Idempotent matrix, with rows forming a basis

Question

I'm taking an introductory class on linear algebra, we don't know anything about rank, kernel, or determinants yet.

I'm asked to show that if $A^2=A$ for a matrix $A$ with elements in a field $F$, and if the rows of $A$ form a basis for the space $F^n$, then $A=I_n$.

My proof went something like this: We cannot have A be the zero matrix, since that wouldn't form a basis with its rows for $F^n$. Now assume that $A\neq I_n$. This implies that $A^2 \neq A$. This is a contradiction, therefore we must have $A=I_n$.

What's wrong with my argument? I don't understand why I can't reason with the inequalities this way. It works in a field like $\mathbb{R}$, why not here? I also know it must be wrong, since there are an infinite number of matrices with this property, but I can only find 0 and "1".

Answer 1

An important comment first: if $A\ne I$ you can't assert that $A^2\ne A$. Try with $$ A=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} $$ Thus, unfortunately, your attempt is not good.

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Suppose $x$ is a row vector and let $a_1,a_2,\dots,a_n$ be the rows of $A$. Then $$ x=\sum_{i=1}^n \alpha_i a_i $$ for a unique choice of the scalars $\alpha_1,\dots,\alpha_n$, which amounts to saying that $$ x=[\alpha_1\ \alpha_2\ \dots\ \alpha_n]A $$ that is, $x=yA$ for a unique row vector $y$. Then $$ x=yA=yA^2=(yA)A=xA $$ Hence $yA=xA$ and the uniqueness of $y$ yields $x=y$. Hence $x=xA$ for every row vector $x$.

Can you finish?

Answer 2

For any x $Ax-x$ is a linear combination of the rows of A so $Ax-x=\sum a_i Ae_i$. Apply A on both sides to get $0=A^{2} x-Ax= A(Ax-x)=\sum a_i A^{2} e_i =\sum a_i Ae_i=Ax-x$. hence $Ax=x$ for all x.

Answer 3

If the rows of $A$ form a basis for the space then the matrix is invertible i.e. exists $A^{-1}$ such that $AA^{-1}=I$.
Then $A^2=A$ leads to $AAA^{-1}=AA^{-1}$ and $A=I$.