00:01
So in this problem, we're given this matrix a, which is in the form a, b, b, b, a, where b is non -zero.
00:08
We're asked to find a matrix that orthogonally diagonalizes this matrix a.
00:15
So what does that mean? what that means is that if this matrix is p, we'll call it, then p transpose a times p is the identity matrix.
00:37
It's a diagonal matrix, not the identity.
00:40
Sorry about that.
00:46
Is a diagonal matrix.
00:51
Diagonal matrix simply means that the elements on the diagonal are non -zero and everything else is zero.
00:59
So this means i have p transpose ap is equal to some diagonal matrix d, which i can write as lambda 1 ,0, 0, lambda 2, okay? or lambda 1, lambda 2 are non -zero.
01:32
And they're also going to be eigenvalues, as it turns out.
01:45
Well, this simply means then that a minus lambda i is zero.
01:59
So if i have a, b, b, a minus lambda 0, lambda, 0, then add these together, a minus lambda, b, b, a minus lambda, this matrix is equal to the zero matrix.
02:32
Which means then that our characteristic equation will have a minus lambda squared minus b squared is zero in other words the determinant of it is zero well this factors into a minus lambda plus b times a minus lambda minus b right because a minus a squared minus b squared minus b squared minus b or a times a plus b.
03:06
Okay, so by zero product property rule, i can set these equal to zero.
03:12
And so this tells me that i have lambda 1 is a plus b, and lambda 2 is a minus b.
03:29
Okay, now i'm going to call this equation 1 here.
03:38
So with lambda 1 equal to a plus b, then equation 1, gives me a minus a plus b b b a minus a plus b is zero i left off x1 and x2 here that's the zero matrix 0 0 okay so well a minus a is b so this gives me b x1 plus bx2 is 0 for the first entry, and then bx1a minus a is minus, right here.
05:11
Minus bx2 is again zero.
05:15
So these two together say the same thing, right? they're exact same equations.
05:20
And they say that x1 is equal to x2...