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.
\(A=\left[\begin{array}{ccc}
-4 & 0 & 4 \\
0 & 1 & -2 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\) is not in reduced row echelon form because the pivots are not all \(1\text{.}\)
\(B=\left[\begin{array}{ccc}
0 & 1 & 2 \\
1 & 0 & -3 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\) is not in reduced row echelon form because the pivots are not descending to the right.
\(C=\left[\begin{array}{ccc}
1 & -4 & 4 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\) is not in reduced row echelon form because not every entry above and below each pivot is zero.
Since the second column is a non-pivot column, we let \(x_2=a\text{.}\) Making this substitution and then solving for \(x_1\text{,}\)\(x_3\text{,}\) and \(x_4\) produces the system
\begin{equation*}
\begin{matrix}
x_1 &=& 1-a \\
x_2 &=& a \\
x_3 &=& 1 \\
x_4 &=& 2 \\
\end{matrix}
\end{equation*}
Thus, the solution set is \(\left\{ \left[\begin{array}{c}
-a + 1 \\
a \\
1 \\
2
\end{array}\right] \,\middle|\, a \in\mathbb R \right\} \text{.}\)
ExampleC.1.5.EV1.
Consider each of these claims about a vector equation.
(a)
\(\left[\begin{array}{c} -13 \\ 3 \\ -15 \end{array}\right]\) is a linear combination of the vectors \(\left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right] ,
\left[\begin{array}{c} 2 \\ 0 \\ 2 \end{array}\right] ,
\left[\begin{array}{c} 3 \\ 0 \\ 3 \end{array}\right] , \text{ and }
\left[\begin{array}{c} -5 \\ 1 \\ -5 \end{array}\right]\text{.}\)
(i)
Write a statement involving the solutions of a vector equation that’s equivalent to this claim.
The bottom row requires \(0=1\text{.}\) Therefore the vector equation has no solutions, so \(\left[\begin{array}{c} -13 \\ 3 \\ -15 \end{array}\right]\) is not a linear combination.
(iii)
If your statement was true, explain and demonstrate how to construct a specific linear combination of \(\left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} 2 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ 3 \end{array}\right] , \text{ and } \left[\begin{array}{c} -5 \\ 1 \\ -5 \end{array}\right]\) that equals \(\left[\begin{array}{c} -13 \\ 3 \\ -15 \end{array}\right]\text{.}\)
Solution.
N/A
(b)
\(\left[\begin{array}{c} -13 \\ 3 \\ -13 \end{array}\right]\)is a linear combination of the vectors \(\left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} 2 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ 3 \end{array}\right] , \text{ and } \left[\begin{array}{c} -5 \\ 1 \\ -5 \end{array}\right]\text{.}\)
(i)
Write a statement involving the solutions of a vector equation that’s equivalent to this claim.
No row requires \(0=1\text{.}\) Therefore vector equation has at least one solution, so \(\left[\begin{array}{c} -13 \\ 3 \\ -13 \end{array}\right]\) is a linear combination.
(iii)
If your statement was true, explain and demonstrate how to construct a specific linear combination of \(\left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} 2 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ 3 \end{array}\right] , \text{ and } \left[\begin{array}{c} -5 \\ 1 \\ -5 \end{array}\right]\) that equals \(\left[\begin{array}{c} -13 \\ 3 \\ -13 \end{array}\right]\text{.}\)
Solution1.
By setting the free variables \(x_2=0\) and \(x_3=0\text{,}\) we obtain the equations \(x_1=2\) and \(x_4=3\text{.}\) Therefore we may construct
has at least one solution for every \(\vec w\in\mathbb R^4\text{.}\)”
(ii)
Explain and demonstrate how to determine whether or not this statement is true.
Solution1.
Note that \(\mathrm{RREF}\, \left[\begin{array}{ccc} 1 & -5 & 12 \\
0 & 1 & -2 \\
1 & -5 & 12 \\
1 & -2 & 6 \end{array}\right] =
\left[\begin{array}{ccc} 1 & 0 & 2 \\
0 & 1 & -2 \\
0 & 0 & 0 \\
0 & 0 & 0 \end{array}\right]\) has a row of zeros that makes a \(0=1\) contradiction possible.
