, so alpha is negative 3x minus 3y divided by 5 plus x, that is distributing these 3 and this negative 1, that is negative 3 inside parentheses, and doing the fraction operation we get 5 denominator, negative 3x minus 3y plus 5x in the numerator, and simplifies 5x minus 3x is 2x minus 3y over 5, and that's alpha value in terms of x and y.
04:14
So given any vector x, y in r2, we have found coefficient alpha and beta such that, alpha times 1 negative 1 and beta times negative 3 negative 2 equal vector v.
04:28
So we can say that v is alpha, that is 2x minus 3y divided by 5 here, times vector 1 negative 1, plus beta, that is negative x plus y divided by 5, plus beta here, times vector negative 3 negative 2.
05:02
So we can use the linearity of t.
05:10
By linearity, we get that t at vector v in r2 is t at the linear combination that is equal to v here, and then the linearity means this is equal to scalar here, 2x minus 3y over 5 times, t at vector 1 negative 1 minus scalar here, x plus y divided by 5 times t at vector negative 3 negative 2, and now we replace the image of t at 1 negative 1 and t at negative 3 negative 2 by the vectors given here.
06:13
So this is equal to 2x minus 3y divided by 5 times vector negative 5, 9 minus x plus y divided by 5 times vector 5 negative 7.
06:45
And so we do this linear combination here, so we will get a vector of two components...