Let $W$ be the set of $3\times 3$ skew-symmetric matrices. Show that $W$ is a subspace of the vector space $V$ of all $3\times 3$ matrices. Then, exhibit a spanning set for $W$.

Proof.

To prove that $W$ is a subspace of $V$, the $3\times 3$ zero matrix $\mathbf{0}$ is the zero vector of $V$, and since $\mathbf{0}$ is clearly skew-symmetric, $\mathbf{0}\in W$. Next, let $A,B\in W$. Then $A^{T}=-A$ and $B^{T}=-B$. Therefore,
\[
(A+B)^{T}
=A^{T}+B^{T}
=-A+(-B)
=-(A+B).
\] Thus $A+B$ is skew-symmetric, and therefore $A+B\in W$. Next, take any $A\in W$ and $r$ in the scalar field. Then
\[
(rA)^{T}
=rA^{T}
=r(-A)
=-rA.
\] Thus $rA$ is skew symmetric, which implies $rA\in W$. Therefore, $W$ is a subspace of $V$.

To exhibit a spanning set for $W$, first note that
\[
W=
\left\{
A\left|
A=
\begin{bmatrix}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{bmatrix}
,\;a,b,c\in\R
\right.\right\}.
\] Let
\[
A_{1}=
\begin{bmatrix}
0 & 1 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}
,\;
A_{2}=
\begin{bmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
-1 & 0 & 0
\end{bmatrix}
,\;
A_{3}=
\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & -1 & 0
\end{bmatrix}
.
\] Then any $A\in W$ can be written as
\[
A=
\begin{bmatrix}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{bmatrix}
=aA_{1}+bA_{2}+cA_{3},
\] which is a linear combination of $A_{1},A_{2},A_{3}$. Thus $\{A_{1},A_{2},A_{3}\}$ is a spanning set for $W$.

Find a basis for $\Span(S)$ where $S=
\left\{
\begin{bmatrix}
1 \\ 2 \\ 1
\end{bmatrix}
,
\begin{bmatrix}
-1 \\ -2 \\ -1
\end{bmatrix}
,
\begin{bmatrix}
2 \\ 6 \\ -2
\end{bmatrix}
,
\begin{bmatrix}
1 \\ 1 \\ 3
\end{bmatrix}
\right\}$.

Solution.

We will first use the leading $1$ method. Consider the system
\[
x_{1}
\begin{bmatrix}
1 \\ 2 \\ 1
\end{bmatrix}
+x_{2}
\begin{bmatrix}
-1 \\ -2 \\ -1
\end{bmatrix}
+x_{3}
\begin{bmatrix}
2 \\ 6 \\ -2
\end{bmatrix}
+x_{4}
\begin{bmatrix}
1 \\ 1 \\ 3
\end{bmatrix}
=
\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}
.\] The augmented matrix for this system is
\[
\left[\begin{array}{cccc|c}
1 & -1 & 2 & 1 & 0 \\
2 & -2 & 6 & 1 & 0 \\
1 & -1 & -2 & 3 & 0
\end{array}\right] \xrightarrow{\substack{R_{2}-2R_{1} \\ R_{3}-R_{1}}}
\left[\begin{array}{cccc|c}
1 & -1 & 2 & 1 & 0 \\
0 & 0 & 2 & -1 & 0 \\
0 & 0 & -4 & 2 & 0
\end{array}\right] \] \[
\to
\left[\begin{array}{cccc|c}
1 & -1 & 0 & 2 & 0 \\
0 & 0 & 1 & -1/2 & 0 \\
0 & 0 & 0 & 0
\end{array}\right].
\] Since the above matrix has leading $1$’s in the first and third columns, we can conclude that the first and third vectors of $S$ form a basis of $\Span(S)$. Thus
\[
\left\{
\begin{bmatrix}
1 \\ 2 \\ 1
\end{bmatrix}
,
\begin{bmatrix}
2 \\ 6 \\ -2
\end{bmatrix}
\right\}
\] is a basis for $\Span(S)$.

Let $A=\begin{bmatrix}
2 & 4 & 6 & 8 \\
1 &3 & 0 & 5 \\
1 & 1 & 6 & 3
\end{bmatrix}$.

(a) Find a basis for the nullspace of $A$.