Therefore the vector equation will not have solutions for every \(\vec w\text{,}\) and thus the set of vectors does not span \(\mathbb{R}^4\text{.}\)
Solution2.
It takes at least \(4\) vectors to span \(\mathbb R^4\text{,}\) so the equation cannot always have solutions and the set cannot span.
ExampleC.1.7.EV3.
Answer the following questions about Euclidean subspaces.
(a)
Consider the following subsets of Euclidean space \(\mathbb R^4\) defined by
\begin{equation*}
U=\left\{ \left[\begin{array}{c} x \\ y \\ z \\ w \end{array}\right] \middle|
y^{2} - 7 \, z^{2} = x \right\} \hspace{1em} \text{and}
\hspace{1em} W=\left\{ \left[\begin{array}{c} x \\ y \\ z \\ w \end{array}\right] \middle|
-5 \, w - 7 \, x - y = -7 \, z \right\}
\end{equation*}
Without writing a proof, explain why only one of these subsets is likely to be a subspace.
Solution.
\(W\) appears to be a subspace as its equation is a linear combination of variables and constant scalars, and \(U\) is likely not due to its equation having squared terms.
(b)
Consider the following subset of Euclidean space \(\mathbb R^3\)
\begin{equation*}
Q=\left\{ \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \middle|
0 = 5 \, y^{2} - 5 \, x + 3 \, z \right\}
\end{equation*}
Prove that \(Q\)is not a subspace.
Solution.
Note that \(\left[\begin{array}{c} 0 \\ 3 \\ -15 \end{array}\right]\) belongs to \(Q\text{,}\) since
Consider the following subset of Euclidean space \(\mathbb R^3\)
\begin{equation*}
R=\left\{ \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \middle|
5 \, x - 5 \, y = -4 \, z \right\}
\end{equation*}
Prove that \(R\)is a subspace.
Solution.
First, note that \(\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\in R\) since \(5(0)-5(0)=0\) and \(-4(0)=0\) as well.
Let \(\left[\begin{array}{c} x \\ y \\ z \end{array}\right]\in R\) so that \(5 \, x - 5 \, y = -4 \, z\text{,}\) and let \(\left[\begin{array}{c} a \\ b \\ c \end{array}\right]\in R\) so that \(5 \, a - 5 \, b = -4 \, c\text{.}\) We may then compute
So \(5(x+a)-5(y+b) = -4(z+c)\) and therefore \(\left[\begin{array}{c} x \\ y \\ z \end{array}\right]+
\left[\begin{array}{c} a \\ b \\ c \end{array}\right]=
\left[\begin{array}{c} x+a \\ y+b \\ z+c \end{array}\right]\in R\text{,}\) showing \(R\) is closed under addition.
Let \(\left[\begin{array}{c} x \\ y \\ z \end{array}\right]\in R\) so that \(5 \, x - 5 \, y = -4 \, z\text{,}\) and let \(k\in\mathbb R\) be a scalar. We may then compute
and therefore \(k\left[\begin{array}{c} x \\ y \\ z \end{array}\right]
=\left[\begin{array}{c} kx \\ ky \\ kz \end{array}\right]\in R\text{,}\) showing \(R\) is closed under scalar multiplication.
Since the RREF matrix has two non-pivot columns (the second and fourth), the solution set has free variables and thus there are more than one solution. This means the set is linearly dependent.
Since the RREF matrix has two non-pivot columns (the fourth and fifth), the solution set has free variables and thus there are more than one solution. This means the set is linearly dependent.
Solution2.
Since these vectors are from \(\mathbb R^4\) and there are more than \(4\) vectors, the equation must have infinitely-many solutions and the set must be linearly dependent.
