Section B.1 Sample Exercises with Solutions
Here we model one exercise and solution for each learning objective. Your solutions should not look identical to those shown below, but these solutions can give you an idea of the level of detail required for a complete solution.Example B.1.1. LE1.
Consider the scalar system of equations
Rewrite this system as a vector equation.
Write an augmented matrix corresponding to this system.
Example B.1.2. LE2.
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For each of the following matrices, explain why it is not in reduced row echelon form.
-
Show step-by-step why
is not in reduced row echelon form because the pivots are not all is not in reduced row echelon form because the pivots are not descending to the right. is not in reduced row echelon form because not every entry above and below each pivot is zero.
Example B.1.3. LE3.
Consider each of the following systems of linear equations or vector equations.
Explain how to find a simpler system or vector equation that has the same solution set for each.
Explain whether each solution set has no solutions, one solution, or infinitely-many solutions. If the set is finite, describe it using set notation.
-
This matrix corresponds to the simpler system
The third equation
indicates that the system has no solutions. The solution set is -
This matrix corresponds to the simpler system
Since there are three variables and two nontrivial equations, the solution set has infinitely-many solutions.
-
This matrix corresponds to the simpler system
This system has one solution. The solution set is
Example B.1.4. LE4.
Consider the following vector equation.
Explain how to find a simpler system or vector equation that has the same solution set.
Explain how to describe this solution set using set notation.
First, we compute
This corresponds to the simpler system
Since the second column is a non-pivot column, we let
Thus, the solution set is
Example B.1.5. VS1.
Let
-
Show that scalar multiplication distributes over vector addition:
Explain why
nonetheless is not a vector space.
-
We compute both sides:
and
Since these are the same, we have shown that the property holds.
-
To show
is not a vector space, we must show that it fails one of the 8 defining properties of vector spaces. We will show that scalar multiplication does not distribute over scalar addition, i.e., there are values such that-
(Solution method 1) First, we compute
Then we compute
Since
when the property fails to hold. -
(Solution method 2) When we let
we may simplify both sides as follows.Since these ordered pairs are different, the property fails to hold.
-
Example B.1.6. VS2.
-
Write a statement involving the solutions of a vector equation that's equivalent to each claim below.
is a linear combination of the vectors is a linear combination of the vectors
Use these statements to determine if each vector is or is not a linear combination. If it is, give an example of such a linear combination.
-
is a linear combination of the vectors exactly when the vector equationhas a solution. To solve this vector equation, we compute
We see that this vector equation has solution set
so is a linear combination; for example, -
is a linear combination of the vectors exactly when the vector equationhas a solution. To solve this vector equation, we compute
This vector equation has no solution, so
is not a linear combination.
Example B.1.7. VS3.
-
Write a statement involving the solutions of a vector equation that's equivalent to each claim below.
The set of vectors
spansThe set of vectors
does not span
Explain how to determine which of these statements is true.
The set of vectors
has a solution for all
We see that for some
Example B.1.8. VS4.
Consider the following two sets of Euclidean vectors.
Explain why one of these sets is a subspace of
To show that
Consider
To see if
Thus
Now consider
Similarly, to check that
and we see that
Now, to show
-
(Solution Method 1) Now let
and so we know that andConsider
To see if
we need to check if We computeand thus
\textbf{only when} Since this is not true in general, is not closed under vector addition, and thus cannot be a subspace. (Solution Method 2) Note that the vector
belongs to since However, the vector does not belong to since Therefore is not closed under scalar multiplication, and thus is not a subspace.
Example B.1.9. VS5.
-
Write a statement involving the solutions of a vector equation that's equivalent to each claim below.
The set of vectors
is linearly independent.The set of vectors
is linearly dependent.
Explain how to determine which of these statements is true.
The set of vectors
has no non-trivial (i.e. nonzero) solutions. The set is linearly dependent when there exists a nontrivial (i.e. nonzero) solution. We compute
Thus, this vector equation has a solution set
Example B.1.10. VS6.
-
Write a statement involving spanning and independence properties that's equivalent to each claim below.
The set of vectors
is a basis ofThe set of vectors
is not a basis of
Explain how to determine which of these statements is true.
The set of vectors
To answer this, we compute
We see that this set of vectors is linearly dependent, so therefore the set of vectors
Example B.1.11. VS7.
Consider the subspace
Explain how to find a basis of
Explain how to find the dimension of
-
Observe that
If we remove the vectors yielding non-pivot columns, the resulting set will span the same vectors while being linearly independent. Therefore
is a basis of
Since this (and thus every other) basis has three vectors in it, the dimension of
is
Example B.1.12. VS8.
