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Linear Algebra : Linear Independence and Rank

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Example Question #11 : Linear Independence And Rank

Consider the following set of three vectors:

$\b{v}_1,\b{v}_2,\b{v}_3$

where $\b{v}_3=2\b{v}_1+7\b{v}_2$

Is the set linearly independent?

Possible Answers:

Yes

Not enough information

Correct answer:

Explanation:

Since $\b{v}_3$ can be written as a linear combination of of $\b{v}_1$ and $\b{v}_2$ then the set cannot be linearly independent.

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Example Question #11 : Linear Independence And Rank

Does the following row reduced echelon form of a matrix represent a linearly independent set?

$\begin{bmatrix} 1 & 0& 0& 0& \\0 &0 &1 &0 & \\ 0 &0 &0 &0 &\end{bmatrix}$

Possible Answers:

Not enough information

Yes

Correct answer:

Explanation:

The set is linearly dependent because there is a row of all zeros.

Notice that having columns of all zeros does not tell if the set is linearly independent or not.

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Example Question #11 : Linear Independence And Rank

Determine the row rank of the matrix

$\begin{pmatrix} 1&2 &3 \\ 2&4 &6 \end{pmatrix}$

Possible Answers:

Correct answer:

Explanation:

To determine the matrix, we turn the matrix into reduced row echelon form

By adding times the first row to the second we get

$\begin{pmatrix} 1&2 &3 \\ 2+(-2)&4+(-4) &6+(-6) \end{pmatrix}=\begin{pmatrix} 1&2 &3 \\ 0&0 &0 \end{pmatrix}$

And find that the row rank is

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Example Question #14 : Linear Independence And Rank

Determine the row rank of the matrix

$\begin{pmatrix} 1&4 &5 \\ 2&8 &10 \\ 3&6 &4 \end{pmatrix}$

Possible Answers:

Correct answer:

Explanation:

To determine the row rank of the matrix we reduce the matrix into reduced echelon form.

First we add times the 1st row to the 2nd row

$\begin{pmatrix} 1&4 &5 \\ 2+(-2)&8+(-8) &10+(-10) \\ 3&6 &4 \end{pmatrix}=\begin{pmatrix} 1&4 &5 \\ 0&0 &0 \\ 3&6 &4 \end{pmatrix}$

add times the 1st row to the 3rd row

$\begin{pmatrix} 1&4 &5 \\ 0&0 &0 \\ 3+(-3)&6+(-12) &4+(-15) \end{pmatrix}=\begin{pmatrix} 1&4 &5 \\ 0&0 &0 \\ 0&-6 &-11 \end{pmatrix}$

Switch the 2nd row and the 3rd row

$\begin{pmatrix} 1&4 &5 \\ 0&-6 &-11 \\ 0&0 &0 \end{pmatrix}$

multiply the 2nd row by $-\frac{1}{6}$

$\begin{pmatrix} 1&4 &5 \\ 0&-6*(-\frac{1}{6}) &-11*(-\frac{1}{6}) \\ 0&0 &0 \end{pmatrix}=\begin{pmatrix} 1& 4& 5\\ 0&1 &\frac{11}{6} \\ 0&0 &0 \end{pmatrix}$

add times the 2nd row to the 1st row

$\begin{pmatrix} 1& 4+(-4)& 5+(-\frac{44}{6})\\ 0&1 &\frac{11}{6} \\ 0&0 &0 \end{pmatrix} = \begin{pmatrix} 1&0 &-\frac{7}{3} \\ 0&1 &\frac{11}{6} \\ 0&0 & 0 \end{pmatrix}$

And we find that the row rank is

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Example Question #15 : Linear Independence And Rank

Consider the following set of vectors

$\b{v}_1= (1,0,0)$

$\b{v}_2= (0,1,0)$

$\b{v}_3= (0,0,1)$

Is the the set linearly independent?

Possible Answers:

Not enough information

No.

Yes.

Correct answer:

Yes.

Explanation:

Yes, the set is linearly independent. There are multiple ways to see this

Way 1) Put the vectors into matrix form,

$\begin{bmatrix} 1 &0 &0 \\0 &1 &0 \\0 &0 &1 \end{bmatrix}$

The matrix is already in reduced echelon form. Notice there are three rows that have a nonzero number in them and we started with 3 vectors. Thus the set is linearly independent.

