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Linear Algebra : Linear Independence and Rank

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Example Questions

Example Question #21 : Linear Independence And Rank

If $A= \begin{bmatrix} 1& 0&1 \\ 1& 1& 1\\ 1& 0 &1 \end{bmatrix}$ , what is rank(A) ?

Possible Answers:

None of the other answers

Correct answer:

Explanation:

rank(A) is equal to the number of linearly independent columns of . The first and third columns are the same, so one of these columns is redundant in the column space of . The second column evidently cannot be a multiple of the first, since the second has two 's, and the first has none. Hence rank(A)=2 .

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Example Question #21 : Linear Independence And Rank

Consider the polynomials

$p_{1}(x) = x^{3}+ x^{2}-x+ 1$

$p_{2}(x) = x^{3}+2 x^{2}+ 2x+ 1$

$p_{3}(x) = x^{2}+ x+ 1$

$p_{4}(x) = x^{2}+ x$

True or false: these four polynomials form a basis for $P_{3}$ , the set of all polynomials with degree less than or equal to 3.

Possible Answers:

True

False

Correct answer:

True

Explanation:

A test to determine whether these matrices form a basis is to set up a matrix with each row comprising the coefficients of one polynomial, and performing row reductions until the matrix is in row-echelon form. $P_{3}$ is a vector space of dimension 4, so these four polynomials will form a basis if and only if the resulting $4 \times 4$ matrix has rank 4. The initial matrix is:

$\begin{bmatrix} 1 & 1 & -1 & 1 \\ 1 & 2 & 2 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 \end{bmatrix}$

Perform the following row operations:

$-R1+ R2 \rightarrow R2$

$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 &1& 3 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 \end{bmatrix}$

$-R2+ R3 \rightarrow R3$

$-R2+ R4 \rightarrow R4$

$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 &1& 3 & 0 \\ 0 & 0 & -2 & 1 \\ 0 & 0 & -2 & 0 \end{bmatrix}$

$-\frac{1}{2} R3 \rightarrow R3$

$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 &1& 3 & 0 \\ 0 & 0 & 1 & -\frac{1}{2}\\ 0 & 0 & -2 & 0 \end{bmatrix}$

$2R3 + R 4 \rightarrow R4$

$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 &1& 3 & 0 \\ 0 & 0 & 1 & -\frac{1}{2}\\ 0 & 0 & 0 & -1\end{bmatrix}$

$-R4 \rightarrow R4$

$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 &1& 3 & 0 \\ 0 & 0 & 1 & -\frac{1}{2}\\ 0 & 0 & 0 &1\end{bmatrix}$

The matrix is now in row-echelon form. Each row has a leading 1, so the matrix has rank 4, the dimension of $P_{3}$ . It follows that the given polynomials comprise a basis of $P_{3}$ .

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Example Question #202 : Operations And Properties

Consider the polynomials:

$p_{1}(x) = x^{3}+ x^{2}+ x+ 1$

$p_{2}(x) = x^{3}+2 x^{2}+ 2x+ 1$

$p_{3}(x) = x^{2}+ x+ 1$

$p_{4}(x) = x^{2}+ x$

True or false: these four polynomials form a basis for $P_{3}$ , the set of all polynomials with degree less than or equal to 3.

Possible Answers:

False

True

Correct answer:

False

Explanation:

Elements of a vector space form a basis if the elements are linearly independent and if they span the space - that is, every element in that space can be uniquely expressed as the sum of the elements.

$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 \end{bmatrix}$

Perform the following row operations:

$-R1+ R2 \rightarrow R2$

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

$-R2+ R3 \rightarrow R3$

$-R2+ R4 \rightarrow R4$

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

The matrix is now in row-echelon form. There is a row of zeroes at the bottom, so the rank of the matrix is 3. Therefore, the four polynomials are not linearly independent, and they do not form a basis for $P_{3}$ .

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Example Question #21 : Linear Independence And Rank

$A = \begin{bmatrix} -1 & 1 \\ 1 & 1 \end{bmatrix}$ , $B= \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$ , $C = \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix}$ , and $D = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$ .

True or false: these four matrices form a basis for the vector space $M_{2 \times 2}$ , the set of $2 \times 2$ matrices.

