if rank(A)=trace(A)=1 than A is a projection

Question

Let A be a complex square matrix. Suppose that rank(A)=trace(A)=1. Prove that A is a projection.

I have no idea how to prove it, so will be thankful for any help.

Answer 1

Rank 1 matrices have the form $A=vw^T$. Note that $\operatorname{tr}(vw^T)=w^Tv$. Therefore $$A^2=vw^Tvw^T=v\cdot 1\cdot w^T=A.$$

Answer 2

Hint: Choose a basis containing a vector in the image of $A$, and write down what form the matrix must take in that basis. Then see what $A^2$ looks like.

Answer 3

Hints:

(1) $A$ is a projection $\iff$ $A^2=A$

(2) If $\text{rank}(A)=1$, then $Ae_i=\lambda_i v$ for all $1\leq i\leq n$ ($e_i$ denotes the standard basis of the vector space of dimension equal to the number of columns/the number of rows of $A$)

(3) If $v=\sum_{i=1}^{n} a_ie_i$, then what's the trace of $A$ given (2) in terms of $a_i,\lambda_i$ for $1\leq i\leq n$?

I hope this helps!

Answer 4

If $\mathrm{rank}(A)=1$ then by the rank nullity theorem $\dim\ker A=n-1$ so let $(e_1,\ldots,e_{n-1})$ a basis for $\ker A$ which we complete to $\mathcal B=(e_1,\ldots,e_{n-1},v)$ a basis to the space and the similar matrix to $A$ written in the basis $\mathcal B$ is $$B=\left(\begin{matrix}\\ 0&0&\cdots&a_1\\ \vdots&&&a_{n-1}\\ 0&&\cdots&1 \end{matrix}\right)$$ with $Bv=a_1 e_1+\cdots+a_{n-1}e_{n-1}+v$ since $\mathrm{tr}(B)=\mathrm{tr}(A)=1$ and we verify easily that $B^2=B$ so $B$ and then $A$ is a projection.