(b) Find a basis for the row space of $A$.

(c) Find a basis for the range of $A$ that consists of column vectors of $A$.

(d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.

Solution.

We first obtain the reduced row echelon form matrix corresponding to the matrix $A$.
We reduce the matrix $A$ as follows:
\begin{align*}
A=\begin{bmatrix}
2 & 4 & 6 & 8 \\
1 &3 & 0 & 5 \\
1 & 1 & 6 & 3
\end{bmatrix}
\xrightarrow{\frac{1}{2}R_1}
\begin{bmatrix}
1 & 2 & 3 & 4 \\
1 &3 & 0 & 5 \\
1 & 1 & 6 & 3
\end{bmatrix}\\[6pt] \xrightarrow[R_3-R_1]{R_2-R_1}
\begin{bmatrix}
1 & 2 & 3 & 4 \\
0 &1 & -3 & 1 \\
0 & -1 & 3 & -1
\end{bmatrix}
\xrightarrow[R_3+R_2]{R_1-2R_2}
\begin{bmatrix}
1 & 0 & 9 & 2 \\
0 &1 & -3 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}.
\end{align*}
The last matrix is in reduced row echelon form. That is,
\[\rref(A)=\begin{bmatrix}
1 & 0 & 9 & 2 \\
0 &1 & -3 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}. \tag{*}\]

(a) Find a basis for the nullspace of $A$.

By the computation above, we see that the general solution of $A\mathbf{x}=\mathbf{0}$ is
\begin{align*}
x_1&=-9x_3-2x_4\\
x_2&=3x_3-x_4,
\end{align*}
where $x_3$ and $x_4$ are free variables.
Thus, the vector form solution to $A\mathbf{x}=\mathbf{0}$ is
\begin{align*}
\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}
=\begin{bmatrix}
-9x_3-2x_4 \\
3x_3-x_4 \\
x_3 \\
x_4
\end{bmatrix}
=x_3\begin{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix}.
\end{align*}
It follows that the nullspace of the matrix $A$ is given by
\begin{align*}
\calN(A)&=\left\{ \mathbf{x}\in \R^4 \quad \middle | \quad \mathbf{x}= x_3\begin{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix}, \text{ for all } x_3, x_4 \in \R^4 \right \}\\[6pt] &= \Span \left\{ \begin{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix} \right \}.
\end{align*}
Thus, the set
\[\left\{ \begin{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix} \right \}\] is a spanning set for the nullspace $\calN(A)$.
It is straightforward to see that this set is linearly independent, and hence it is a basis for $\calN(A)$.

(b) Find a basis for the row space of $A$.

Recall that the nonzero rows of $\rref(A)$ form a basis for the row space of $A$.
Thus,
\[\left\{\begin{bmatrix}
1 \\
0 \\
9 \\
2
\end{bmatrix}, \quad \begin{bmatrix}
0 \\
1 \\
-3 \\
1
\end{bmatrix} \right \}\] is a basis for the row space of $A$.

(c) Find a basis for the range of $A$ that consists of column vectors of $A$.

Recall that by the leading 1 method, the columns of $A$ corresponding to columns of $\rref(A)$ that contain leading 1 entries form a basis for the range $\calR(A)$ of $A$.
From (*), we see that the first and the second columns contain the leading 1 entries. Thus,
\[\left\{\begin{bmatrix}
2 \\
1 \\
1
\end{bmatrix}, \begin{bmatrix}
4 \\
3 \\
1
\end{bmatrix}\right \}\] is a basis for the range $\calR(A)$ of $A$.

(d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.

Let us write $A_1, A_2, A_3$, and $A_4$ for the column vectors of the matrix $A$.
In part (c), we showed that $\{A_1, A_2\}$ is a basis for the range $\calR(A)$.
Thus, we need to express the vectors $A_3$ and $A_4$ as a linear combination of $A_1$ and $A_2$, respectively.