Since the RREF matrix has all pivot columns, the solution set lacks free variables and thus there is exactly one solution. This means the set is linearly independent.
has exactly one solution for every \(\vec w\in\mathbb R^4\text{.}\)”
(ii)
Explain and demonstrate how to determine whether or not this statement is true.
Solution1.
Since \(\mathrm{RREF}\, \left[\begin{array}{cccc} 1 & 2 & 3 & 5 \\ -2 & -4 & -6 & -5 \\ 0 & 0 & 0 & -2 \\ 1 & 2 & 3 & 4 \end{array}\right] = \left[\begin{array}{cccc} 1 & 2 & 3 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\text{,}\) we see from the zero row that there are some vectors \(\vec w\) for which the equation is not true, so the set fails to span and therefore fails to be a basis.
Solution2.
Since \(\mathrm{RREF}\, \left[\begin{array}{cccc} 1 & 2 & 3 & 5 \\ -2 & -4 & -6 & -5 \\ 0 & 0 & 0 & -2 \\ 1 & 2 & 3 & 4 \end{array}\right] = \left[\begin{array}{cccc} 1 & 2 & 3 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\text{,}\) we see from the non-pivot column that there are some vectors \(\vec w\) for which the equation has infinitely-many solutions, so the set is linearly dependent and therefore fails to be a basis.
has exactly one solution for every \(\vec w\in\mathbb R^4\text{.}\)”
(ii)
Explain and demonstrate how to determine whether or not this statement is true.
Solution1.
Since \(\mathrm{RREF}\, \left[\begin{array}{ccc} 1 & -1 & 0 \\ 3 & -3 & 1 \\ 4 & -4 & 3 \\ -4 & 4 & -3 \end{array}\right] = \left[\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\) we see from the zero row that there are some vectors \(\vec w\) for which the equation is not true, so the set fails to span and therefore fails to be a basis.
Solution2.
The set has only three vectors, so the set cannot span and there must be vectors for which the equation has no solutions. Therefore the set is not a basis.
Answer the following questions about transformations.
Consider the following maps of Euclidean vectors \(P:\mathbb R^3\rightarrow\mathbb R^3\) and \(Q:\mathbb R^3\rightarrow\mathbb R^3\) defined by
\begin{equation*}
P\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)=
\left[\begin{array}{c} 3 \, x - y + z \\ 2 \, x - 2 \, y + 4 \, z \\ -2 \, x - 2 \, y - 3 \, z \end{array}\right]
\hspace{1em} \text{and} \hspace{1em} Q\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)=
\left[\begin{array}{c} y - 2 \, z \\ -3 \, x - 4 \, y + 12 \, z \\ 5 \, x y + 3 \, z \end{array}\right].
\end{equation*}
Without writing a proof, explain why only one of these maps is likely to be a linear transformation.
Consider the following map of Euclidean vectors \(S:\mathbb R^2\rightarrow\mathbb R^2\)
\begin{equation*}
S\left( \left[\begin{array}{c} x \\ y \end{array}\right] \right)=\left[\begin{array}{c} x + 2 \, y \\ -3 \, x y \end{array}\right].
\end{equation*}
Prove that \(S\)is not a linear transformation.
Consider the following map of Euclidean vectors \(T:\mathbb R^2\rightarrow\mathbb R^2\)
\begin{equation*}
T\left( \left[\begin{array}{c} x \\ y \end{array}\right] \right)=\left[\begin{array}{c} -4 \, x - 5 \, y \\ 2 \, x - 4 \, y \end{array}\right].
\end{equation*}
Prove that \(T\)is a linear transformation.
Solution.
A linear map between Euclidean spaces must consist of linear polynomials in each component. All three components of \(P\) are linear so \(P\) is likely to be linear; however, the third component of \(Q\) contains the nonlinear term \(xy\text{,}\) so \(Q\) is unlikely to be linear.
We need to show either that \(S\) fails to preserve either vector addition or that \(S\) fails to preserve scalar multiplication.
