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 \(\left\{ \left[\begin{array}{c}
-a + 1 \\
a \\
1 \\
2
\end{array}\right] \,\middle|\, a \in\mathbb R \right\} \text{.}\)
ExampleB.1.5.EV1.
Write a statement involving the solutions of a vector equation that’s equivalent to each claim below.
\(\left[\begin{array}{c}
-13 \\
3 \\
-13
\end{array}\right]\)is a linear combination of the vectors \(\left[\begin{array}{c}
1 \\
0 \\
1
\end{array}\right] , \left[\begin{array}{c}
2 \\
0 \\
2
\end{array}\right] , \left[\begin{array}{c}
3 \\
0 \\
3
\end{array}\right] , \text{ and } \left[\begin{array}{c}
-5 \\
1 \\
-5
\end{array}\right]\text{.}\)
\(\left[\begin{array}{c}
-13 \\
3 \\
-15
\end{array}\right]\)is a linear combination of the vectors \(\left[\begin{array}{c}
1 \\
0 \\
1
\end{array}\right] , \left[\begin{array}{c}
2 \\
0 \\
2
\end{array}\right] , \left[\begin{array}{c}
3 \\
0 \\
3
\end{array}\right] , \text{ and } \left[\begin{array}{c}
-5 \\
1 \\
-5
\end{array}\right]\text{.}\)
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.
We see that this vector equation has solution set \(\left\{\left[\begin{array}{c}2-2a-3b \\ a \\ b \\ 3 \end{array}\right]\ \middle|\ a,b \in \mathbb{R}\right\}\text{,}\) so \(\left[\begin{array}{c}
-13 \\
3 \\
-13
\end{array}\right]\) is a linear combination; for example, \(2 \left[\begin{array}{c}
1 \\
0 \\
1
\end{array}\right] + 3 \left[\begin{array}{c}
-5 \\
1 \\
-5
\end{array}\right] = \left[\begin{array}{c}
-13 \\
3 \\
-13
\end{array}\right]\)
\(\left[\begin{array}{c}
-13 \\
3 \\
-15
\end{array}\right]\) is a linear combination of the vectors \(\left[\begin{array}{c}
1 \\
0 \\
1
\end{array}\right] , \left[\begin{array}{c}
2 \\
0 \\
2
\end{array}\right] , \left[\begin{array}{c}
3 \\
0 \\
3
\end{array}\right] , \text{ and } \left[\begin{array}{c}
-5 \\
1 \\
-5
\end{array}\right]\) exactly when the vector equation
has a solution for all\(\vec{v} \in \mathbb{R}^4\text{.}\) If there is some vector \(\vec{v} \in \mathbb{R}^4\) for which this vector equation has no solution, then the set does not span \(\mathbb{R}^4\text{.}\) To answer this, we compute
We see that for some \(\vec{v} \in \mathbb{R}^4\text{,}\) this vector equation will not have a solution, so the set of vectors \(\left\{ \left[\begin{array}{c}
1 \\
-1 \\
2 \\
0
\end{array}\right] , \left[\begin{array}{c}
3 \\
-2 \\
3 \\
3
\end{array}\right] , \left[\begin{array}{c}
10 \\
-7 \\
11 \\
9
\end{array}\right] , \left[\begin{array}{c}
-6 \\
3 \\
-3 \\
-9
\end{array}\right] \right\}\) does not span \(\mathbb{R}^4\text{.}\)
ExampleB.1.7.EV3.
Consider the following two sets of Euclidean vectors.
\begin{equation*}
W = \setBuilder{\left[\begin{array}{c} x \\ y \\ z \\ w \end{array}\right] }{x+y=3z+2w}
\hspace{3em}
U = \setBuilder{\left[\begin{array}{c} x \\ y \\ z \\ w \end{array}\right]}{x+y=3z+w^2}
\end{equation*}
Explain why one of these sets is a subspace of \(\IR^3\text{,}\) and why the other is not.
To show that \(W\) is a subspace, let \(\vec v=\left[\begin{array}{c} x_1 \\y_1 \\ z_1 \\ w_1 \end{array}\right]\in W\) and \(\vec w=\left[\begin{array}{c} x_2 \\y_2 \\ z_2 \\ w_2 \end{array}\right] \in W
\text{,}\) so we know that \(x_1+y_1=3z_1+2w_1\) and \(x_2+y_2=3z_2+2w_2\text{.}\)
and we see that \(c\vec v\in W\text{,}\) so \(W\) is closed under scalar multiplication. Therefore \(W\) is a subspace of \(\IR^3\text{.}\)
Now, to show \(U\) is not a subspace, we will show that it is not closed under vector addition.
(Solution Method 1) Now let \(\vec v=\left[\begin{array}{c} x_1 \\y_1 \\ z_1 \\ w_1 \end{array}\right]\in U\) and \(\vec w=\left[\begin{array}{c} x_2 \\y_2 \\ z_2 \\ w_2 \end{array}\right] \in U
\text{,}\) so we know that \(x_1+y_1=3z_1+w_1^2\) and \(x_2+y_2=3z_2+w_2^2\text{.}\)
and thus \(\vec v+\vec w\in U\) \textbf{only when} \(w_1^2+w_2^2=(w_1+w_2)^2\text{.}\) Since this is not true in general, \(U\) is not closed under vector addition, and thus cannot be a subspace.
(Solution Method 2) Note that the vector \(\vec v=\left[\begin{array}{c} 0\\1\\0\\1\end{array}\right] \) belongs to \(U\) since \(0+1=3(0)+1^2\text{.}\) However, the vector \(2\vec v=\left[\begin{array}{c} 0\\2\\0\\2\end{array}\right] \) does not belong to \(U\) since \(0+2\not=3(0)+2^2\text{.}\) Therefore \(U\) is not closed under scalar multiplication, and thus is not a subspace.