Possible Answers:

False

True

Correct answer:

True

Explanation:

A test to determine whether these matrices form a basis is to set up a matrix with each row comprising the coefficients of one polynomial, and performing row reductions until the matrix is in row-echelon form. $M_{2 \times 2}$ is a vector space of dimension 4, so these four polynomials will form a basis if and only if the resulting $4 \times 4$ matrix has rank 4. The initial matrix is:

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

Perform the following row operations:

$-R1 \rightarrow R1$

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

$-R1 + R2 \rightarrow R2$

$-R1 + R3 \rightarrow R3$

$-R1 + R4 \rightarrow R4$

$\begin{bmatrix} 1 &- 1 &- 1 & -1 \\ 0& 0& 2 & 2 \\ 0 & 2&0 &2 \\ 0 &2 & 2 & 0 \end{bmatrix}$

$R2 \leftrightarrow R3$

$\begin{bmatrix} 1 &- 1 &- 1 & -1 \\ 0 & 2&0 &2 \\ 0& 0& 2 & 2 \\0 &2 & 2 & 0 \end{bmatrix}$

$\frac{1}{2} R2 \rightarrow R2$

$\begin{bmatrix} 1 &- 1 &- 1 & -1 \\ 0 & 1&0 &1 \\ 0& 0& 2 & 2 \\0 &2 & 2 & 0 \end{bmatrix}$

$-2R2 + R4 \rightarrow R4$

$\begin{bmatrix} 1 &- 1 &- 1 & -1 \\ 0 & 1&0 &1 \\ 0& 0& 2 & 2 \\0 &0 & 2 & -2 \end{bmatrix}$

$\frac{1}{2} R3 \rightarrow R3$

$\begin{bmatrix} 1 &- 1 &- 1 & -1 \\ 0 & 1&0 &1 \\ 0& 0& 1 & 1 \\0 &0 & 2 & -2 \end{bmatrix}$

$-2R3 + R4 \rightarrow R4$

$\begin{bmatrix} 1 &- 1 &- 1 & -1 \\ 0 & 1&0 &1 \\ 0& 0& 1 & 1 \\0 &0 &0 & -4 \end{bmatrix}$

$-\frac{1}{4} R4 \rightarrow R4$

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

The matrix is now in row-echelon form. Each row has a leading 1, so the matrix has rank 4, the dimension of $M_{2 \times 2}$ . It follows that the four matrices , , , and comprise a basis of $M_{2 \times 2}$ .

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Example Question #21 : Linear Independence And Rank

In a 5 dimensional vector space, what is the maximum number of vectors you can have in a linearly dependent set?

Possible Answers:

No limit

Five

Zero

Ten

One

Correct answer:

No limit

Explanation:

Linearly dependent sets have no limit to the number of vectors they can have.

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Example Question #21 : Linear Independence And Rank

$\textbf{u} = (1, 1, -1)$ ; $\textbf{v} = (2,0,2)$ ; $\textbf{w} = (1,1,1)$ .

Does the set $\left \{ \textbf{u} ,\textbf{v} ,\textbf{w} \right \}$ form a basis for $\mathbb{R}^{3}$ , and if not, why?

Possible Answers:

$\left \{ \textbf{u} ,\textbf{v} ,\textbf{w} \right \}$ forms a basis of $\mathbb{R}^{3}$ .

$\left \{ \textbf{u} ,\textbf{v} ,\textbf{w} \right \}$ does not form a basis of $\mathbb{R}^{3}$ because the vectors are not linearly independent, nor do they form a spanning set.

$\left \{ \textbf{u} ,\textbf{v} ,\textbf{w} \right \}$ does not form a basis of $\mathbb{R}^{3}$ because the vectors do not form a spanning set.

$\left \{ \textbf{u} ,\textbf{v} ,\textbf{w} \right \}$ does not form a basis of $\mathbb{R}^{3}$ because the vectors are not linearly independent.

Correct answer:

$\left \{ \textbf{u} ,\textbf{v} ,\textbf{w} \right \}$ forms a basis of $\mathbb{R}^{3}$ .

Explanation:

A set of elements in a vector space forms a basis of the space if and only if satisfies two conditions:

1) It must be a spanning set, so that every other vector in the space can be written as a linear combination of those elements; and,

2) It must be a linearly independent set, so each vector can be so written uniquely.

We can test for both conditions by forming a matrix with the vectors and determining the rank of the matrix by way of row operations. The matrix is

$\begin{bmatrix} 1 & 1 & -1 \\ 2 & 0 & 2 \\ 1 & 1 & 1 \end{bmatrix}$

Perform the following row operations:

$-2R 1 + R2 \rightarrow R2$

$-R1 + R3 \rightarrow R3$

$\begin{bmatrix} 1 & 1 & -1 \\0 & -2 & 4 \\ 0 &0& 2\end{bmatrix}$

$-\frac{1}{2} R2 \rightarrow R2$

$\begin{bmatrix} 1 & 1 & -1 \\0 & 1 & -2 \\ 0 &0& 2\end{bmatrix}$

$\frac{1}{2} R3 \rightarrow R3$

$\begin{bmatrix} 1 & 1 & -1 \\0 & 1 & -2 \\ 0 &0& 1\end{bmatrix}$

We now have a matrix in row-echelon form. All three rows have nonzero entries, so the rank of the matrix is equal to 3, the dimension of $\mathbb{R}^{3}$ . $\left \{ \textbf{u} ,\textbf{v} ,\textbf{w} \right \}$ does form a basis of $\mathbb{R}^{3}$ .

