Vector Space
Rank and Nullity

Linearly Independent Sets and Basis

Example 1.

 

Suppose we are given the set U =\{\left (\begin{array}{c}1 \\ 1 \\ 1\end{array}\right ) ,\left (\begin{array}{c}2 \\ 2 \\ 2\end{array}\right ) ,\left (\begin{array}{c}1 \\ -1 \\ 3\end{array}\right )\}. We would like to determine if this set is linearly independent. We know that a set \{\mathbf{u}_{1} ,\mathbf{u}_{2} , . . . ,\mathbf{u}_{n}\} is linearly independent if

(1)   \begin{equation*}c_{1}\mathbf{u}_{1} +c_{2}\mathbf{u}_{2} + . . . +c_{n}\mathbf{u}_{n} =0 \end{equation*}

implies

(2)   \begin{equation*}c_{1} =c_{2} = . . . =c_{n} =0 \end{equation*}

Also note that if any of the constants c_{i} is not equal to zero, say c_{1}, one can write

(3)   \begin{equation*}\frac{c_{2}}{c_{1}}\mathbf{u}_{2} +\frac{c_{3}}{c_{1}}\mathbf{u}_{3} + . . . +\frac{c_{n}}{c_{1}}\mathbf{u}_{n} =\mathbf{u}_{1} \end{equation*}

Of course, this means that when a set is linearly dependent, then at least one member can be written as a linear combination of other members. In this example, we see

(4)   \begin{equation*}\left (\begin{array}{c}1 \\ 1 \\ 1\end{array}\right ) =\frac{1}{2}\left (\begin{array}{c}2 \\ 2 \\ 2\end{array}\right ) +0\left (\begin{array}{c}1 \\ -1 \\ 3\end{array}\right ) \end{equation*}

And therefore, the set U is linearly dependent.

 

Example 2.

 

Suppose we are given the set U =\{\left (\begin{array}{cc}1 & -1 \\ 4 & 5\end{array}\right ) ,\left (\begin{array}{cc}4 & 3 \\ -2 & 3\end{array}\right ) ,\left (\begin{array}{cc}1 & -8 \\ 22 & 23\end{array}\right )\}. We would like to determine if U is linearly independent. Write

(5)   \begin{equation*}a\left (\begin{array}{cc}1 & -1 \\ 4 & 5\end{array}\right ) +b\left (\begin{array}{cc}4 & 3 \\ -2 & 3\end{array}\right ) +c\left (\begin{array}{cc}1 & -8 \\ 22 & 23\end{array}\right ) =\left (\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}\right ) \end{equation*}

This implies

(6)   \begin{equation*}\left (\begin{array}{cc}a +4b +c & -a +3b -8c \\ 4a -2b +22c & 5a +3b +23c\end{array}\right ) =\left (\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}\right ) \end{equation*}

(7)   \begin{align*}a +4b +c =0 \\ -a +3b -8c =0 \\ 4a -2b +22c =0 \\ 5a +3b +23c =0\end{align*}

Of course a =0 ,b =0 ,c =0 is a solution. If this is the only case, then the set U is linearly independent; otherwise dependent. As you get from the first and second equations

(8)   \begin{equation*}a =c \end{equation*}

From the last two equations

(9)   \begin{align*}a +b +c =0 \\ b = -2c\end{align*}

And finally

(10)   \begin{align*}30c =0 \\ c =0 \\ b =0 \\ a =0\end{align*}

Therefore, set U is linearly independent.

 

Example 3.

 

Suppose we are given U =\{x^{2} ,x^{2} +1\} in the vector space of all polynomials with degree less than 3 and would like to know if the set U is linearly dependent. Start with writing

(11)   \begin{align*}a\left (x^{2}\right ) +b\left (x^{2} +1\right ) =0 \\ \left (a +b\right )x^{2} +b\left (1\right ) =0\end{align*}

Note that this is true for every x . We get

(12)   \begin{align*}\left (a +b\right )\left (x^{2}\right ) =0 \\ b\left (1\right ) =0\end{align*}

which implies a =b =0. Also note that since we only have two vectors in U , we can quickly check that no vector is a constant multiple of another and therefore, the set U is linearly independent.

 

Example 4.

 

Suppose we are given U =\{2 -x ,2x -x^{2} ,x^{2} -5x +6\} in the vector space of all polynomials with degree less than 3 and would like to know if the set U is linearly dependent. Start with writing

(13)   \begin{align*}a\left (2 -x\right ) +b\left (2x -x^{2}\right ) +c\left (x^{2} -5x +6\right ) =0 \\ \left (2a +6c\right )\left (1\right ) =0 \\ \left ( -a +2b -5c\right )x =0 \\ \left ( -b +c\right )x^{2} =0\end{align*}

This implies

(14)   \begin{align*}a +3c =0 \\ -a +2b -5c =0 \\ -b +c =0\end{align*}

We can also write this system as followed

(15)   \begin{equation*}\left (\begin{array}{ccc}1 & 0 & 3 \\ -1 & 2 & -5 \\ 0 & -1 & 1\end{array}\right )\left (\begin{array}{c}a \\ b \\ c\end{array}\right ) =\left (\begin{array}{c}0 \\ 0 \\ 0\end{array}\right ) \end{equation*}

Determinant of the coefficient matrix is zero which means, the option of having a unique solution is out. Since there is always the solution a =0 ,y =0 ,z =0 we see immediately that there are infinitely many solutions and therefore the set U is linearly dependent. This also implies one of the polynomials in U can be written as a linear combination of other members in U . Can you find at least one such case?