Way 2) Consider the equation $\alpha \b{v}_1 + \beta \b{v}_2 + \gamma \b{v}_3 = 0$

If when we solve the equation, we get $\alpha = \beta = \gamma = 0$ then it is linearly independent. Let's solve the equation and see what we get.

$\alpha \b{v}_1 + \beta \b{v}_2 + \gamma \b{v}_3 =\b{0}$

$\alpha (1, 0, 0) +\beta (0,1,0) +\gamma(0,0,1) = (0,0,0)$

Distribute the scalar constants to get

$(\alpha, 0, 0) + (0,\beta,0) +(0,0,\gamma) = (0,0,0)$

Thus we get a system of 3 equations

$\alpha = 0$

$\beta = 0$

$\gamma = 0$

Since $\alpha = \beta = \gamma = 0$ the vectors are linearly independent.

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Example Question #14 : Linear Independence And Rank

Consider the following set of vectors

$\b{v}_1= (1,0,0)$

$\b{v}_2= (0,1,0)$

$\b{v}_3= (0,0,1)$

$\v_4 = (2, 1, 5)$ $\b{v}_4 = (2,1,5)$

Is the the set linearly independent?

Possible Answers:

Not enough information

Yes

Correct answer:

Explanation:

The vectors have dimension 3. Therefore the largest possible size for a linearly independent set is 3. But there are 4 vectors given. Thus, the set cannot be linearly independent and must be linearly dependent

Another way to see this is by noticing that $\b{v}_4$ can be written as a linear combination of the other vectors:

$\b{v}_4 = 2 \b{v}_1 + 1 \b{v}_2 + 5 \b{v}_3$

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Example Question #12 : Linear Independence And Rank

In a vector space with dimension 5, what is the maximum number of vectors that can be in a linearly independent set?

Possible Answers:

There is no limit

Five

Not enough information

Two

Ten

Correct answer:

Five

Explanation:

The dimension of a vector space is the maximum number of vectors possible in a linearly independent set. (notice you can have linearly independent sets with 5 or less, but never more than 5)

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Example Question #13 : Linear Independence And Rank

What is the dimension of the space spanned by the following vectors:

$\b{v}_1 = (1,0,0,0,0)$

$\b{v}_2 = (0,1,0,0,0)$

$\b{v}_3 = (0,0,0,0,1)$

Possible Answers:

One

Five

Three

Six

Not enough information

Correct answer:

Three

Explanation:

Since there are three linearly independent vectors, they span a 3 dimensional space.

Notice that the vectors each have 5 coordinates to them. Therefore they actually span a 3 dimensional subspace of a 5 dimensional space.

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Example Question #14 : Linear Independence And Rank

What is the dimension of the space spanned by the following vectors:

$\b{v}_1 = (1,1,0,0,0)$

$\b{v}_2 = (0,0,1,1,0)$

$\b{v}_3 = (0,0,0,0,1)$

Possible Answers:

Five

Three

Not enough information

Two

One

Correct answer:

Three

Explanation:

Since there are three linearly independent vectors, they span a 3 dimensional space.

Notice that the vectors each have 5 coordinates to them. Therefore they actually span a 3 dimensional subspace of a 5 dimensional space.

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Example Question #18 : Linear Independence And Rank

True or False: If a $5 \times 5$ matrix has linearly independent columns, then nullity(A) =2 .

Possible Answers:

True

False

Correct answer:

True

Explanation:

Since is a $5 \times 5$ matrix, rank(A)+nullity(A) = 5 . Since has three linearly independent columns, it must have a column space (and hence row space) of dimension , causing rank(A)=3 by the definition of rank. Hence nullity(A)=2 .

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Linear Algebra : Linear Independence and Rank

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #11 : Linear Independence And Rank

Example Question #11 : Linear Independence And Rank

Example Question #11 : Linear Independence And Rank

Example Question #14 : Linear Independence And Rank

Example Question #15 : Linear Independence And Rank

Example Question #14 : Linear Independence And Rank

Example Question #12 : Linear Independence And Rank

Example Question #13 : Linear Independence And Rank

Example Question #14 : Linear Independence And Rank

Example Question #18 : Linear Independence And Rank

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