Vector Algebra and Spaces: Learning Objective 3
Question 1: Let $S$ be a vector space. A set of vectors $\textbf{v}_1, \dots, \textbf{v}_n \in S$ is said to be linearly independent if the equation $a_1\textbf{v}_1 + \dots + a_n\textbf{v}_n = \textbf{0}$ can only be satisfied by $a_i = 0$ for all $i = 1,\dots,n$. Determine whether the following vectors are linearly independent:
- $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and $\mathbf{v}_2 = \begin{bmatrix} 2 \\ 4 \end{bmatrix}$
- $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$
For each pair, try to find non-zero scalars $c_1, c_2$ such that $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 = \mathbf{0}$. If you can, they are dependent. If the only solution is $c_1=c_2=0$, they are independent.
- Linearly dependent.
- Linearly independent.
a) We check if one vector is a scalar multiple of the other. Notice that $\mathbf{v}_2 = \begin{bmatrix} 2 \\ 4 \end{bmatrix} = 2 \begin{bmatrix} 1 \\ 2 \end{bmatrix} = 2\mathbf{v}_1$. We can write a non-trivial linear combination that equals the zero vector: $2\mathbf{v}_1 - \mathbf{v}_2 = \mathbf{0}$. Since the coefficients (2 and -1) are not all zero, the vectors are linearly dependent.
b) We set up the equation $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 = \mathbf{0}$: \[ c_1\begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] This implies that $c_1=0$ and $c_2=0$. Since this is the only solution, the vectors are linearly independent.
b) We set up the equation $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 = \mathbf{0}$: \[ c_1\begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] This implies that $c_1=0$ and $c_2=0$. Since this is the only solution, the vectors are linearly independent.
Question 2: Let $\mathbf{a}_1, \mathbf{a}_2,$ and $\mathbf{a}_3$ be $3\times1$ vectors defined as follows:
\[ \mathbf{a}_1 = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, \quad \mathbf{a}_2 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, \quad \mathbf{a}_3 = \begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix} \]
Determine whether the vectors $\mathbf{a}_1$, $\mathbf{a}_2$, and $\mathbf{a}_3$ are linearly dependent.
Look for a simple relationship between any two of the vectors. Can one be written as a multiple of another?
Yes, they are linearly dependent.
The set is linearly dependent because one vector can be expressed as a linear combination of the others. Specifically, vector $\mathbf{a}_3$ is a scalar multiple of vector $\mathbf{a}_2$:
\[ \mathbf{a}_3 = 2\mathbf{a}_2 \]
We can rearrange this to form a linear combination that equals the zero vector:
\[ 2\mathbf{a}_2 - \mathbf{a}_3 = \mathbf{0} \]
To include all three vectors, we can write:
\[ 0\mathbf{a}_1 + 2\mathbf{a}_2 - 1\mathbf{a}_3 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]
Since the coefficients $(0, 2, -1)$ are not all zero, the set of vectors is linearly dependent.
Question 3: Let $\mathbf{b}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix}, \mathbf{b}_2 = \begin{bmatrix} 3 \\ 6 \end{bmatrix}, \mathbf{b}_3 = \begin{bmatrix} 4 \\ 8 \end{bmatrix}$ be vectors. We can define the matrix $B$ to contain the vectors as columns such that $B = \begin{bmatrix} 1 & 3 & 4\\ 2 & 6 & 8 \end{bmatrix}$.
- Are the columns of $B$ linearly dependent?
- The rank of a matrix is the maximum number of linearly independent rows or columns. What is $rank(B)$?
For part (a), check if any column can be written as a combination of the others. For part (b), determine the dimension of the column space (or row space).
- Yes, the columns are linearly dependent.
- The rank of B is 1.
a) Yes. We can see by inspection that the second column is 3 times the first column ($\mathbf{b}_2 = 3\mathbf{b}_1$) and the third column is 4 times the first column ($\mathbf{b}_3 = 4\mathbf{b}_1$). A specific non-trivial linear combination that gives the zero vector is $3\mathbf{b}_1 - \mathbf{b}_2 + 0\mathbf{b}_3 = \mathbf{0}$.
b) The rank is the maximum number of linearly independent columns. Since all columns are scalar multiples of the first column, any set of two or more columns will be linearly dependent. The largest set of linearly independent columns contains only one vector (e.g., $\{\mathbf{b}_1\}$). Therefore, the rank of B is 1.
b) The rank is the maximum number of linearly independent columns. Since all columns are scalar multiples of the first column, any set of two or more columns will be linearly dependent. The largest set of linearly independent columns contains only one vector (e.g., $\{\mathbf{b}_1\}$). Therefore, the rank of B is 1.
