If $M, N$ are finite dimensional vector spaces with same dimension, then if $M$ is subset of $N$, then $M=N$.

Question

If $M, N$ are finite dimensional vector spaces with same dimension, then if $M$ is subset of $N$, then $M=N$.

I think i need to show that both vector spaces are spanned by the same bases in order to do this or to prove $N$ is subset of $M$?

But i am not sure how to do this.

Thanks

Answer 1

Let me reformulate the problem:

If $M$ is a (finite dimensional) subspace of a finite dimensional vector space $N$ and $\dim M=\dim N$, then $M=N$.

The clause that $M$ is finite dimensional is redundant, but it's not important, as you have it in your assignment.

Suppose $M\ne N$ and let $v\in N$, $v\notin M$. If $\{v_1,\dots,v_n\}$ is a basis for $M$, then $\{v_1,\dots,v_n,v\}$ is a linearly independent set in $N$: this is impossible, because a linearly independent sets in $N$ has at most $n=\dim N=\dim M$ elements.

Answer 2

Hint:

Let $\;B:=\{m_1,...,m_k\}\;$ be a basis of $\;M\;$ . Then $\;B\;$ is a linearly independent set of $\;N\;$ as $\;M\le N\;$ . But we're also given $\;k=\dim M=\dim N\;$ , so...complete here.

Answer 3

I think the notation etc can get over-engineered and technical. Here is an attempt to look at things more straightforwardly.

Here are two facts you should know about vector spaces:

Every basis of a finite dimensional vector space has the same number of elements.

Any linearly independent subset of a vector space can be extended to a basis.

You know that $M$ is contained in $N$, and they have the same dimension $r$. A basis $B$ of $M$ therefore has $r$ elements, and as a basis is a linearly independent subset of $M$ and hence of $N$ (because $N$ contains $M$).

$B$, as a basis of $M$, can therefore be extended to a basis of $N$, which will have $r$ elements - so $B$ is already a basis of $N$ as well as of $M$.

Since $B$ is a basis of $N$, any element $v\in N$ can be written as a linear combination of the basis vectors. But this is just an expression of $v$ as an element of $M$. So every element of $N$ is also an element of $M$ and the two are equal.