Question

how do you find a matrix A given its eigenvectors and eigenvalues?? please helpp Dx

Answer 1

By definition, a vector v is an eigenvector of matrix A with eigenvalue \( \lambda \) if
\[ Av = \lambda v \]
Hence
\[ (A - \lambda I)v = 0 \]
and this is equivalent to saying
\[ \det(A - \lambda I) = 0 \]
This is therefore the equation you solve to find the eigenvalues.

Once you have found the eigenvalues, \( \lambda_i \), you want to find the null space of the operator
\[ A - \lambda_i I \]
for each \( i \). This corresponds to the eigenspace for \( \lambda_i \).

This all probably seems a bit abstract, but is absolutely correct. I STRONGLY recommend you go through you lecture notes and/or text book to find some worked examples. Once you understand them, attempt your particular problem.

Answer 2

ok thx for the advice !

Answer 3

@JamesJ i can get to the part where you use matrix multiplication to get like 4 separate equations..2 with a, b and 2 with c, d...do you just solve the 4 unknowns and put it in as the matrix??sorry

Answer 4

You want to solve for lambda

Answer 5

If you look at an example, you would see that

Answer 6

yeah but in this problem its given so do you just work backwards to get the original matrix?

Answer 7

Then for each eigenvector Av = lambda.v

Let v1, v2, ..., vn be basis vectors for each of the eigenspaces. If we treat each of those as column vectors, then you can see that
A [ v1 v2 ... vn ] = [ lambda1.v1 lambda2.v2 .... lambdan.vn ]
where here lambdaj is the eigenvector corresponding to vj. Call this matrix
V = [ v1 v2 ... vn ]
Then
V^-1 A V = V^-1 [ lambda1.v1 lambda2.v2 .... lambdan.vn ] = D
where D is a diagonal matrix with the lambdaj down the diagonal.

Hence

A = V D V^-1

So that's how you recover A from the lambda and the vj.

Again, read an example!