Question 4: A matrix is classified as full rank if the rank is equal to the maximum possible rank for a matrix of its dimensions. For an $m\times n$ matrix, the rank is at most $\min(m,n)$. Consider the matrix $A = \begin{bmatrix} 1 & 2 & 3\\ 2 & 4 & 6\\3&6&9 \end{bmatrix}$. Find $rank(A)$, and determine if $A$ is full rank.
Observe the relationship between the columns of the matrix. The maximum possible rank for this $3 \times 3$ matrix is 3.
$rank(A) = 1$. Since the rank (1) is less than the maximum possible rank (3), the matrix is not full rank.
Let's examine the columns of the matrix:
Column 2 is $2 \times$ Column 1.
Column 3 is $3 \times$ Column 1.
Since all columns are scalar multiples of the first row, there is only one linearly independent columns. Therefore, $rank(A)=1$.
For a $3 \times 3$ matrix, the maximum possible rank is $\min(3,3)=3$. Since $1 < 3$, the matrix is not full rank (it is rank-deficient).
Column 2 is $2 \times$ Column 1.
Column 3 is $3 \times$ Column 1.
Since all columns are scalar multiples of the first row, there is only one linearly independent columns. Therefore, $rank(A)=1$.
For a $3 \times 3$ matrix, the maximum possible rank is $\min(3,3)=3$. Since $1 < 3$, the matrix is not full rank (it is rank-deficient).
Question 5: Find the rank of the following matrices, and determine if they are full rank.
- $A = \begin{bmatrix} 1 & -2 & 3\\ 2 & 4 & -6\\5 & 1 & -1 \end{bmatrix}$
- $B = \begin{bmatrix} 1 & 1 & 3 & 2\\ 2 & 1 & 4 & 2\\ 1 & 3 & 7 & 6 \end{bmatrix}$
To find the rank, you can check for linear dependencies between the columns or rows. For a square matrix, you can also check if its determinant is non-zero.
- $rank(A) = 3$. It is full rank.
- $rank(B) = 2$. It is not full rank.
a) For matrix A, let's check for linear dependence of the columns $\mathbf{c}_1, \mathbf{c}_2, \mathbf{c}_3$. We can test this by seeing if the determinant is non-zero.
\[ \det(A) = 1(4(-1) - (-6)(1)) - (-2)(2(-1) - (-6)(5)) + 3(2(1) - 4(5)) \]
\[ = 1(-4+6) + 2(-2+30) + 3(2-20) = 1(2) + 2(28) + 3(-18) = 2 + 56 - 54 = 4 \]
Since $\det(A) \neq 0$, the three columns are linearly independent. The rank is the maximum number of linearly independent columns, so $rank(A)=3$. Since the maximum possible rank is 3, A is full rank.
b) For matrix B, let the columns be $\mathbf{c}_1, \mathbf{c}_2, \mathbf{c}_3, \mathbf{c}_4$. Let's check for dependencies. Is $\mathbf{c}_3$ a linear combination of $\mathbf{c}_1$ and $\mathbf{c}_2$? We look for scalars $k_1, k_2$ such that $k_1\mathbf{c}_1 + k_2\mathbf{c}_2 = \mathbf{c}_3$. \[ k_1\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} + k_2\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \\ 7 \end{bmatrix} \implies \begin{cases} k_1+k_2=3 \\ 2k_1+k_2=4 \\ k_1+3k_2=7 \end{cases} \] Subtracting the first equation from the second gives $k_1=1$. Substituting into the first gives $1+k_2=3 \implies k_2=2$. Checking the third equation: $1+3(2)=7$. It holds. So, $\mathbf{c}_3 = \mathbf{c}_1 + 2\mathbf{c}_2$.
