# Thread: matrix multiplied by its transpose on the left or right = same eigenvalues?

1. ## matrix multiplied by its transpose on the left or right = same eigenvalues?

If P is an m x n matrix

does P.P^T have the same eigenvalues as P^T.P?

this is what I seem to be getting when I do my calculations although I sometimes end up with a 0 eigenvalue in one and not the other.. is there an explanation of this?

many thanks

2. We can easily show it holds for the case of matrixes P of $\displaystyle 2 \times 2$, however, it doesn't hold in general ( take $\displaystyle P = \left( {\begin{array}{*{20}c} 3 \\ 1 \\ \end{array} } \right)$ you'll find that one of the matrixes has an additional eigenvalue)

Let's see the 2x2 matrixes. Remember that the characteristic polynomial of a matrix $\displaystyle A$ in this case is given by: $\displaystyle \chi _A \left( \lambda \right) = \lambda ^2 - {\text{tr}}\left( A \right) \cdot \lambda + \det \left( A \right) = \det \left( {A - \lambda \cdot I} \right)$

For any 2 matrixes $\displaystyle A$ and $\displaystyle B$ we have: $\displaystyle {\text{tr}}\left( {AB} \right) = {\text{tr}}\left( {BA} \right)$

So if $\displaystyle P$ is a 2x2 matrix we have: $\displaystyle {\text{tr}}\left( {P^t P} \right) = {\text{tr}}\left( {PP^t } \right)$ and $\displaystyle {\text{det}}\left( {P^t P} \right) = {\text{det}}\left( {PP^t } \right)$ and therefore $\displaystyle \chi _{P^t P} \left( \lambda \right) = \chi _{PP^t } \left( \lambda \right)$ for all $\displaystyle \lambda$.

Thus the assertion is valid when P is a 2x2 matrix

3. Originally Posted by James0502
If P is an m x n matrix

does P.P^T have the same eigenvalues as P^T.P?

this is what I seem to be getting when I do my calculations although I sometimes end up with a 0 eigenvalue in one and not the other.. is there an explanation of this?
It is true more generally that if A is an m×n matrix and B is an n×m matrix then AB and BA have the same nonzero eigenvalues.

In fact, suppose that $\displaystyle \lambda\ (\ne0)$ is an eigenvalue for AB, with eigenvector x. Then $\displaystyle BA(Bx) = B(ABx) = B(\lambda x) = \lambda(Bx)$. Thus $\displaystyle \lambda$ is also an eigenvector for BA, with eigenvector Bx.

The argument breaks down if $\displaystyle \lambda = 0$, because of the possibility that Bx = 0 in that case.

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# what is eigen value of product of A and its transpose

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