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Linear Algebra : Linear Algebra

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Example Questions

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 #291 : Linear Algebra

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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Example Question #31 : Linear Independence And Rank

The nullity of a matrix is 3; the nullity of $A^{T }$ is also 3.

True, false or undetermined: is a square matrix.

Possible Answers:

True

False

Correct answer:

True

Explanation:

The sum of the rank and the nullity of a matrix is equal to the number of columns in the matrix. Therefore, the number of columns in is equal to $\textup{Rnk}(A) + 3$ , and the number of columns in $A^{T}$ - the number of rows in - is $\textup{Rnk}(A^{T}) + 3$ .

Furthermore, the rank of a matrix is equal to its transpose, so

$\textup{Rnk}(A) = \textup{Rnk}(A^{T})$ ,

and

$\textup{Rnk}(A) + 3 = \textup{Rnk}(A^{T}) + 3$

This makes a matrix with the same number of rows as columns. It is square, and the statement is true.

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Example Question #31 : Linear Independence And Rank

$A = \begin{bmatrix} 1 & 4 & 2 & -3 & 3 \\ 0 & 3 & 2 & 1 & -2 \\ 2 & 11 & 6 & -5 & 4 \end{bmatrix}$

What are the rank and nullity of ?

Possible Answers:

Rank 5; nullity 0

Rank 2; nullity 1

Rank 2; nullity 3

Rank 3; nullity 0

Rank 3; nullity 2

Correct answer:

Rank 2; nullity 3

Explanation:

The rank of a matrix can be found by performing row operations until it is reduced to row-echelon form. On matrix , we want to have leading 1's in each nonzero row, we want the leading 1's to go from upper left to lower right, and we want zero rows gathered at the bottom. Beginning with :

$\begin{bmatrix} 1 & 4 & 2 & -3 & 3 \\ 0 & 3 & 2 & 1 & -2 \\ 2 & 11 & 6 & -5 & 4 \end{bmatrix}$

Perform the following row operations:

$-2R1+R3 \rightarrow R3$

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

$\frac{1}{3} R2 \rightarrow R2$

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

$-3R2 +R3 \leftarrow R3$

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

The matrix is now in row-echelon form. There are two nonzero rows, so the rank is 2. The sum of the rank and the nullity is equal to the number of columns, of which there are 5; this makes the nullity 3.

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Example Question #31 : Linear Independence And Rank

$A = \begin{bmatrix} 1 & 0 & 2 \\4 & 3 & 11 \\ 2 & 2 & 6 \\-3 & 1 & -5 \\ 3 & -2 & 4\end{bmatrix}$

What are the rank and nullity of ?

Possible Answers:

Rank 2; nullity 3

Rank 3; nullity 0

Rank 5; nullity 0

Rank 3; nullity 2

Rank 2, nullity 1

Correct answer:

Rank 2, nullity 1

Explanation:

$A = \begin{bmatrix} 1 & 0 & 2 \\4 & 3 & 11 \\ 2 & 2 & 6 \\-3 & 1 & -5 \\ 3 & -2 & 4\end{bmatrix}$

The following four operations can be performed simultaneously:

$-4R1 + R2 \rightarrow R2$

$-2R1 + R 3 \rightarrow R3$

$3R1 + R4 \rightarrow R4$

$- 3R1 + R5 \rightarrow R5$

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

$\frac{1}{3}R2 \rightarrow R2$

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

The next three can be performed simultaneously:

$-2R2 + R3 \rightarrow R3$

$- R2 + R4 \rightarrow R4$

$2R2 + R5 \rightarrow R5$

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

The matrix is now in reduced row-echelon form. There are two nonzero rows, so the rank is 2. The sum of the rank and the nullity is equal to the number of columns, of which there are 3, so the nullity is 1.

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Example Question #31 : Linear Independence And Rank

Let $m, n, N \in \mathbb{N}$ .

Define $M_{m \times n }$ as the set of all $m \times n$ matrices, and $P_{N}$ as the set of all polynomials in of dimension or less.

True or false: $M_{2 \times 2 }$ and $P_{4}$ are isomorphic linear spaces.

