Question

Which one of the following statements is ALWAYS correct about a collection of \(p\) column vectors, each having \(n\) real-valued entries?

• A. If \(p>n\), then the column vectors must be linearly dependent
• B. If \(p>n\), then the column vectors must be linearly independent
• C. If \(p=n\), then the column vectors must be orthogonal
• D. If \(p<n\), then the column vectors must be linearly independent

Hint:

Compare the number of vectors to the dimension of the space to decide whether independence is even possible.

Answer 1

The Correct Option is A

Step 1: Set the space these vectors live in.
Each vector has $n$ real entries, so every vector sits inside $R^n$, a space of dimension $n$.

Step 2: Use the rank bound.
Stack the $p$ vectors as columns of an $n\times p$ matrix $M$. The rank of $M$ can be at most $n$, since rank cannot exceed the number of rows.

Step 3: Apply this to option (A).
If $p>n$, then rank$(M)\le n<p$, so the $p$ columns cannot all be independent, since independence of all columns would force rank$(M)=p$. So at least one column is a combination of the rest, and the set is linearly dependent whenever $p>n$.

Step 4: Rule out option (B) using the same bound.
Independence of all $p$ columns needs rank$(M)=p$, but we just showed rank$(M)\le n<p$. So (B) can never hold when $p>n$.

Step 5: Rule out option (C) with a counterexample.
Take $n=p=2$ and the vectors $(1,0)$ and $(1,1)$. These are independent, since neither is a scalar multiple of the other, but their dot product is $1\times1+0\times1=1\neq0$, so they are not orthogonal. This case alone breaks the claim that they must be orthogonal.

Step 6: Rule out option (D) with a counterexample.
Take $n=3$, $p=2$, and both vectors equal to $(1,1,1)$. Here $p<n$, yet the two vectors are identical, so they are dependent, not independent. This breaks the claim in (D).

Step 7: Conclude.
Only the rank bound argument in option (A) survives every case.
\[ \boxed{\text{If } p>n\text{, the column vectors must be linearly dependent}} \]