Since column 3 is dependent on columns 1 and 2, the rank is at most 2. Columns 1 and 2 are not scalar multiples of each other, so they are linearly independent. Thus, the maximum number of linearly independent columns is 2. $rank(B)=2$. The maximum possible rank is $\min(3,4)=3$. Since $2 < 3$, B is not full rank.
b) For matrix B, let the columns be $\mathbf{c}_1, \mathbf{c}_2, \mathbf{c}_3, \mathbf{c}_4$. Let's check for dependencies. Is $\mathbf{c}_3$ a linear combination of $\mathbf{c}_1$ and $\mathbf{c}_2$? We look for scalars $k_1, k_2$ such that $k_1\mathbf{c}_1 + k_2\mathbf{c}_2 = \mathbf{c}_3$. \[ k_1\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} + k_2\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \\ 7 \end{bmatrix} \implies \begin{cases} k_1+k_2=3 \\ 2k_1+k_2=4 \\ k_1+3k_2=7 \end{cases} \] Subtracting the first equation from the second gives $k_1=1$. Substituting into the first gives $1+k_2=3 \implies k_2=2$. Checking the third equation: $1+3(2)=7$. It holds. So, $\mathbf{c}_3 = \mathbf{c}_1 + 2\mathbf{c}_2$.
Since column 3 is dependent on columns 1 and 2, the rank is at most 2. Columns 1 and 2 are not scalar multiples of each other, so they are linearly independent. Thus, the maximum number of linearly independent columns is 2. $rank(B)=2$. The maximum possible rank is $\min(3,4)=3$. Since $2 < 3$, B is not full rank.
Question 6: For an $n \times n$ square matrix $A$, $det(A) = 0$ if and only if the columns of $A$ are linearly dependent. Let $n=2$ and $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
- Write the conditions on $a,b,c$ and $d$ for the columns of $A$ to be linearly dependent.
- Show that if $det(A) = 0$, the columns of matrix $A$ are linearly dependent.
- Show that if the columns of $A$ are linearly dependent, then $det(A) = 0$.
- We can view the rows of the matrix $A$ as vectors. For any matrix, the number of linearly dependent rows equals the number of linearly dependent columns. Show that for a $2 \times 2$ matrix $A$, if the columns are linearly dependent, then the rows are also linearly dependent.
Linear dependence means one vector is a scalar multiple of the other. Show that the algebraic condition for this is the same as the condition for the determinant being zero.
- $ad = bc$.
- If $det(A)=ad-bc=0$, then $ad=bc$, which is the condition for linear dependence.
- If columns are dependent, $ad=bc$, which implies $ad-bc=0$, so $det(A)=0$.
- The condition for row dependence is also $ad=bc$.
Let the columns be $\mathbf{v}_1 = \begin{bmatrix} a \\ c \end{bmatrix}$ and $\mathbf{v}_2 = \begin{bmatrix} b \\ d \end{bmatrix}$. Let the rows be $\mathbf{r}_1 = \begin{bmatrix} a & b \end{bmatrix}$ and $\mathbf{r}_2 = \begin{bmatrix} c & d \end{bmatrix}$.
a) The columns are linearly dependent if one is a scalar multiple of the other, i.e., $\mathbf{v}_2 = k \mathbf{v}_1$. This gives $b = ka$ and $d = kc$. Assuming $a \neq 0$, $k=b/a$, so $d=(b/a)c \implies ad=bc$.
b) If $det(A) = ad-bc = 0$, then $ad=bc$. This is the algebraic condition for the columns to be linearly dependent, as shown in part (a).
c) If the columns are linearly dependent, then from (a) we know $ad=bc$. This implies $ad-bc=0$, which means $det(A)=0$.
d) The rows are linearly dependent if one is a scalar multiple of the other, i.e., $\mathbf{r}_2 = k \mathbf{r}_1$. This gives $c=ka$ and $d=kb$. Assuming $a \neq 0$, $k=c/a$, so $d=(c/a)b \implies ad=bc$. This is the exact same condition derived for column dependence. Therefore, if the columns are dependent, the rows must also be dependent.
a) The columns are linearly dependent if one is a scalar multiple of the other, i.e., $\mathbf{v}_2 = k \mathbf{v}_1$. This gives $b = ka$ and $d = kc$. Assuming $a \neq 0$, $k=b/a$, so $d=(b/a)c \implies ad=bc$.
b) If $det(A) = ad-bc = 0$, then $ad=bc$. This is the algebraic condition for the columns to be linearly dependent, as shown in part (a).
c) If the columns are linearly dependent, then from (a) we know $ad=bc$. This implies $ad-bc=0$, which means $det(A)=0$.
d) The rows are linearly dependent if one is a scalar multiple of the other, i.e., $\mathbf{r}_2 = k \mathbf{r}_1$. This gives $c=ka$ and $d=kb$. Assuming $a \neq 0$, $k=c/a$, so $d=(c/a)b \implies ad=bc$. This is the exact same condition derived for column dependence. Therefore, if the columns are dependent, the rows must also be dependent.