ExampleB.1.8.EV4.
Write a statement involving the solutions of a vector equation that’s equivalent to each claim below.
The set of vectors \(\left\{ \left[\begin{array}{c}
1 \\
3 \\
4 \\
-4
\end{array}\right] , \left[\begin{array}{c}
-1 \\
-3 \\
-4 \\
4
\end{array}\right] , \left[\begin{array}{c}
0 \\
1 \\
3 \\
-3
\end{array}\right] \right\}\) is linearly independent.
The set of vectors \(\left\{ \left[\begin{array}{c}
1 \\
3 \\
4 \\
-4
\end{array}\right] , \left[\begin{array}{c}
-1 \\
-3 \\
-4 \\
4
\end{array}\right] , \left[\begin{array}{c}
0 \\
1 \\
3 \\
-3
\end{array}\right] \right\}\) is linearly dependent.
Explain how to determine which of these statements is true.
Thus, this vector equation has a solution set \(\left\{ \left[\begin{array}{c}a \\ a \\ 0 \end{array}\right]\ \middle|\ a \in \mathbb{R}\right\}\text{.}\) Since there are nontrivial solutions, we conclude that the set of vectors \(\left\{ \left[\begin{array}{c}
1 \\
3 \\
4 \\
-4
\end{array}\right] , \left[\begin{array}{c}
-1 \\
-3 \\
-4 \\
4
\end{array}\right] , \left[\begin{array}{c}
0 \\
1 \\
3 \\
-3
\end{array}\right] \right\}\) is linearly dependent.
ExampleB.1.9.EV5.
Write a statement involving spanning and independence properties that’s equivalent to each claim below.
The set of vectors \(\left\{ \left[\begin{array}{c}
1 \\
3 \\
4 \\
-4
\end{array}\right] , \left[\begin{array}{c}
0 \\
1 \\
3 \\
-3
\end{array}\right] , \left[\begin{array}{c}
3 \\
11 \\
18 \\
-18
\end{array}\right] , \left[\begin{array}{c}
-2 \\
-7 \\
-11 \\
11
\end{array}\right] \right\}\) is a basis of \(\mathbb{R}^4\text{.}\)
The set of vectors \(\left\{ \left[\begin{array}{c}
1 \\
3 \\
4 \\
-4
\end{array}\right] , \left[\begin{array}{c}
0 \\
1 \\
3 \\
-3
\end{array}\right] , \left[\begin{array}{c}
3 \\
11 \\
18 \\
-18
\end{array}\right] , \left[\begin{array}{c}
-2 \\
-7 \\
-11 \\
11
\end{array}\right] \right\}\) is not a basis of \(\mathbb{R}^4\text{.}\)
Explain how to determine which of these statements is true.
The set of vectors \(\left\{ \left[\begin{array}{c}
1 \\
3 \\
4 \\
-4
\end{array}\right] , \left[\begin{array}{c}
0 \\
1 \\
3 \\
-3
\end{array}\right] , \left[\begin{array}{c}
3 \\
11 \\
18 \\
-18
\end{array}\right] , \left[\begin{array}{c}
-2 \\
-7 \\
-11 \\
11
\end{array}\right] \right\}\) is a basis of \(\mathbb{R}^4\) exactly when it is linearly independent and the set spans \(\mathbb{R}^4\text{.}\) If it is either linearly dependent, or the set does not span \(\mathbb{R}^4\text{,}\) then the set is not a basis.
We see that this set of vectors is linearly dependent, so therefore the set of vectors \(\left\{ \left[\begin{array}{c}
1 \\
3 \\
4 \\
-4
\end{array}\right] , \left[\begin{array}{c}
0 \\
1 \\
3 \\
-3
\end{array}\right] , \left[\begin{array}{c}
3 \\
11 \\
18 \\
-18
\end{array}\right] , \left[\begin{array}{c}
-2 \\
-7 \\
-11 \\
11
\end{array}\right] \right\}\) is not a basis.
But note that \(S(f(x))=3xf(x)\) and \(S(g(x))=3xg(x)\text{,}\) so we have \(S(f(x)+g(x))=S(f(x))+S(g(x))\text{.}\)
For the second part, we compute
\begin{align*}
S\left(cf(x)\right) &= 3x\left(cf(x)\right) & \text{by definition of } \\\\
&= 3cxf(x) & \text{rewriting the multiplication.}
\end{align*}
But note that \(cS(f(x))=c(3xf(x))=3cxf(x)\) as well, so we have \(S(cf(x))=cS(f(x))\text{.}\) Now, since \(S\) respects both addition and scalar multiplication, we can conclude \(S\) is a linear transformation.
(Solution method 1) As for \(T\text{,}\) we compute
\begin{align*}
T(f(x)+g(x))& =3 (f(x)+g(x))'(f(x)+g(x)) &\text{by definition of }\\
&= 3(f'(x)+g'(x))(f(x)+g(x)) & \text{since the derivative is linear}\\
&= 3f(x)f'(x)+3f(x)g'(x)+3f'(x)g(x)+3g(x)g'(x) &\text{by distributing}
\end{align*}
However, note that \(T(f(x))+T(g(x))=3f'(x)f(x)+3g'(x)g(x)\text{,}\) which is not always the same polynomial (for example, when \(f(x)=g(x)=x\)). So we see that \(T(f(x)+g(x)) \neq T(f(x))+T(g(x))\text{,}\) so \(T\) does not respect addition and is therefore not a linear transformation.
(Solution method 2) As for \(T\text{,}\) we may choose the polynomial \(f(x)=x\) and scalar \(c=2\text{.}\) Then
\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 \(T\text{.}\)
ExampleB.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.
ExampleB.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*}
\(AC\) 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:
