Question

Let V be a nonzero subspace of the complex vector space $M_7\left(C\right)$ such that every nonzero matrix in V is invertible. Then, the dimension of V over C is

1. 1
2. 2
3. 7
4. 49

Answer: 1

Solution :-

We have a result that say: If every non-zero matrix of order $n$ in a complex vector space $V\left(ℂ\right)$ is invertible, then that complex vector space $V\left(ℂ\right)$ must consist of scalar multiples of a single invertible matrix of order $n$ .
$i\mathrm{.}e\mathrm{.},V\left(ℂ\right)=\{\lambda A_{n\times n}:A\text{is any fixed invertible matrix},0\ne \lambda \in ℂ\}\cup \left\{0_{n\times n}\right\}$ .

Now, according to question, V be a nonzero subspace of the complex vector space $M_7\left(C\right)$ such that every nonzero matrix in V is invertible then V(C) must consist of scalar multiples of a single invertible matrix of order 7.
Therefore, Basis of V(C) consist a single invertible matrix of order 7.

Hence, dim(V(C)) = 1.

So, the correct answer is option-(A), $\to$ 1

Proof of above result:

Let $V\left(ℂ\right)$ is a vector space and every nonzero matrix of order n is invertible.

Let us assume that $V\left(ℂ\right)$ consist a matrix which is not scalar multiple of a single invertible matrix ${\prime }^{}{A}^{\prime }$ of order n. i.e., there exist two linearly independent vectors A and B in $V\left(ℂ\right)$ .
i.e., there exist $B\ne \lambda A\forall 0\ne \lambda \in ℂ$ . since, $A$ and $B$ are in $V\left(ℂ\right)$ this implies $det\left(A\right)\ne 0,det\left(B\right)\ne 0$ .

Now, if this is result hold for $n=2$ then also hold for forall $n$ .

Now, $V\left(ℂ\right)$ is a vector space and $A,B$ are two vectors of $V\left(ℂ\right)$ then every linear combination of $A$ and $B$ must also in $V\left(ℂ\right)$ .
i.e., $\alpha A+\beta B\in V\left(ℂ\right),\text{where}\alpha ,\beta \in ℂ,$ not both $\alpha ,\beta$ zero.

Let’s check there exist $\alpha$ and $\beta$ for which $det\left(\alpha A+\beta B\right)=0$ or not.

Hence, $det\left(\alpha A+\beta B\right)={\alpha }^{2}i+\alpha \beta j+{\beta }^{2}k$ .
Now, $det\left(\alpha A+\beta B\right)=0⟹{\alpha }^{2}i+\alpha \beta j+{\beta }^{2}k=0$

Therefore, there exist $\alpha$ and $\beta$ for which $det\left(\alpha A+\beta B\right)=0$ for some $\alpha ,\beta \in ℂ,\alpha A+\beta B\notin V\left(ℂ\right)$ . So, $V\left(ℂ\right)$ is not a vector space if two vector $A$ and $B$ in $V\left(ℂ\right)$ are linearly independent. i.e., if $B\ne \lambda A\text{for some}\lambda \in ℂ$ then $V\left(ℂ\right)$ not a vector space.

Now we have to prove that if $V\left(ℂ\right)$ is a vector space and if $A\in V\left(ℂ\right)$ then $\forall B\in V\left(ℂ\right),B=\lambda A$ , for some $0\ne \lambda \in ℂ$ .

Since, $V\left(ℂ\right)$ is a vector space and $A,B\in V\left(ℂ\right)$ , then $det\left(A\right)\ne 0$ and also $det\left(B\right)\ne 0$ and every linear combination of $A$ and $B$ must also in $V\left(ℂ\right)$ .
i.e., $\alpha A+\beta B\in V\left(ℂ\right),$ for all $\alpha ,\beta \in ℂ,\text{not both}\alpha ,\beta \text{zero}$ .

We can write $\alpha A+\beta B=\alpha A+\beta \lambda A=\left(\alpha +\beta \lambda \right)A$ , where $\left(\alpha +\beta \lambda \right)$ must be non-zero.

Now, $det\left(\alpha A+\beta B\right)=det\left(\left(\alpha +\beta \lambda \right)A\right)={\left(\alpha +\beta \lambda \right)}^{n}det\left(A\right)$ . Here, $\left(\alpha +\beta \lambda \right)\ne 0$ and also $det\left(A\right)\ne 0$ .
So, $det\left(\alpha A+\beta B\right)\ne 0$ and hence $\left(\alpha A+\beta B\right)\in V\left(ℂ\right)$ , $\forall \alpha ,\beta \in ℂ,\text{not both}\alpha ,\beta \text{zero}$ .

Now, final conclusion,

If $V\left(ℂ\right)$ is complex vector space and every nonzero matrix of order n is invertible then $V\left(ℂ\right)$ must consist of scalar multiple of a single invertible matrix of order n.