Define $u=(u_1,u_2,\dots,u_n)^t$, $v=(v_1,v_2,\dots,v_n)^t$ where $u_i,v_j\in$ {$1,-1$}, $\forall i,j$ and $A=[a_{ij}]$, $a_{i,j}=u_iv_j \forall i,j$.

Question

Let $n\in\mathbb N$ and $n\geq 4$ define $u=(u_1,u_2,\dots,u_n)^t$, $v=(v_1,v_2,\dots,v_n)^t$ where $u_i,v_j\in$ {$1,-1$}, $\forall i,j$ and let $A=[a_{ij}]$, $a_{i,j}=u_iv_j \forall i,j$. Then mention true or false for the following:

$a)$. $0$ is an eigenvalue of $A$ for every $u_i,v_j$.

b). If $n$ is odd then $A$ is diagonalizable.

c). $\forall n$ either $A$ is nilpotent or diagonalizable.

Instead of directly jumping on $n$ I tried for $n=4$ firstly. Each cases that I took for $u_i,v_j$ , $0$ is an eigenvalue for sure. Referring to some particular instance likewise the cases I formed for $n=5$ (odd), A came out to be diagonalizable (here I was looking at the algebraic and geometric multiplicities of the eigenvalues and both were coming out to be equal). The last one is also given true but how to conclude anything about nilpotency of $A$? My thoughts over the same are, as we can only talk about the possible eigenvalues and for A to be nilpotent it can have only $0$ as its eigenvalue(couldn't find such $A$). Also that a nilpotent matrix is diagonalizable iff it is a zero matrix. How do I conclude $c).$ to be true as matrix A need not be $0$ matrix and both nilpotency and diagonalizability can't happen together ( this much is sufficient to claim c correct?):/

Correct me if I am wrong. Moreover, generalizing over particular n is fine or not? Thanks.

Answer 1

We can see that for any $x \in V$, where $V$ denotes the vector space we're dealing with, we have

$$A(x) = \langle v,x\rangle u$$ where $(a,b) \mapsto \langle a,b\rangle$ denotes the usual inner product.

From there it is easy to see that:

(a) $0$ is an eigenvalue of $A$ as the dimension of the null space of $A$ is equal to $n-1$. Namely, that null space is the kernel of the linear form $x \mapsto \langle v,x\rangle$.

(b) Suppose that $\langle v,u\rangle \neq 0$, which is the case for $n$ odd. Hence $A(u) = \langle v , u \rangle u$, meaning that $u$ is a non-zero eigenvalue of $A$. As noted above, we also know that $n-1$ linearly independent vectors form a basis of $\ker A$. Added with $u$, we get a basis of $V$ of eigenvectors. Therefore $A$ is diagonalizable.

(c) We just covered at (b) the case $\langle v,u\rangle \neq 0$, proving that then $A$ is diagonalizable. So let's suppose now that $\langle v,u\rangle = 0$. In that case, the image of $A$ is equal to $\mathbb F u$ and $u$ is in the null space of $A$ which is of dimension $n-1$. Therefore $A$ is nilpotent with $A^2=0$.

Note: the only place where we used the hypothesis $u_i,v_j \in \{-1,1\}$ was to infer that $\langle u, v \rangle \neq 0$ for $n$ odd.

Answer 2

These are very interesting exercises and I don't want to spoil them, but I can give some tips.

For part (a), it is useful to note that all rows are multiples of each other. Can you prove that? What does that tell about the null space of $A$, in particular about its dimension?

For (b), you may want to use the fact that the sum of the eigenvalues is the trace of the matrix. What are the possible values for the trace for $n$ even and for $n$ odd?

Combining this with (a), you will find that there is a non-zero eigenvalue if $n$ is odd. What can the dimension of its eigenspace be?

To attack (c), suppose $A$ is not diagonalisable. What are the possible eigenvalues for $A$?

I hope this gets you started. If not, feel free to ask follow-up questions.