Identity matrix in span

Question

Is this true?

Let $M_{n\times n}(\mathbb{F})$ be the vector space of all $n\times n$ matrices over the field $\mathbb{F}$, and consider $\textbf{A}\in M_{n\times n}(\mathbb{F})$.

Let $U$ be a subspace of $M_{n\times n}(\mathbb{F})$, satisfying that $U = Span(\mathbf{B})$, for a given $\mathbf{B}\in M_{n\times n}(\mathbb{F})$.

$\mathbf{I}_{n\times n}, \mathbf{A} \in U \Rightarrow rank(\mathbf{A})=n$,

where $\mathbf{I}_{n\times n}$ is the $n\times n$ identity matrix

I have no lead on this, any help appreciated.

Answer 1

A would necessarily be a scalar multiple of the identity (because of your definition of U). If A isn't the zero matrix, it will have rank n, just like the identity.

Answer 2

If $I$$\in$$U$ $\implies$ I=$\lambda_1$B from here we can get that $det(B)$$\neq$$0$

So for every matrix in U we know A=$\lambda_2$B and det(A)$\neq$$0$ unless $\lambda_2=0$