For example, for a scalar \(c \in \IR\) and a vector \(\left[\begin{array}{c}x \\y \end{array}\right] \in \IR^2\text{,}\) we can compute
\begin{equation*}
cS\left(\left[\begin{array}{c} x \\ y \end{array} \right]\right) =
c\left[\begin{array}{c}x+2y \\ -3xy \end{array} \right] =
\left[\begin{array}{c}cx+2cy \\ -3cxy \end{array} \right].
\end{equation*}
Since \(-3c^2xy \neq -3cxy\text{,}\) we see that \(S\left(c\left[\begin{array}{c} x \\ y \end{array} \right]\right) \neq cS\left(\left[\begin{array}{c} x \\ y \end{array} \right]\right)\text{,}\) so \(S\) fails to preserve scalar multiplication and cannot be a linear transformation.
Alternatively, we could instead take two vectors \(\left[\begin{array}{c}x_1 \\y_1 \end{array}\right], \left[\begin{array}{c}x_2 \\y_2 \end{array}\right] \in \IR^2\) and compute
Since \(-3(x_1+x_2)(y_1+y_2) \neq -3x_1y_1-3x_2y_2 \text{,}\) we see that \(S \left( \left[\begin{array}{c}x_1 \\y_1 \end{array}\right] + \left[\begin{array}{c}x_2 \\y_2 \end{array}\right]\right) \neq
S \left( \left[\begin{array}{c}x_1 \\y_1 \end{array}\right]\right) +S\left( \left[\begin{array}{c}x_2 \\y_2 \end{array}\right]\right)
\text{,}\) so \(S\) fails to preserve addition and cannot be a linear transformation.
We need to show that \(T\) preserves both vector addition and that \(T\) preserves scalar multiplication.
First, let us take two vectors \(\left[\begin{array}{c}x_1 \\y_1 \end{array}\right], \left[\begin{array}{c}x_2 \\y_2 \end{array}\right] \in \IR^2\) and compute
So we see that \(T \left( \left[\begin{array}{c}x_1 \\y_1 \end{array}\right] + \left[\begin{array}{c}x_2 \\y_2 \end{array}\right]\right) =
T \left( \left[\begin{array}{c}x_1 \\y_1 \end{array}\right]\right) +T\left( \left[\begin{array}{c}x_2 \\y_2 \end{array}\right]\right)
\text{,}\) so \(T\) preserves addition.
Now, take a scalar \(c \in \IR\) and a vector \(\left[\begin{array}{c}x \\y \end{array}\right] \in \IR^2\text{,}\) and compute
\begin{equation*}
cT\left(\left[\begin{array}{c} x \\ y \end{array} \right]\right) =
c\left[\begin{array}{c}-4x-5y \\ 2x-4y \end{array} \right] =
\left[\begin{array}{c}-4cx-5cy \\ 2cx-4cy\end{array} \right].
\end{equation*}
We see that \(T\left(c\left[\begin{array}{c} x \\ y \end{array} \right]\right) = cT\left(\left[\begin{array}{c} x \\ y \end{array} \right]\right)\text{,}\) so \(T\) preserves scalar multiplication.
Since \(T\) preserves both addition and scalar multiplication, we have proven that \(T\) is a linear transformation.
ExampleC.1.13.AT2.
Find the standard matrix for the linear transformation \(T: \IR^3\rightarrow \IR^4\) given by
\begin{equation*}
T\left(\left[\begin{array}{c} x \\ y \\ z \\ \end{array}\right] \right) = \left[\begin{array}{c} -x+y \\ -x+3y-z \\ 7x+y+3z \\ 0 \end{array}\right].
\end{equation*}
Let \(S: \IR^4 \rightarrow \IR^3\) be the linear transformation given by the standard matrix
\begin{equation*}
\ker T = \setBuilder{\left[\begin{array}{c}-5a+9b \\ a-2b \\ a \\ b \end{array}\right]}{a,b \in \IR}.
