Diagonalizing a 3x3 matrix using eigenvectors

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[daniel_an]

As several comments say, there is a mistake. Please watch the following video instead:

[shironoyami]

First you find λ from determinant equation

det(A-λI)=0

And after getting λ1, λ2, etc. variables you substitute each λ values in

(A-λI)*v=0,

this will be done to find v1, v2, etc.
Where you try to find it by trying different numbers you can dot product to get the zero metric.

No there is no formula as far as i care you have to do it manually.

After getting both v
Each v take corresponding columns in
Q={v1 v2 }
After that you try to get the diagonalization, Λ for A=QΛQ^(T).

To get diagonalization, D but in this case, Λ is the diagonalization of
{λ1, λ2, ...} matrix
For example in 2*2matrix
Your Λis
λ1 0
0 λ2 i hope you get the idea.
Ill summarize it here;

First find λ(eigenvalues)from determinant equation: det(A-λI)=0
Then find v(eigenvectors )from
Equation : (A-λI)v=0
Then find Q=[ v1 v2]
Once you get Λ you can make this equation :
A=QΛQ^(T)

[ifekalejaiye]

He made a mistake at lamder= -3, while he was trying to get the eigen vectors