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We need to solve this one:

What is the eigenvalues of the $n\times n$ matrix $xy^T$ where $x,y$ be non zero $n\times 1$ vectors?

well, let $\lambda$ be an eigen value of the matrix so $xy^Tu=\lambda u$ for some $u$ vector, so $u^T yx^T=\lambda u^T$, $\lambda=yx^T$?

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No. An eigenvalue is a number --- $xy^t$ is an $n\times n$ matrix. –  Gerry Myerson Oct 27 '12 at 5:33
    
Why don't you make up an example? Write down some $3\times1$ vectors, say, $x$ and $y$, and work out the eigenvalues of $xy^t$ --- you do know how to find eigenvalues of a matrix of numbers, yes? So find the eigenvalues of your example, and then think about why they worked out to be what they did. –  Gerry Myerson Oct 27 '12 at 5:35

3 Answers 3

up vote 2 down vote accepted

$xy^T$ is non-invertible. In fact rank $1$. So $0$ is an eigenvalue and the eigenspace of $\, 0$ is the nullity of the matrix $xy^T$. So by rank nullity theorem the geometric multiplicity of $0$ is $n-1$. Argue using the fact that the geometric multiplicity is at most the algebraic multiplicity of an eigenvalue. If $xy^T$ is the nilpotent then the $n$ th eigenvalue is $0$ as well. Else it's something non-zero. At least $n-1$ eigenvalues being $0$, The trace gives the last eigenvalue and completes the problem! (Find the trace of $xy^T$)

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Just to make it explicit, it might be that the algebraic multiplicity of $0$ is $n$ although the geometric multiplicity is only $n-1$. E.g. $x = (1 \; 0)$ and $y = (0 \; 1)$. Or both multiplicities of $0$ could be $n$, or there could be a nonzero eigenvalue (with multiplicity $1$). –  hardmath May 28 '13 at 12:47

One is a scalar, one is a matrix, so they aren't the same (except in the $n=1$ case, effectively). They act the same only on vectors in the eigenspace of $xy^T$ corresponding to $\lambda$, but they're different things.

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Your equation $xy^Tu=\lambda u$ needs no transposition, but you should know how to read it: $x,y,u$ are vectors, which makes $xy^T$ a matrix, $y^Tu$ is a scalar, and $xy^Tu$ is a vector that you can either view as the matrix $xy^T$ applied to the vector $u$ (which is how you obtained it) or as the vector $x$ multiplied by the scalar $y^Tu$. The latter point is useful, since it means that in order for the equation to be solved for some $\lambda\neq0$, one needs $u$ to be a scalar multiple of $x$. Since $u$ is an eigenvector it must be nonzero, and we don't care about nonzero scalar multiples, we might as well take $u=x$ in this case, and so $\lambda=y^Tu=y^Tx$, which must be nonzero (if not, then no eigenvalue $\lambda\neq0$ exists). On the other hand the equation is solved with $\lambda=0$ for any vector $u$ such that $y^Tu=0$. This gives two types of eigenvectors; you should be able to check that the dimension of the eigenspaces for $\lambda=y^Tx\neq0$ is $1$, and the dimension of of the eigenspace for $\lambda=0$ is $n-1$, and these are all solutions.

So $xy^T$ is diagonalisable whenever $y^Tx\neq0$, since the eigenvector $x$ for $\lambda=y^Tx\neq0$ is necessarily independent of the $n-1$-dimensional eigenspace for $0$. What if $y^Tx=0$ (which means that the vectors $x,y$ are orthogonal for the standard inner product)? Well, as indicated there are no non-zero eigenvalues, and the eigenspace for $\lambda=0$ is only of dimension $n-1$, so $xy^T$ is not diagonalisable in this case (you may check that its square is zero, so it is nilpotent).

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