# Every Basis of a Subspace Has the Same Number of Vectors

## Problem 577

Let $V$ be a subspace of $\R^n$.
Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k\}$ is a basis of the subspace $V$.

Prove that every basis of $V$ consists of $k$ vectors in $V$.

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## Hint.

You may use the following fact:

Fact.
If $S=\{\mathbf{v}_1, \dots, \mathbf{v}_m\}$ is a spanning set of a subspace $V$ of $\R^n$, then any set of $m+1$ or more vectors of $V$ is linearly dependent.

For a proof of this fact, see the post ↴
If there are More Vectors Than a Spanning Set, then Vectors are Linearly Dependent

## Proof.

Let $B’=\{\mathbf{w}_1, \mathbf{w}_2, \dots, \mathbf{w}_l\}$ be an arbitrary basis of the subspace $V$.
Our goal is to show that $l=k$.

As $B$ is a basis, it is a spanning set for $V$ consisting of $k$ vectors.
By the fact stated above, a set of $k+1$ or more vectors of $V$ must be linearly dependent.
Since $B’$ is a basis, it is linearly independent.
It follows that $l\leq k$.

We now change the roles of $B$ and $B’$.
As $B’$ is a basis, it is a spanning set for $V$ consisting of $l$ vectors.
So it follows from Fact that a set of $l+1$ or more vectors must be linearly dependent.
Since $B$ is a basis, it is linearly independent.
Hence $k \leq l$.

Therefore we have $l\leq k$ and $k \leq l$, and it yields that $l=k$, as required.

### 1 Response

1. 10/02/2017

[…] these into one matrix equation, we obtain [VA=U,] where Every Basis of a Subspace Has the Same Number of Vectors [A=begin{bmatrix} a_{1 1} & a_{1 2} & cdots & a_{1 k} \ a_{2 1} & a_{2 2} […]

##### If there are More Vectors Than a Spanning Set, then Vectors are Linearly Dependent

Let $V$ be a subspace of $\R^n$. Suppose that $S=\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_m\}$ is a spanning set for $V$. Prove...

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