Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $u^T yx^T=\lambda u^T$, $\lambda=yx^T$?

share|cite|improve this question
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
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 $xy^T$)

share|cite|improve this answer
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.

share|cite|improve this answer

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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.