\end{equation*}
Since \(\Im(T) = \vspan\left\{\left[\begin{array}{c}1 \\ 2 \\ 1 \end{array}\right], \left[\begin{array}{c} 3 \\ 4 \\ 6 \end{array}\right],
\left[\begin{array}{c} 2 \\ 6 \\ -1 \end{array}\right], \left[\begin{array}{c} -3 \\ -10 \\ 3 \end{array}\right]\right\}\text{,}\) we simply need to find a linearly independent subset of these four spanning vectors. So we compute
Since the first two columns are pivot columns, they form a linearly independent spanning set, so a basis for \(\Im T\) is \(\setList{\left[\begin{array}{c}1\\2\\1 \end{array}\right], \left[\begin{array}{c}3\\4\\6 \end{array}\right]}.\)
To find a basis for the kernel, note that
\begin{equation*}
\ker T = \setBuilder{\left[\begin{array}{c}-5a+9b \\ a-2b \\ a \\ b \end{array}\right]}{a,b \in \IR}
\end{equation*}
The dimension of the image (the rank) is \(2\text{,}\) the dimension of the kernel (the nullity) is \(2\text{,}\) and the dimension of the domain of \(T\) is \(4\text{,}\) so we see \(2+2=4\text{,}\) which verifies that the sum of the rank and nullity of \(T\) is the dimension of the domain of \(T\text{.}\)
ExampleC.1.15.AT4.
Let \(T: \IR^4 \rightarrow \IR^3\) be the linear transformation given by the standard matrix \(\left[\begin{array}{cccc} 1 & 3 & 2 & -3 \\ 2 & 4 & 6 & -10 \\ 1 & 6 & -1 & 3 \end{array}\right]\text{.}\)
Note that the third and fourth columns are non-pivot columns, which means \(\ker T\) contains infinitely many vectors, so \(T\) 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 \(\IR^3\)), so \(T\) is not surjective.
ExampleC.1.16.AT5.
Let \(V\) be the set of all pairs of numbers \((x,y)\) of real numbers together with the following operations:
Since these are the same, we have shown that the property holds.
To show \(V\) 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
\begin{equation*}
(c+d)\odot(x,y) \neq c \odot(x,y) \oplus d\odot(x,y)
\end{equation*}
Explain which two may be multiplied and why. Then show how to find their product.
Solution.
\(AC\) is the only one that can be computed, since \(C\) corresponds to a linear transformation \(\mathbb{R}^3 \rightarrow \mathbb{R}^2\) and \(A\) corresponds to a linear transfromation \(\mathbb{R}^2 \rightarrow \mathbb{R}^2\text{.}\) Thus the composition \(AC\) corresponds to a linear transformation \(\mathbb{R}^3 \rightarrow \mathbb{R}^2\) with a \(2\times 3\) standard matrix. We compute
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.
Using the techniques from section Section 4.3, and letting \(M = \left[\begin{array}{ccc}
1& 2& 1\\
0& 0& 2\\
1& 1& 1\\
\end{array}\right]\text{,}\) we find \(M^{-1} = \left[\begin{array}{ccc}
-1& -1/2& 2\\
1& 0& -1\\
0& 1/2& 0\\
\end{array}\right]\text{.}\) Our equation can be written as \(M\vec{v} = \left[\begin{array}{c}4\\ -2 \\ 2 \end{array}\right]\text{,}\) and may therefore be solved via
Give a \(4\times 4\) matrix \(P\) that may be used to perform the row operation \({R_3} \to R_3+4 \, {R_1} \text{.}\)
Give a \(4\times 4\) matrix \(Q\) that may be used to perform the row operation \({R_1} \to -4 \, {R_1}\text{.}\)
Use matrix multiplication to describe the matrix obtained by applying \({R_3} \to 4 \, {R_1} + {R_3}\) and then \({R_1} \to -4 \, {R_1}\) to \(A\) (note the order).
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 \(2\times 2\) determinants:
Here is another possible solution, using row and column operations to first reduce the determinant to a \(3\times 3\) matrix and then applying a formula:
