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Linear Algebra : Linear Algebra

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Example Questions

Example Question #201 : Operations And Properties

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:

Yes

Not enough information

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 #281 : Linear Algebra

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

Possible Answers:

Five

Not enough information

There is no limit

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 #282 : Linear Algebra

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:

Five

Six

Three

One

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 #204 : Operations And Properties

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

Not enough information

Three

One

Two

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 #205 : Operations And Properties

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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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:

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. $P_{3}$ 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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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #201 : Operations And Properties

Example Question #281 : Linear Algebra

Example Question #282 : Linear Algebra

Example Question #204 : Operations And Properties

Example Question #205 : Operations And Properties

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

All Linear Algebra Resources

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