Linear Independence and Rank - Linear Algebra

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What is the dimension of the space spanned by the following vectors:

$\b{v}_1 = (1,1,0,0,0)$

$\b{v}_2 = (0,0,1,1,0)$

$\b{v}_3 = (0,0,0,0,1)$

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Since there are three linearly independent vectors, they span a 3 dimensional space.

Notice that the vectors each have 5 coordinates to them. Therefore they actually span a 3 dimensional subspace of a 5 dimensional space.

In a 5 dimensional vector space, what is the maximum number of vectors you can have in a linearly dependent set?

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Linearly dependent sets have no limit to the number of vectors they can have.

Consider the following set of three vectors:

$\b{v}_1,\b{v}_2,\b{v}_3$

where $\b{v}_3=2\b{v}_1+7\b{v}_2$

Is the set linearly independent?

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Since $\b{v}_3$ can be written as a linear combination of of $\b{v}_1$ and $\b{v}_2$ then the set cannot be linearly independent.

Does the following row reduced echelon form of a matrix represent a linearly independent set?

$\begin{bmatrix} 1 & 0& 0& 0& \0 &0 &1 &0 & \ 0 &0 &0 &0 &\end{bmatrix}$

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The set is linearly dependent because there is a row of all zeros.

Notice that having columns of all zeros does not tell if the set is linearly independent or not.

Does the following row reduced echelon form of a matrix represent a linearly independent set?

$\begin{bmatrix} 1 & 0& 0& 0& \0 &0 &1 &0 & \end{bmatrix}$

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The set must be linearly independent because there are no rows of all zeros. There are columns of all zeros, but columns do not tell us if the set is linearly independent or not.

Consider a set of 3 vectors from a 3 dimensional vector space.

Is the set linearly independent?

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It depends on what the vectors are.

For example, if

$\b{v}_1 = (1,0,0)$

$\b{v}_2 = (0,1,0)$

$\b{v}_3 = (0,0,1)$

Then the set is linearly independent.

However if the vectors were

$\b{v}_1 = (1,0,0)$

$\b{v}_2 = (0,1,0)$

$\b{v}_3 = (2,0,0)$

then the set would be linearly dependent.

Consider a set of 3 vectors from a 2 dimensional vector space.

Is the set linearly independent?

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Since the dimension of the space is 2, a linearly independent set can have at most two vectors. Since the set in consideration has 3 and 3>2, the set must be linearly dependent.

Determine whether the following vectors in Matrix form are Linearly Independent.

$A=\begin{bmatrix} 2 & 4 & 10\ 3 & -7 & 11\ -1 & 4 & 10 \end{bmatrix}$

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To figure out if the matrix is independent, we need to get the matrix into reduced echelon form. If we get the Identity Matrix, then the matrix is Linearly Independent.

$\begin{bmatrix} 2 & 4 & 10\ 3 & -7 & 11\ -1 & 4 & 10 \end{bmatrix}$

$$\frac{2}{3}$R_2-R_1\rightarrow R_2$

$\begin{bmatrix} 2 & 4 & 10\ 0 & -$\frac{26}{3}$ & -$\frac{8}{3}$\ -1 & 4 & 10 \end{bmatrix}$

$-2 R_3+R_1\rightarrow R_3$

$\begin{bmatrix} 2 & 4 & 10\ 0 & -$\frac{26}{3}$ & -$\frac{8}{3}$\ 0 & -4 & -10 \end{bmatrix}$