A shortcut is to note that the entries of third column vector of $\rref(A)$ give the coefficients of the linear combination for $A_3$. That is, we have
\[A_3=9A_1-3A_2.\] Similarly, the entries of the fourth column of $\rref(A)$ yield
\[A_4=2A_1+A_2.\] Click here if solved 176The post How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix first appeared on Problems in Mathematics.]]> https://yutsumura.com/how-to-find-a-basis-for-the-nullspace-row-space-and-range-of-a-matrix/feed/ 0 6926 https://yutsumura.com/can-we-reduce-the-number-of-vectors-in-a-spanning-set/ https://yutsumura.com/can-we-reduce-the-number-of-vectors-in-a-spanning-set/#respond Mon, 26 Feb 2018 13:09:40 +0000 https://yutsumura.com/?p=6923 Suppose that a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^3$. Is it possible that $S_2=\{\mathbf{v}_1\}$ is a spanning set for $V$?   Solution. Yes, in...

Suppose that a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^5$. If $\mathbf{v}_4$ is another vector in $V$, then is the set
\[S_2=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}\] still a spanning set for $V$? If so, prove it. Otherwise, give a counterexample.

Proof.

We prove that $S_2$ is also a spanning set for $V$, that is, we prove that
\[\Span(S_2)=V.\]

Prove $\Span(S_2) \subset V$

We first show that $\Span(S_2)$ is contained in $V$. Let $\mathbf{x}$ be an element in $\Span(S_2)$. Then there exist scalars $c_1, c_2, c_3, c_4$ such that
\[\mathbf{x}=c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3 \mathbf{v}_3 + c_4 \mathbf{v}_4.\] Since $\Span(S_1)=V$, we know that $c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3 \mathbf{v}_3$ is a vector in $V$. As $\mathbf{v}_4\in V$, we have $c_4\mathbf{v}_4 \in V$.
Since $V$ is a vector space, the sum of two elements in $V$ is in $V$.
So, \[\mathbf{x}=(c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3 \mathbf{v}_3) + (c_4 \mathbf{v}_4) \in V.\] This proves that $\Span(S_2) \subset V$.

Prove $\Span(S_2) \supset V$

Note that since $S_1$ is a spanning set for $V$, every element of $S_1$ can be written as a linear combination of the vectors $\mathbf{v}_1, \mathbf{v}_2$, and $\mathbf{v}_3$.
That is, for any $\mathbf{v}\in V$, there exist scalars $c_1, c_2, c_3$ such that
\[\mathbf{v}=c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3.\] Observe that this can be written as follows.
\[\mathbf{v}=c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3+0\mathbf{v}_4.\] This tells us that $\mathbf{v}$ is a linear combination of $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$, and $\mathbf{v}_4$.
Hence, any vector in $V$ can be written as a linear combination of the vectors in $S_2$.
Thus, $V\subset \Span(S_2)$.

Putting these inclusion together yields that $V=\Span(S_2)$, and hence $S_2$ is a spanning set for $V$.

Let $A=\begin{bmatrix}
2 & 4 & 6 & 8 \\
1 &3 & 0 & 5 \\
1 & 1 & 6 & 3
\end{bmatrix}$.
(a) Find a basis for the nullspace of $A$.

(b) Find a basis for the row space of $A$.

(c) Find a basis for the range of $A$ that consists of column vectors of $A$.

(d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.

Solution.

We first obtain the reduced row echelon form matrix corresponding to the matrix $A$.
We reduce the matrix $A$ as follows:
\begin{align*}
A=\begin{bmatrix}
2 & 4 & 6 & 8 \\
1 &3 & 0 & 5 \\
1 & 1 & 6 & 3
\end{bmatrix}
\xrightarrow{\frac{1}{2}R_1}
\begin{bmatrix}
1 & 2 & 3 & 4 \\
1 &3 & 0 & 5 \\
1 & 1 & 6 & 3
\end{bmatrix}\\[6pt] \xrightarrow[R_3-R_1]{R_2-R_1}
\begin{bmatrix}
1 & 2 & 3 & 4 \\
0 &1 & -3 & 1 \\
0 & -1 & 3 & -1
\end{bmatrix}
\xrightarrow[R_3+R_2]{R_1-2R_2}
\begin{bmatrix}
1 & 0 & 9 & 2 \\
0 &1 & -3 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}.
\end{align*}
The last matrix is in reduced row echelon form. That is,
\[\rref(A)=\begin{bmatrix}
1 & 0 & 9 & 2 \\
0 &1 & -3 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}. \tag{*}\]

(a) Find a basis for the nullspace of $A$.

