The basis theorem statement

Question

According to the Basis theorem "Let $V$ be a $p$-dimensional vector space, $p \geq 1$. Any linearly independent set of exactly $p$ elements in $V$ is automatically a basis for $V $. Any set of exactly $p$ elements that spans $V$ is automatically a basis for $V$ ."

Now, let us consider the set of three vectors $\{(1, 0, 3), (0, 1, 2), (1, 1, 5)\}$. These vectors span ${\mathbb R}^ 3$, but they do not form the basis of the vector space ${\mathbb R}^ 3$ because the third vector is the sum of the first two vectors. So, isn't the second sentence of the basis theorem "Any set of exactly $p$ elements that spans $V$ is automatically a basis for $V$." false or contradicting?

Answer 1

The sum of the first two vectors is $(1,1,5)$, and not $(0,1,5)$. So they do form a basis. In fact, $$ \det\begin{pmatrix} 1 & 0 & 3 \cr 0 & 1 & 2 \cr 0 & 1 & 5\end{pmatrix}=3, $$ which is non-zero in the real numbers.

Answer 2

This might clarify things up:

If you had two vectors, say they are $v_1$ and $v_2$ which are linearly dependent (i.e. $v_1=3v_2$). Then, the $dim(span(\lbrace v_1, v_2 \rbrace))$ does not have to necessarily equal two. However, if you had two vectors, $v_1$ and $v_2$ which are linearly independent. Then, the $dim(span(\lbrace v_1, v_2 \rbrace))=2$. The dimension of $\mathbb{R}^3=3$, because the $dim(span( \lbrace (1, 0, 0) , (0, 1, 0), (0, 0, 1) \rbrace))=3$ as there are three vectors (independent) and we know that the $span( \lbrace (1, 0, 0) , (0, 1, 0), (0, 0, 1) \rbrace)=\mathbb{R}^3$.

The statement "any set of exactly p elements that spans V is automatically a basis for V." for all sets of vectors whether linearly dependent or independent" is wrong.

For it to be correct, the statement would need to be the following: "any set of vectors with exactly p elements that spans V is automatically a basis for V if the p vectors are independent."

Answer 3

Try and express $(0,0,1)$ as a linear combination of those three vectors.

You’ll soon discover you can’t.

A linearly dependent set with three vectors cannot span a three-dimensional vector space. That set spans a two-dimensional subspace.

The theorem you quote is correctly stated. A spanning set consisting of $p$ vectors in a $p$-dimensional vector space is a basis.

From every spanning set you can extract a basis: if your set is not a basis, you'd find a basis of $V$ with less elements than $p$. The statement “if a set of $p$ elements is a spanning set of a $p$-dimensional vector space, then it is a basis” is correct.