Possible Answers:

True

False

Correct answer:

False

Explanation:

Two linear spaces are isomorphic if and only if they are of the same dimension, or, equivalently, the size of a basis of each includes the same number of elements.

$M_{2 \times 2}$ is the set of $2 \times 2$ matrices; each such matrix includes 4 elements, so it has four elements in one of its bases; for example, one such basis is

$\left \{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \right \}$

$P_{4}$ is the set of all polynomials in of degree 4 or less - that is, all polynomials of the form

$p(x) = a_{0}+ a_{1}x+ a_{2}x^{2} + a_{3}x^{3}+ a_{4}x^{4}$

Each polynomial is defined by five coefficients, so the dimension of $P_{4}$ is 5. Each basis of $P_{4}$ has five elements; for example, one such basis is

$\left \{ 1, x, x^{2},x^{3}, x^{4}\right \}$ .

Since $\dim (M_{2 \times 2}) \ne \dim (P_{4})$ , the spaces are not isomorphic.

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Example Question #32 : Linear Independence And Rank

Define two vectors in $\mathbb{R}^{2}$ as follows:

$\textbf{u} = (\ln 9, \ln 4 )$

$\textbf{v} =\left ( \ln \frac{1}{27} , \ln \frac{1}{8} \right )$

Does $\left \{ \textbf{u}, \textbf{v} \right \}$ form a basis for $\mathbb{R}^{2}$ , and if not, why or why not?

Possible Answers:

No; the set is a linearly independent set, but not a spanning set.

Yes; the set is both a linearly independent set and a spanning set.

No; the set is a spanning set, but not a linearly independent set.

No; the set is neither a linearly independent set nor a spanning set.

Correct answer:

No; the set is neither a linearly independent set nor a spanning set.

Explanation:

$\mathbb{R}^{2}$ is a vector space with dimension 2, so any basis for $\mathbb{R}^{2}$ must include exactly two vectors. This makes it possible that $\left \{ \textbf{u}, \textbf{v} \right \}$ forms a basis for $\mathbb{R}^{2}$ ; also, this makes their linear independence necessary and sufficient for them to form a basis, since, if they are as such, they are a spanning set as well.

We can test for linear independence by forming the following matrix with their entries:

$\begin{bmatrix} \ln 9 & \ln 4 \\ \\ \ln \frac{1}{27} & \ln \frac{1}{8} \end{bmatrix}$

The rows (and the original vectors) are linearly independent if and only if

$\begin{vmatrix} \ln 9 & \ln 4 \\ \\ \ln \frac{1}{27} & \ln \frac{1}{8} \end{vmatrix} \ne 0$ ,

so calculate the determinant:

$D =\left ( \ln 9 \right ) \left (\ln \frac{1}{8} \right ) - \left ( \ln 4 \right ) \left (\ln \frac{1}{27} \right )$

Applying the rules of logarithms:

$D =\left ( \ln 3^{2 }\right ) \left (\ln 2^{-3} \right ) - \left ( \ln 2^{2 } \right ) \left (\ln 3^{-3} \right )$

$D =\left ( 2 \ln 3 \right ) \left (-3 \ln 2 \right ) - \left ( 2\ln 2 \right ) \left (-3\ln 3 \right )$

$D =-6 \left ( \ln 3 \right ) \left ( \ln 2 \right ) -(-6) \left ( \ln 2 \right ) \left ( \ln 3 \right )$

D= 0

The rows of the matrix, and, consequently, the vectors $\textbf{u}$ and $\textbf{v}$ , are linearly dependent. Furthermore, $\textbf{u}$ and $\textbf{v}$ do not span $\mathbb{R}^{2}$ .

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #21 : Linear Independence And Rank

Example Question #291 : Linear Algebra

Example Question #23 : Linear Independence And Rank

Example Question #22 : Linear Independence And Rank

Example Question #25 : Linear Independence And Rank

Example Question #31 : Linear Independence And Rank

Example Question #31 : Linear Independence And Rank

Example Question #31 : Linear Independence And Rank

Example Question #31 : Linear Independence And Rank

Example Question #32 : Linear Independence And Rank

All Linear Algebra Resources

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