By the computation above, we see that the general solution of $A\mathbf{x}=\mathbf{0}$ is
\begin{align*}
x_1&=-9x_3-2x_4\\
x_2&=3x_3-x_4,
\end{align*}
where $x_3$ and $x_4$ are free variables.
Thus, the vector form solution to $A\mathbf{x}=\mathbf{0}$ is
\begin{align*}
\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}
=\begin{bmatrix}
-9x_3-2x_4 \\
3x_3-x_4 \\
x_3 \\
x_4
\end{bmatrix}
=x_3\begin{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix}.
\end{align*}
It follows that the nullspace of the matrix $A$ is given by
\begin{align*}
\calN(A)&=\left\{ \mathbf{x}\in \R^4 \quad \middle | \quad \mathbf{x}= x_3\begin{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix}, \text{ for all } x_3, x_4 \in \R^4 \right \}\\[6pt] &= \Span \left\{ \begin{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix} \right \}.
\end{align*}
Thus, the set
\[\left\{ \begin{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix} \right \}\] is a spanning set for the nullspace $\calN(A)$.
It is straightforward to see that this set is linearly independent, and hence it is a basis for $\calN(A)$.

(b) Find a basis for the row space of $A$.

Recall that the nonzero rows of $\rref(A)$ form a basis for the row space of $A$. (The row space method.)

Thus,
\[\left\{\begin{bmatrix}
1 \\
0 \\
9 \\
2
\end{bmatrix}, \quad \begin{bmatrix}
0 \\
1 \\
-3 \\
1
\end{bmatrix} \right \}\] is a basis for the row space of $A$.

(c) Find a basis for the range of $A$ that consists of column vectors of $A$.

Recall that by the leading 1 method, the columns of $A$ corresponding to columns of $\rref(A)$ that contain leading 1 entries form a basis for the range $\calR(A)$ of $A$.
From (*), we see that the first and the second columns contain the leading 1 entries. Thus,
\[\left\{\begin{bmatrix}
2 \\
1 \\
1
\end{bmatrix}, \begin{bmatrix}
4 \\
3 \\
1
\end{bmatrix}\right \}\] is a basis for the range $\calR(A)$ of $A$.

(d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.

Let us write $A_1, A_2, A_3$, and $A_4$ for the column vectors of the matrix $A$.
In part (c), we showed that $\{A_1, A_2\}$ is a basis for the range $\calR(A)$.
Thus, we need to express the vectors $A_3$ and $A_4$ as a linear combination of $A_1$ and $A_2$, respectively.

A shortcut is to note that the entries of third column vector of $\rref(A)$ give the coefficients of the linear combination for $A_3$. That is, we have
\[A_3=9A_1-3A_2.\] Similarly, the entries of the fourth column of $\rref(A)$ yield
\[A_4=2A_1+A_2.\] Click here if solved 131The post Find a Basis for Nullspace, Row Space, and Range of a Matrix first appeared on Problems in Mathematics.]]> https://yutsumura.com/find-a-basis-for-nullspace-row-space-and-range-of-a-matrix/feed/ 0 6906 https://yutsumura.com/if-mathbfv-mathbfw-are-linearly-independent-vectors-and-a-is-nonsingular-then-amathbfv-amathbfw-are-linearly-independent/ https://yutsumura.com/if-mathbfv-mathbfw-are-linearly-independent-vectors-and-a-is-nonsingular-then-amathbfv-amathbfw-are-linearly-independent/#respond Mon, 12 Feb 2018 16:08:12 +0000 https://yutsumura.com/?p=6874 Let $A$ be an $n\times n$ nonsingular matrix. Let $\mathbf{v}, \mathbf{w}$ be linearly independent vectors in $\R^n$. Prove that the vectors $A\mathbf{v}$ and $A\mathbf{w}$ are linearly independent.   Proof. Suppose that we have a...

Let
\[A=\begin{bmatrix}
-4 & -6 & -12 \\
-2 &-1 &-4 \\
2 & 3 & 6
\end{bmatrix}, \quad \mathbf{u}=\begin{bmatrix}
6 \\
5 \\
-3
\end{bmatrix}, \quad \mathbf{v}=\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}, \quad \text{ and } \mathbf{w}=\begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}.\]

(a) Express the vector $\mathbf{u}$ as a linear combination of $\mathbf{v}$ and $\mathbf{w}$.

