Question about eigenvalues

[Leo321]

We have vectors x,y of size n and a matrix A of size nxn.
Is it true that the matrix xy^TA has at most one non zero eigenvalue? Why is it so?

 

[Leo321]

Ok, I think I got it.
The matrix xy^T has a rank of 1.
It is known that rank(AB)<=min(rank(A),rank(B))
Thus rank(xy^TA)<=1
It is also known that the number of non-zero eigenvalues of a matrix is less or equal to the matrix rank. Thus the number of non-zero eigenvalues of xy^TA is at most 1.
Right?

 

[khemist]

xy(transpose) will yield a scalar correct?

That means the maximum number of eigenvalues is n i believe

 

[Outlined]

xy(transpose) will yield a scalar correct?

no, x^Ty is a scalar

 

[blue_raver22]

Yes, it is true that the matrix xyTA has at most one non-zero eigenvalue. This is because the matrix xyTA is a product of two matrices, xyT and A, which means that its eigenvalues are a combination of the eigenvalues of xyT and A. Since xyT is a vector of size n, it can have at most n non-zero eigenvalues. Similarly, A is a matrix of size nxn, so it can also have at most n non-zero eigenvalues. Therefore, the product of these two matrices, xyTA, can have at most n non-zero eigenvalues. Since n is the maximum number of non-zero eigenvalues that can be present, it follows that xyTA can have at most one non-zero eigenvalue.