00:01
We are given a matrix and we are asked to orthogonally diagonalize this matrix.
00:08
The matrix is 822 -2 -2 -6 -9, 282 -6 -9, 228, negative 6 -9, negative 6 -6 - negative 6 - negative 6 -2 -249 -99 -99 -9 -9 -9 -9 -9 -221.
00:34
And i could write out a transpose and show that it's it's equal to a.
00:42
But i think just by looking at this matrix, it's pretty clear that the entries opposite each other on the diagonal, across the diagonal, are equal.
00:52
So it's clear that a transpose is equal to a, which means that a is a symmetric matrix, and therefore follows that a is orthoginally diagonalizable.
01:09
So this problem makes sense.
01:14
To orthogonally diagonalize a, first we'll find the eigenvalues of a, characteristic polynomial, of a is the determinant of a minus lambda i and this is a five by five matrix so calculating the determinant could take a very long time instead i'll use a computer program to do this and i want to do the same and pause here so using a computer algebra system i calculated that the characteristic polynomial is in factored form lambda plus 30 times lambda minus 30 times lambda minus 15 times lambda minus 6 squared.
02:16
You want to find the solutions to the characteristic equation, which is when this equals zero, so the roots of the polynomial.
02:24
And so we have their eigenvalues are lambda 1 equals negative 30, lambda 2 equals positive 30, lambda 3 equals 15, and lambda 4 is equal to lambda 5 is equal to 6.
02:41
So it has a multiplicity of 2.
02:53
If v is an eigenvector associated with lambda 1, it satisfies the equation a plus 30i times v equals the zero vector.
03:14
To find v, let's reduce the matrix a plus 30i.
03:19
Pause, so you can do this yourself.
03:26
After row reduction, we obtained the matrix 1 -0 -0 -0 -4 or 1 -1 -0 -4, or 1 -1 -1.
03:36
1 4th, i mean.
03:39
01014, 0 .0 is 1 0 1 4th, and 0 .01 1 4th.
04:07
This wrap, last row is actually just the 0 row.
04:13
And so we obtain the equations.
04:17
V1 is equal to negative 1 4, which is also equal to v2, which is equal to v3.
04:30
And so if we take v4 to be the parameter negative 4 t, then we obtain the general form of the eigenvector v is negative 1 .4 times negative 4 is 1 t, t, t, t, negative 4 t, t, or t times the vector 1 -1 -1 -1 - negative -4, i mean.
05:01
If we take t to be 1, we obtain the eigenve 4.
05:03
Vector v1, which is 1 -1 -1 -94.
05:12
The norm of v1, well, we have that the unit vector u1 is v1 over the norm of v1, which is 1 plus 1 plus 1 plus 16 square rooted, which is root 19.
05:29
So this is going to be 1 over root 19, 1 over root 19, 1 over root 19, 1 over root 19, 1 over root 19 and negative 4 over root 19.
06:14
Actually didn't make a mistake.
06:15
This should be 1 .1 4th here as well.
06:22
And then another row of zeros.
06:28
And so this should actually be that this is all equal to negative 1 4v.
06:40
And so if we take v5 to be the parameter negative 4t, we obtain t -t -t -t negative 4t, which is 1 -1.
06:52
1, 1, negative 4.
07:07
Now the norm is 16 plus 1, 17, 18, 19, square to 20, which is going to be 2 root 5.
07:18
This is 1 over 2 root 5, 1 over 2 root 5, 1 over 2 root 5, and negative 4 over 2 root 5, which is negative 2 over root 5.
07:46
Likewise, if v is an eigenvector associated with lambda 2, it satisfies a minus 30i v equals the zero vector.
07:57
To find v, we'll have to re -reduce a minus 30 i.
08:03
Might want to pause here.
08:09
This time we obtain matrix 1 .0130 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0.
08:43
And 0 .01, 1 3rd, 0 ,000, 1, and a row of zeros.
09:06
And so we obtain the system of equations.
09:10
V1 is equal to negative 1 3rd, v2 is also equal to negative 1 3rd v4, as is v3, and we have that v5 is equal to 0.
09:37
So if we take v4 to be the parameter negative 3t, then we have the general eigenvector v is going to be t t t t negative 3t 0, or t times the vector 1 -1 -1 -negative 3 -0.
10:13
And if we take t to be 1, we obtain the eigenvector v2, which is 1 -1 -1 -negative 3 -0...