(b) Compute $A^5\mathbf{v}$.

(c) Compute $A^5\mathbf{w}$.

(d) Compute $A^5\mathbf{u}$.

Solution.

(a) Express the vector $\mathbf{u}$ as a linear combination of $\mathbf{v}$ and $\mathbf{w}$.

Our goal here is to find scalars $c_1, c_2$ such that
\[\mathbf{u}=c_1\mathbf{v}+c_2\mathbf{w}.\] This is the same as the matrix equation
\[\begin{bmatrix}
-2 & -2 \\
0 & -1 \\
1 &1
\end{bmatrix}
\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}=\begin{bmatrix}
6 \\
5 \\
-3
\end{bmatrix}.\] To solve this, we reduced the augmented matrix as follows:
\begin{align*}
\left[\begin{array}{rr|r}
-2 & -2 & 6 \\
0 &-1 &5 \\
1 & 1 & -3
\end{array}\right] \xrightarrow{R_1 \leftrightarrow R_3}
\left[\begin{array}{rr|r}
1 & 1 & -3 \\
0 &-1 &5 \\
-2 & -2 & 6
\end{array}\right]\\[6pt] \xrightarrow[-R_2]{R_3+2R_1}
\left[\begin{array}{rr|r}
1 & 1 & -3 \\
0 &1 &-5 \\
0 & 0 & 0
\end{array}\right] \xrightarrow{R_1-R_2}
\left[\begin{array}{rr|r}
1 & 0 & 2 \\
0 &1 &-5 \\
0 & 0 & 0
\end{array}\right].
\end{align*}
This yields the solution $c_1=2$ and $c_2=-5$.
Hence, we have the linear combination
\[\mathbf{u}=2\mathbf{v}-5\mathbf{w}.\]

(b) Compute $A^5\mathbf{v}$.

We first compute $A\mathbf{v}$. We have
\[A\mathbf{v}= \begin{bmatrix}
-4 & -6 & -12 \\
-2 &-1 &-4 \\
2 & 3 & 6
\end{bmatrix}
\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}
=\begin{bmatrix}
-4 \\
0 \\
2
\end{bmatrix}=2\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}=2\mathbf{v}.
\] Using this relation $A\mathbf{v}=2\mathbf{v}$, we obtain
\[A^2\mathbf{v}=AA\mathbf{v}=A(2\mathbf{v})=2A\mathbf{v}=2(2\mathbf{v})=2^2\mathbf{v}.\] Next, we have
\[A^3\mathbf{v}=AA^2\mathbf{v}=A(2^2\mathbf{v})=2^2A\mathbf{v}=2^2(a\mathbf{v})=2^3\mathbf{v}.\] Repeating this process, we see that $A^5\mathbf{v}=2^5\mathbf{v}$.
Or, we can find this by computing as follows:
\begin{align*}
A^5\mathbf{v}=A^2A^3\mathbf{v}=A^2(2^3\mathbf{v})=2^3A^2\mathbf{v}=2^3(2^2\mathbf{v})=2^5\mathbf{v}.
\end{align*}
In summary, we have
\[A^5\mathbf{v}=2^5\mathbf{v}=32\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}=\begin{bmatrix}
-64 \\
0 \\
32
\end{bmatrix}.\]

(c) Compute $A^5\mathbf{w}$.

First, we note that
\[A\mathbf{w}= \begin{bmatrix}
-4 & -6 & -12 \\
-2 &-1 &-4 \\
2 & 3 & 6
\end{bmatrix} \begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}=\begin{bmatrix}
2 \\
1 \\
-1
\end{bmatrix}=-\begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}=-\mathbf{w}.\] Using this relation $A\mathbf{w}=-\mathbf{w}$ as in part (a), we obtain
\[A^5\mathbf{w}=(-1)^5\mathbf{w}=-\begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}=\begin{bmatrix}
2 \\
1 \\
-1
\end{bmatrix}.\]

(d) Compute $A^5\mathbf{u}$.