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Example Question #21 : Linear Independence And Rank

The rank and the nullity of a matrix with four rows and six columns are the same. What number do they share?

Possible Answers:

Correct answer:

Explanation:

The sum of the rank and the nullity of a matrix is equal to the number of columns in the matrix. Since the matrix in question has six columns, for the rank and the nullity to be equal, they must each be 3.

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Example Question #23 : Linear Independence And Rank

$M_{2 \times 2}$ refers to the set of $2\times 2$ matrices with real entries.

$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ , $B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ , $C = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$ , $D = \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}$

Does the set $\left \{ A, B, C, D\right \}$ form a basis for $M_{2 \times 2}$ , and if not, why not?

Possible Answers:

$\left \{ A, B, C, D\right \}$ does not form a basis of $M_{2 \times 2}$ , because the matrices are not linearly independent, nor do they form a spanning set.

$\left \{ A, B, C, D\right \}$ does not form a basis of $M_{2 \times 2}$ , because the matrices do not form a spanning set.

$\left \{ A, B, C, D\right \}$ forms a basis of $M_{2 \times 2}$ .

$\left \{ A, B, C, D\right \}$ does not form a basis of $M_{2 \times 2}$ , because the matrices are not linearly independent.

Correct answer:

$\left \{ A, B, C, D\right \}$ forms a basis of $M_{2 \times 2}$ .

Explanation:

A set of elements in a vector space forms a basis of the space if and only if satisfies two conditions:

1) It must be a spanning set, so that every other vector in the space can be written as a linear combination of those elements; and,

2) It must be a linearly independent set, so each vector can be so written uniquely.

We can test for both conditions by forming a $4 \times 4$ matrix with the entries of the four matrices and determining the rank of the matrix by way of row operations. The matrix is

$\begin{bmatrix} 1 & 2 & 3 & 4 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}$

Perform the following row operations in order to get the matrix to a row-equivalent row-echelon form, and to therefore determine its rank:

$R1 \leftrightarrow R2$

$\begin{bmatrix} 1 & 0 & 0 & 0 \\1 & 2 & 3 & 4 \\ 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}$

$-R1 + R2 \rightarrow R2$

$-R1 + R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0 & 0 & 0 \\0 & 2 & 3 & 4 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}$

$\frac{1}{2} R2 \rightarrow R2$

$\begin{bmatrix} 1 & 0 & 0 & 0 \\0 &1 & \frac{3}{2} & 2 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}$

$-R3 + R4 \rightarrow R4$

$\begin{bmatrix} 1 & 0 & 0 & 0 \\0 &1 & \frac{3}{2} & 2 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$

We now have a matrix in row-echelon form. All four rows have nonzero entries, so the rank of the matrix is equal to 4, the dimension of $M_{2 \times 2}$ . The set $\left \{ A, B, C, D\right \}$ forms a basis of $M_{2 \times 2}$ .

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Example Question #22 : Linear Independence And Rank

A $4 \times 4$ matrix has rank 3. Which of the following is true?

Possible Answers:

$A^{-1}$ may exist, and if it does, it must have rank 1.

$A^{-1}$ does not exist.

$A^{-1}$ may exist, and if it does, it must have rank 3.

$A^{-1}$ may exist, and can have any rank.

Correct answer:

$A^{-1}$ does not exist.

Explanation:

A $n \times n$ matrix is nonsingular if and only if its rank is . This is not the case with , so $A^{-1}$ does not exist.

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Example Question #25 : Linear Independence And Rank

The rank and nullity of a matrix are 4 and 2, respectively. The nullity of $A^{T }$ is 3. What are the dimensions of ?

Possible Answers:

Insufficient information is given to determine the dimensions of the matrix.

is a $4 \times 5$ matrix.

is a $6 \times 7$ matrix.

is a $5 \times 4$ matrix.

is a $7 \times 6$ matrix.

Correct answer:

is a $7 \times 6$ matrix.

Explanation:

The sum of the rank and the nullity of any matrix is the number of columns in the matrix, so , whose rank and nullity are 4 and 2, respectively, must have 6 columns. Also, the rank of the transpose $A^{T }$ is always equal to that of itself. $A^{T }$ therefore has rank 4 and nullity 3, so it has 7 columns - the number of rows in . This makes a matrix with seven rows and six columns - a $7 \times 6$ matrix.

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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 #21 : Linear Independence And Rank

Example Question #21 : Linear Independence And Rank

Example Question #202 : Operations And Properties

Example Question #21 : Linear Independence And Rank

Example Question #21 : Linear Independence And Rank

Example Question #21 : Linear Independence And Rank

Example Question #21 : Linear Independence And Rank

Example Question #23 : Linear Independence And Rank

Example Question #22 : Linear Independence And Rank

Example Question #25 : Linear Independence And Rank

All Linear Algebra Resources

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