Proving $ab^T$ is diagonalizable $\iff a^T b \neq 0$

Question

Assuming $a$ and $b$ as two $n$-element vectors, how can we show that $$a b^T$$ is diagonalizable, if and only if their dot product is nonzero ( = they are not orthogonal)?

I've tried to prove $ab^T$ has $n$ linearly independent eigenvectors and then, automatically as a result, the beginning statement would be true, but I've not been successful in proving it.

Any help would be appreciated.

Answer 1

The claim is not true. If $a=0$ or $b=0$ then $ab^T$ is trivially diagonalizable.

If $a$ and $b$ are non-zero but $a^Tb=0$ then the matrix $ab^T$ is a non-zero nilpotent matrix, hence not diagonalizable.

Now assume that $a$ and $b$ are both non-zero, $a^Tb\ne0$, and $n\ge2$. Then the matrix has rank one but is not invertible. So $\lambda=0$ is an eigenvalue. Since the matrix has rank one, the eigenspace of $\lambda=0$ is $n-1$ dimensional. In order that $ab^T$ is diagonalizable there must be a non-zero eigenvalue. Its corresponding eigenvector has to be an element of the column space of the matrix. Hence it has to be a multiple of $a$. It is now easy to verify that $a$ is an eigenvector to the eigenvalue $a^Tb\ne0$. This gives us a basis of eigenvectors, and the matrix is diagonalizable.

Answer 2

(It's understood that $a$ and $b$ are nonzero, right?)

Here are some ideas to get you started: Nonzero multiples of $a$ are the only possible eigenvectors with $\lambda \neq 0$. The eigenspace belonging to $\lambda=0$ is the orthogonal complement of $b$.

Answer 3

The matrix $ab^T$ has a very specific structure. If the coordinates of $a$ and $b$ are $a_i$ and $b_i$ then $$ab^T=\begin{pmatrix} a_1b_1 & a_1b_2 & \cdots & a_1b_n \\ a_2b_1 & a_2b_2 & \cdots & a_2b_n \\ \vdots & \vdots & \vdots & \vdots \\ a_nb_1 & a_nb_2 & \cdots & a_nb_n \end{pmatrix}$$ Since the rows are all multiple of one another, it shows that $0$ is an eigenvalue with high multiplicity, try finding the corresponding eigenspace. Then find the remaining eigenvectors.

Answer 4

Let me give very simple solution. Let $A=ab^T$. Since $Ax=ab^Tx=(b^Tx)a$, $ImA=span\{a\}$. So rank of $A$ is 1 and by rank nullity, dimension of the null space of $A$ is $n-1$. What is kernel of $A$?

Kernel of $A$ is obviously $b^\perp$. So if $a$ and $b$ are orthogonal, then $a$ is also in kernel. So $ImA \subset KerA$. Thus the only eigenvalue is zero and since the zero eigenspace (kernel) is $n-1$ dimensional, $A$ is not diagonalizable.

Assume that $a$ and $b$ are NOT orthogonal. So $a$ is not in kernel. In this case, eigenvalues are $0$ and $b^Ta$. Since the zero eigenspace $n-1$ dimensional and $b^Ta$ eigenspace 1 dimensional and their sum is $n$, $A$ is diagonalizable.