Using the linear combination $\mathbf{u}=2\mathbf{v}-5\mathbf{w}$ obtained part (a), we compute
\begin{align*}
A^5\mathbf{w}&=A^5(2\mathbf{v}-5\mathbf{w})\\
&=2A^5\mathbf{v}-5A^5\mathbf{w}\\[6pt] &=2\begin{bmatrix}
-64 \\
0 \\
32
\end{bmatrix}-5\begin{bmatrix}
2 \\
1 \\
-1
\end{bmatrix} &&\text{by (b), (c)}\\[6pt] &=\begin{bmatrix}
-138 \\
-5 \\
69
\end{bmatrix}.
\end{align*}

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

One common pitfall is to compute $A^5$, but this is time consuming and it is very likely to make a mistake by hand computaiton.

In this problem, we use the following vectors in $\R^2$.
\[\mathbf{a}=\begin{bmatrix}
1 \\
0
\end{bmatrix}, \mathbf{b}=\begin{bmatrix}
1 \\
1
\end{bmatrix}, \mathbf{c}=\begin{bmatrix}
2 \\
3
\end{bmatrix}, \mathbf{d}=\begin{bmatrix}
3 \\
2
\end{bmatrix}, \mathbf{e}=\begin{bmatrix}
0 \\
0
\end{bmatrix}, \mathbf{f}=\begin{bmatrix}
5 \\
6
\end{bmatrix}.\] For each set $S$, determine whether $\Span(S)=\R^2$. If $\Span(S)\neq \R^2$, then give algebraic description for $\Span(S)$ and explain the geometric shape of $\Span(S)$.

(a) $S=\{\mathbf{a}, \mathbf{b}\}$
(b) $S=\{\mathbf{a}, \mathbf{c}\}$
(c) $S=\{\mathbf{c}, \mathbf{d}\}$
(d) $S=\{\mathbf{a}, \mathbf{f}\}$
(e) $S=\{\mathbf{e}, \mathbf{f}\}$
(f) $S=\{\mathbf{a}, \mathbf{b}, \mathbf{c}\}$
(g) $S=\{\mathbf{e}\}$

Solution.

By definition, the subspace $\Span(S)$ spanned by $S$ is the set of all linear combinations of vectors in $S$. Thus, $\Span(S)$ is a subset in $\R^2$. The question is whether all of the vectors in $\R^2$ are linear combinations of vectors in $S$ or not.

(a) $S=\{\mathbf{a}, \mathbf{b}\}$

Let $\begin{bmatrix}
a \\
b
\end{bmatrix}$ be an arbitrary vector in $\R^2$. We determine whether it is a linear combination of $\mathbf{a}$ and $\mathbf{b}$:
\[\begin{bmatrix}
a \\
b
\end{bmatrix}=c_1\mathbf{a}+c_2\mathbf{b}.\] The augmented matrix is
\begin{align*}
\left[\begin{array}{rr|r}
1 & 1 & a \\
0 & 1 & b
\end{array} \right] \xrightarrow{R_1-R_2}
\left[\begin{array}{rr|r}
1 & 0 & a-b \\
0 & 1 & b
\end{array} \right].
\end{align*}
Hence, the solution is $c_1=a-b$, $c_2=b$.
Thus, we have
\[\begin{bmatrix}
a \\
b
\end{bmatrix}=(a-b)\mathbf{a}+b\mathbf{b},\] and this implies that every vector in $\R^2$ is a linear combination of $\mathbf{a}$ and $\mathbf{b}$.
Hence, $\Span(S)=\R^2$.

(b) $S=\{\mathbf{a}, \mathbf{c}\}$

Just like part (a), let $\begin{bmatrix}
a \\
b
\end{bmatrix}$ be any vector in $\R^2$.
Are there $c_1, c_2$ satisfying
\[\begin{bmatrix}
a \\
b
\end{bmatrix}=c_1\mathbf{a}+c_2\mathbf{c}?\] The augmented matrix is
\begin{align*}
\left[\begin{array}{rr|r}
1 & 2 & a \\
0 & 3 & b
\end{array} \right] \xrightarrow{\frac{1}{3} R_2}
\left[\begin{array}{rr|r}
1 & 2 & a \\
0 & 1 & b/3
\end{array} \right] \xrightarrow{R_1-2R_2}
\left[\begin{array}{rr|r}
1 & 0 & a-\frac{2}{3}b \\
0 & 1 & b/3
\end{array} \right].
\end{align*}
Hence, we have $c_1=a-\frac{2}{3}b$, $c_2=b/3$.
Thus,
\[\begin{bmatrix}
a \\
b
\end{bmatrix}=\frac{2b}{3} \mathbf{a}+\frac{b}{3}\mathbf{c}.\] This yields that every vector in $\R^2$ is in $\Span(S)$. Hence $\Span(S)=\R^2$.