-
Given the set
write a statement involving the solutions to a polynomial equation that's equivalent to each claim below.
The set of polynomials is linearly independent.
The set of polynomials is linearly dependent.
Explain how to determine which of these statements is true.
The set of polynomials
is linearly independent exactly when the polynomial equation
has no nontrivial (i.e. nonzero) solutions. The set is linearly dependent when this equation has a nontrivial (i.e. nonzero) solution.
To solve this equation, we distribute and then collect coefficients to obtain
These polynomials are equal precisely when their coefficients are equal, leading to the system
To solve this, we compute
The system has (infintely many) nontrivial solutions, so we that the set of polynomials is linearly dependent.
Example B.1.13. VS9.
Consider the homogeneous system of equations
Find the solution space of the system.
Find a basis of the solution space.
-
Observe that
Letting
and (since those correspond to the non-pivot columns), this is equivalent to the systemThus, the solution set is
-
Since we can write
a basis for the solution space is
Example B.1.14. AT1.
Consider the following maps of polynomials
Explain why one of these maps is a linear transformation, and why the other map is not.
To show
To show
But note that
For the second part, we compute
But note that
-
(Solution method 1) As for
we computeHowever, note that
which is not always the same polynomial (for example, when ). So we see that so does not respect addition and is therefore not a linear transformation. -
(Solution method 2) As for
we may choose the polynomial and scalar ThenBut on the other hand,
Since this isn't the same polynomial,
does not preserve multiplication and is therefore not a linear transformation.
Example B.1.15. AT2.
-
Find the standard matrix for the linear transformation
given by -
Let
be the linear transformation given by the standard matrixCompute
-
Since
the standard matrix for
is -
Example B.1.16. AT3.
Let
Explain how to find the image of
and the kernel ofExplain how to find a basis of the image of
and a basis of the kernel ofExplain how to find the rank and nullity of T, and why the rank-nullity theorem holds for T.
-
To find the image we compute
-
The kernel is the solution set of the corresponding homogeneous system of equations, i.e.
So we compute
Then, letting
and we have -
Since
we simply need to find a linearly independent subset of these four spanning vectors. So we computeSince the first two columns are pivot columns, they form a linearly independent spanning set, so a basis for
isTo find a basis for the kernel, note that
so a basis for the kernel is
The dimension of the image (the rank) is
the dimension of the kernel (the nullity) is and the dimension of the domain of is so we see which verifies that the sum of the rank and nullity of is the dimension of the domain of
Example B.1.17. AT4.
Let
Explain why
is or is not injective.Explain why
is or is not surjective.
Compute
Note that the third and fourth columns are non-pivot columns, which means
contains infinitely many vectors, so is not injective.Since there are only two pivots, the image (i.e. the span of the columns) is a 2-dimensional subspace (and thus does not equal
), so is not surjective.
Example B.1.18. MX1.
Of the following three matrices, only two may be multiplied.
Explain which two may be multiplied and why. Then show how to find their product.
Thus
Example B.1.19. MX2.
Let
Give a
matrix that may be used to perform the row operationGive a
matrix that may be used to perform the row operationUse matrix multiplication to describe the matrix obtained by applying
and then to (note the order).
Example B.1.20. MX3.
Explain why each of the following matrices is or is not invertible by disussing its corresponding linear transformation. If the matrix is invertible, explain how to find its inverse.
We compute
We see
Similarly, we solve
Similarly, we solve
Similarly, we solve
Combining these, we obtain
We compute
We see
Example B.1.21. GT1.
Let
Let
be the matrix obtained from by applying the row operation What isLet
be the matrix obtained from by applying the row operation What isLet
be the matrix obtained from by applying the row operation What is
Adding a multiple of one row to another row does not change the determinant, so
Scaling a row scales the determinant by the same factor, so so
Swaping rows changes the sign of the determinant, so
Example B.1.22. GT2.
Show how to compute the determinant of the matrix
Here is one possible solution, first applying a single row operation, and then performing Laplace/cofactor expansions to reduce the determinant to a linear combination of
Here is another possible solution, using row and column operations to first reduce the determinant to a
Example B.1.23. GT3.
Explain how to find the eigenvalues of the matrix
Compute the characteristic polynomial:
The eigenvalues are the roots of the characteristic polynomial, namely
Example B.1.24. GT4.
Explain how to find a basis for the eigenspace associated to the eigenvalue
The eigenspace associated to
Thus we see the kernel is
which has a basis of