Question:
Which one of the following statements is ALWAYS correct about a collection of \(p\) column vectors, each having \(n\) real-valued entries?
Which one of the following statements is ALWAYS correct about a collection of \(p\) column vectors, each having \(n\) real-valued entries?
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Compare the number of vectors to the dimension of the space to decide whether independence is even possible.
Updated On: Jul 20, 2026
- If \(p>n\), then the column vectors must be linearly dependent
- If \(p>n\), then the column vectors must be linearly independent
- If \(p=n\), then the column vectors must be orthogonal
- If \(p<n\), then the column vectors must be linearly independent
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The Correct Option is A
Solution and Explanation
Step 1: Recall what determines independence of a set of vectors.
A set of \(p\) vectors, each with \(n\) real entries, lives in \(R^n\). The largest number of vectors that can be linearly independent in \(R^n\) is \(n\), because the rank of any \(n\times p\) matrix built from these vectors can never exceed \(n\).
Step 2: Test option (A).
If \(p>n\), we have more vectors than the dimension of the space they live in. Since at most \(n\) of them can be independent, the remaining \(p-n\) vectors must be a combination of the others. So the full set cannot be linearly independent, which means it is linearly dependent. This holds for every choice of \(p>n\) vectors, so option (A) is always true.
Step 3: Test option (B).
Option (B) claims the opposite of what Step 2 shows. Since \(p>n\) always forces dependence, the vectors can never be independent in this case. So option (B) is false.
Step 4: Test option (C).
When \(p=n\), the vectors can form a basis of \(R^n\) if they are independent, but nothing forces them to be orthogonal. For example, the two vectors \((1,0)\) and \((1,1)\) in \(R^2\) are independent but not orthogonal, and they form a valid collection with \(p=n=2\). So option (C) is false.
Step 5: Test option (D).
When \(p<n\), the vectors could be independent, but they do not have to be. Take two copies of the same nonzero vector in \(R^3\), so \(p=2\) and \(n=3\). Here \(p<n\), but the two vectors are identical and hence dependent. So option (D) is false.
Step 6: Final conclusion.
Only the statement about \(p>n\) forcing dependence holds in every case.
\[ \boxed{\text{If } p>n\text{, the column vectors must be linearly dependent}} \]
A set of \(p\) vectors, each with \(n\) real entries, lives in \(R^n\). The largest number of vectors that can be linearly independent in \(R^n\) is \(n\), because the rank of any \(n\times p\) matrix built from these vectors can never exceed \(n\).
Step 2: Test option (A).
If \(p>n\), we have more vectors than the dimension of the space they live in. Since at most \(n\) of them can be independent, the remaining \(p-n\) vectors must be a combination of the others. So the full set cannot be linearly independent, which means it is linearly dependent. This holds for every choice of \(p>n\) vectors, so option (A) is always true.
Step 3: Test option (B).
Option (B) claims the opposite of what Step 2 shows. Since \(p>n\) always forces dependence, the vectors can never be independent in this case. So option (B) is false.
Step 4: Test option (C).
When \(p=n\), the vectors can form a basis of \(R^n\) if they are independent, but nothing forces them to be orthogonal. For example, the two vectors \((1,0)\) and \((1,1)\) in \(R^2\) are independent but not orthogonal, and they form a valid collection with \(p=n=2\). So option (C) is false.
Step 5: Test option (D).
When \(p<n\), the vectors could be independent, but they do not have to be. Take two copies of the same nonzero vector in \(R^3\), so \(p=2\) and \(n=3\). Here \(p<n\), but the two vectors are identical and hence dependent. So option (D) is false.
Step 6: Final conclusion.
Only the statement about \(p>n\) forcing dependence holds in every case.
\[ \boxed{\text{If } p>n\text{, the column vectors must be linearly dependent}} \]
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