(c) $S=\{\mathbf{c}, \mathbf{d}\}$

The strategy for this problem is the exactly same as before. So let us consider
\begin{align*}
\left[\begin{array}{rr|r}
2 & 3 & a \\
3 & 2 & b
\end{array} \right] \xrightarrow{R_2-R_1}
\left[\begin{array}{rr|r}
2 & 3 & a \\
1 & -1 & b-a
\end{array} \right] \xrightarrow{R_1\leftrightarrow R_2}
\left[\begin{array}{rr|r}
1 & -1 & b-a \\
2 & 3 & a
\end{array} \right]\\[6pt] \xrightarrow{R_2- 2 R_1}
\left[\begin{array}{rr|r}
1 & -1 & b-a \\
0 & 5 & 3a-2b
\end{array} \right] \xrightarrow{\frac{1}{5}R_2}
\left[\begin{array}{rr|r}
1 & -1 & b-a \\
0 & 1 & \frac{3}{5}a-\frac{2}{5}b
\end{array} \right] \\[6pt] \xrightarrow{R_1+R_2}
\left[\begin{array}{rr|r}
1 & 0 & -\frac{2}{5}a+\frac{3}{5}b \\
0 & 1 & \frac{3}{5}a-\frac{2}{5}b
\end{array} \right].
\end{align*}
It follows that any vector $\begin{bmatrix}
a \\
b
\end{bmatrix}\in \R^2$ can be written as a linear combination
\[\begin{bmatrix}
a \\
b
\end{bmatrix}=\left(-\frac{2}{5}a+\frac{3}{5}b \right)\mathbf{c}+\left( \frac{3}{5}a-\frac{2}{5}b \right) \mathbf{d}.\] Hence, $\Span(S)=\R^2$.

(d) $S=\{\mathbf{a}, \mathbf{f}\}$

This is also similar to previous problems. For any vector $\begin{bmatrix}
a \\
b
\end{bmatrix}\in \R^2$, we have
\begin{align*}
\left[\begin{array}{rr|r}
1 & 5 & a \\
0 & 6 & b
\end{array} \right] \xrightarrow{\frac{1}{6}R_2}
\left[\begin{array}{rr|r}
1 & 5 & a \\
0 & 1 & b/6
\end{array} \right] \xrightarrow{R_1-5R_2}
\left[\begin{array}{rr|r}
1 & 0 & a-\frac{5}{6}b \\
0 & 1 & b/6
\end{array} \right].
\end{align*}
This yields that
\[\begin{bmatrix}
a \\
b
\end{bmatrix}=\left(a-\frac{5b}{6} \right)\mathbf{a}+\left(\frac{b}{6}\right)\mathbf{f}.\] Hence, $\Span(S)=\R^2$.

(e) $S=\{\mathbf{e}, \mathbf{f}\}$

Note that as the vector $\mathbf{e}$ is the zero vector, any linear combination of $\mathbf{e}$ and $\mathbf{f}$ is just a scalar multiple of $\mathbf{f}$:
\[c_1\mathbf{e}+c_2\mathbf{f}=c_2\mathbf{f}.\] Thus, the algebraic description is
\begin{align*}
\Span(S)=\Span(\mathbf{e}, \mathbf{f})=\Span(\mathbf{f})=\{\mathbf{x}\in \R^2 \mid \mathbf{x}=c\mathbf{f} \text{ for some } c\in \R\}.
\end{align*}
Geometrically, the span is the line parametrized by $c\mathbf{f}=c\begin{bmatrix}
5 \\
6
\end{bmatrix}$ for any $c\in \R$.