 

Example 5.

 

Suppose we are given the set U =\{1 ,x ,3x ,5x^{2} +1\}. We know the set V =\{1 ,x ,x^{2}\} is a basis for the vector space P_{2} of all polynomials of degree 2 or less. Therefore, any other basis for P_{2} must obey the minimum condition of having 3 elements (even before we check to see if the candidate set spans the space and is linearly independent.). However, set U has 4 members and so it cannot be a basis for P_{2} .

 

Example 6.

 

Suppose we are given the set U =\{\left (\begin{array}{cc}2 & 0 \\ 0 & 3\end{array}\right ) ,\left (\begin{array}{cc}1 & 4 \\ 0 & 1\end{array}\right ) ,\left (\begin{array}{cc}0 & 1 \\ 3 & 2\end{array}\right ) ,\left (\begin{array}{cc}0 & 1 \\ 2 & 0\end{array}\right )\}. We want to check if set U is a basis for the vector space of all 2 by 2 matrices. Two conditions to be checked.

 

1. Does U span the space of 2 by 2 matrices?

 

2. Is U linearly independent.

 

In this case, we know dimension of the vector space of 2 by 2 matrices is 4 and so any one out of the two conditions is enough to show U is a basis. We show both here (but 1 is enough as explained.)

 

Take an arbitrary 2 by 2 matrix

(16)   \begin{equation*}\left (\begin{array}{cc}a & b \\ c & d\end{array}\right ) \end{equation*}

and write

(17)   \begin{equation*}x\left (\begin{array}{cc}2 & 0 \\ 0 & 3\end{array}\right ) +y\left (\begin{array}{cc}1 & 4 \\ 0 & 1\end{array}\right ) +z\left (\begin{array}{cc}0 & 1 \\ 3 & 2\end{array}\right ) +w\left (\begin{array}{cc}0 & 1 \\ 2 & 0\end{array}\right ) =\left (\begin{array}{cc}a & b \\ c & d\end{array}\right ) \end{equation*}

This implies

(18)   \begin{align*}2x +y =a \\ 4y +z +w =b \\ 3z +2w =c \\ 3x +y +2z =d\end{align*}

The associated coefficient matrix is

(19)   \begin{equation*}A =\left (\begin{array}{cccc}2 & 1 & 0 & 0 \\ 0 & 4 & 1 & 1 \\ 0 & 0 & 3 & 2 \\ 3 & 1 & 2 & 0\end{array}\right ) \end{equation*}

Determinant of this matrix is equal to -15 and so the associate system has a unique solution. Therefore U spans the vector space of 2 by 2 matrices.

 

Next, when a =b =c =0 , we see that x =y =z =w =0 is the only solution and so U is linearly independent

 

Example 7.

 

Suppose we are given U =\{\left (\begin{array}{c}3 \\ -2\end{array}\right ) ,\left (\begin{array}{c}4 \\ 5\end{array}\right )\}. Observe that given any vector \left (\begin{array}{c}a \\ b\end{array}\right ) in \mathbb{R}^{2}, we can write

(20)   \begin{align*}x\left (\begin{array}{c}3 \\ -2\end{array}\right ) +y\left (\begin{array}{c}4 \\ 5\end{array}\right ) =\left (\begin{array}{c}a \\ b\end{array}\right ) \\ 3x +4y =a \\ -2x +5y =b\end{align*}

This system has a unique solution since

(21)   \begin{equation*}\det \left (\begin{array}{cc}3 & 4 \\ -2 & 5\end{array}\right ) =23 \neq 0 \end{equation*}

Therefore, U spans \mathbb{R}^{2}. It is easy to see that the two members of U are not constant multiple of each other. And so U is a basis for \mathbb{R}^{2} . Like the previous example, in the case that dimension of the vector space is known, only one of the conditions will do the job.

 

Example 8.

 

Suppose we are given the set U =\{\left (s +4t ,t ,s ,2s -t\right ) :s \in \mathbb{R} ,t \in \mathbb{R}\}. We want to find a basis for this space and in result the dimension of the subspace U of \mathbb{R}^{4} . We start by writing a typical member of this vector space

(22)   \begin{align*}\left (s +4t ,s ,t ,2s -t\right ) =\left (s ,0 ,s ,2s\right ) +\left (4t ,t ,0 , -t\right ) \\ =s\left (1 ,0 ,1 ,2\right ) +t\left (4 ,1 ,0 , -1\right )\end{align*}

Therefore the set \{\left (1 ,0 ,1 ,2\right ) ,\left (4 ,1 ,0 , -1\right )\} spans U . We check if this set is linearly independent. Note that the two members of this set are not constant multiple of one another. Therefore we know this set is linearly independent. You can try other ways to conclude this. We see that the set \{\left (1 ,0 ,1 ,2\right ) ,\left (4 ,1 ,0 , -1\right )\} is a basis for U and since it has two members, U has dimension 2.

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