(f) $S=\{\mathbf{a}, \mathbf{b}, \mathbf{c}\}$

Note that $\Span(\mathbf{a}, \mathbf{b})$ is contained in $\Span(\mathbf{a}, \mathbf{b}, \mathbf{c})$ because any linear combination of $\mathbf{a}$ and $\mathbf{b}$ is a linear combination of $\mathbf{a}, \mathbf{b}$, and $\mathbf{c}$ by taking the coefficient of $\mathbf{c}$ to be $0$.
We already saw in part (a) that $\Span(\mathbf{a}, \mathbf{b})=\R^2$. Hence, we must have $\Span(\mathbf{a}, \mathbf{b}, \mathbf{c})=\R^2$ as well.

(g) $S=\{\mathbf{e}\}$

Note that $\Span(\mathbf{e})$ is the set of all linear combination of $\mathbf{e}$, which is just the zero vector. So,
\[\Span(\mathbf{e})=\left\{\begin{bmatrix}
0 \\
0
\end{bmatrix}\right\}.\]Geometrically, this is just one point: the origin in the $x$-$y$ plane.

Let $A$ be a $3\times 3$ matrix and let
\[\mathbf{v}=\begin{bmatrix}
1 \\
2 \\
-1
\end{bmatrix} \text{ and } \mathbf{w}=\begin{bmatrix}
2 \\
-1 \\
3
\end{bmatrix}.\] Suppose that $A\mathbf{v}=-\mathbf{v}$ and $A\mathbf{w}=2\mathbf{w}$.
Then find the vector
\[A^5\begin{bmatrix}
-1 \\
8 \\
-9
\end{bmatrix}.\]

Solution.

Note that the given information is only about the vectors $\mathbf{v}$ and $\mathbf{w}$.
Thus, we first need to relate the vector $\begin{bmatrix}
-1 \\
8 \\
-9
\end{bmatrix}$ with $\mathbf{v}$ and $\mathbf{w}$.
So let us find a linear combination:
\[\begin{bmatrix}
-1 \\
8 \\
-9
\end{bmatrix}=a\mathbf{v}+b\mathbf{w}.\] The augmented matrix for this is
\begin{align*}
\left[\begin{array}{rr|r}
1 & 2 & -1 \\
2 &-1 &8 \\
-1 & 3 & -9
\end{array}\right] \xrightarrow[R_3+R_1]{R_2-2R_1}
\left[\begin{array}{rr|r}
1 & 2 & -1 \\
0 &-5 &10 \\
0 & 5 & -10
\end{array}\right] \xrightarrow{-\frac{1}{5}R_2}\\[6pt] \left[\begin{array}{rr|r}
1 & 2 & -1 \\
0 &1 &-2 \\
0 & 5 & -10
\end{array}\right] \xrightarrow[R_3-R_2]{R_1-2R_2}
\left[\begin{array}{rr|r}
1 & 0 & 3 \\
0 &1 &-2 \\
0 & 0 & 0
\end{array}\right].
\end{align*}
It yields that the solution is $a=3$ and $b=-2$.
Thus, we have
\[\begin{bmatrix}
-1 \\
8 \\
9
\end{bmatrix}=3\mathbf{v}-2\mathbf{w}.\]

Then, we compute
\begin{align*}
A^5\begin{bmatrix}
-1 \\
8 \\
9
\end{bmatrix}&=A^5(3\mathbf{v}-2\mathbf{w})=3A^5\mathbf{v}-2A^5\mathbf{w} \tag{*}
\end{align*}

Now, by assumption we have $A\mathbf{v}=-\mathbf{v}$.
Using this repeatedly we see that $A^5\mathbf{v}=(-1)^5\mathbf{v}=-\mathbf{v}$.
Similarly, we have $A^5\mathbf{w}=2^5\mathbf{w}=32\mathbf{w}$ using $A\mathbf{w}=2\mathbf{w}$.
Combining these with (*) yields,
\begin{align*}
A^5\begin{bmatrix}
-1 \\
8 \\
9
\end{bmatrix}&=3(-\mathbf{v})-2(32\mathbf{w})\\[6pt] &=-3\begin{bmatrix}
1 \\
2 \\
-1
\end{bmatrix}-64\begin{bmatrix}
2 \\
-1 \\
3
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
-3-128 \\
-6+64 \\
3-192
\end{bmatrix}
=\begin{bmatrix}
-131 \\
58 \\
-189
\end{bmatrix}.
\end{align*}