$\-3R_2\rightarrow R_2 \-R_3\rightarrow R_3$

$\begin{bmatrix} 2 & 4 & 10\ 0 & 26 & 8\ 0 & 4 & 10 \end{bmatrix}$

$$\frac{13}{2}$ R_3-R_2\rightarrow R_3$

$\begin{bmatrix} 2 & 4 & 10\ 0 & 26 & 8\ 0 & 0 & 57 \end{bmatrix}$

$$\frac{13}{2}$ R_3-R_2\rightarrow R_3$

$\begin{bmatrix} 2 & 4 & 10\ 0 & 26 & 8\ 0 & 0 & 57 \end{bmatrix}$

$\$\frac{1}{2}$R_1\rightarrow R_1 \ \ $\frac{1}{2}$R_2\rightarrow R_2$

$\begin{bmatrix} 1 & 2 & 5\ 0 & 13 & 4\ 0 & 0 & 57 \end{bmatrix}$

$$\frac{4}{57}$ R_3-R_2\rightarrow R_2$

$\begin{bmatrix} 1 & 2 & 5\ 0 & -9 & 0\ 0 & 0 & 57 \end{bmatrix}$

$$\frac{5}{57}$ R_3-R_1\rightarrow R_1$

$\begin{bmatrix} 4 & 3 & 0\ 0 & -9 & 0\ 0 & 0 & 57 \end{bmatrix}$

$R_2+3R_1\rightarrow R_1$

$\begin{bmatrix} 12 & 0 & 0\ 0 & -9 & 0\ 0 & 0 & 57 \end{bmatrix}$

$\$\frac{1}{12}$R_1\rightarrow R_1 \ \ -$\frac{1}{9}$R_2\rightarrow R_2 \ \ $\frac{1}{57}$R_3\rightarrow R_3$

$\begin{bmatrix} 1 & 0 & 0\ 0 & 1& 0\ 0 & 0 & 1 \end{bmatrix}$

Since we got the Identity Matrix, we know that the matrix is Linearly Independent.

Find the rank of the following matrix.

$\begin{bmatrix} 10 & 2 \ 4 & 5\ 5& 10 \end{bmatrix}$

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We need to get the matrix into reduced echelon form, and then count all the non all zero rows.

$R_2-$\frac{2}{5}$R_1\rightarrow R_2$

$\begin{bmatrix} 10 & 2 \ 0 & $\frac{21}{5}$\ 5& 10 \end{bmatrix}$

$R_3-$\frac{1}{2}$R_1\rightarrow R_3$

$\begin{bmatrix} 10 & 2 \ 0 & $\frac{21}{5}$\ 0& 9 \end{bmatrix}$

$R_2 \leftrightarrow R_3$

$\begin{bmatrix} 10 & 2 \ 0& 9 \ 0 & $\frac{21}{5}$\end{bmatrix}$

$R_3-$\frac{7}{15}$R_2\rightarrow R_3$

$\begin{bmatrix} 10 & 2 \ 0& 9 \ 0 & 0\end{bmatrix}$

$$\frac{1}{9}$R_2\rightarrow R_2$

$\begin{bmatrix} 10 & 2 \ 0& 1 \ 0 & 0\end{bmatrix}$

$R_1-2R_2\rightarrow R_1$

$\begin{bmatrix} 10 & 0 \ 0& 1 \ 0 & 0\end{bmatrix}$

$$\frac{1}{10}$R_1\rightarrow R_1$

$\begin{bmatrix} 1 & 0 \ 0& 1 \ 0 & 0\end{bmatrix}$

The rank is 2, since there are 2 non all zero rows.

Calculate the Rank of the following matrix

$\begin{bmatrix} 1&2 &3 \ 1& 2& 3\ 1&2 &3 \end{bmatrix}$

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We need to put the matrix into reduced echelon form, and then count all the non-zero rows.

$\R_1-R_2\rightarrow R_2 \R_1-R_3\rightarrow R_3$

$\begin{bmatrix} 1&2 &3 \ 0& 0& 0\ 0&0 &0 \end{bmatrix}$

Since there is only 1 non-zero row, the Rank is 1.

Determine if the following matrix is linearly independent or not.

$A=\begin{bmatrix} 5 & 3\ 10 & 6 \end{bmatrix}$

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Since the matrix is $2\times2$ , we can simply take the determinant. If the determinant is not equal to zero, it's linearly independent. Otherwise it's linearly dependent.

$\det(A)=5(6)-(10)(3)=30-30=0$

Since the determinant is zero, the matrix is linearly dependent.

If matrix A is a 5x8 matrix with a two-dimensional null space, what is the rank of A?

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Given that rank A + dimensional null space of A = total number of columns, we can determine rank A = total number of columns-dimensional null space of A. Using the information given in the question we can solve for rank A:

If matrix A is a 10x12 matrix with a three-dimensional null space, what is the rank of A?

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Given that rank A + dimensional null space of A = total number of columns, we can determine rank A = total number of columns-dimensional null space of A. Using the information given in the question we can solve for rank A:

What is the dimension of the space spanned by the following vectors:

$\b{v}_1 = (1,0,0,0,0)$

$\b{v}_2 = (0,1,0,0,0)$

$\b{v}_3 = (0,0,0,0,1)$

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Since there are three linearly independent vectors, they span a 3 dimensional space.

Notice that the vectors each have 5 coordinates to them. Therefore they actually span a 3 dimensional subspace of a 5 dimensional space.

Determine the row rank of the matrix

$\begin{pmatrix} 1&2 &3 \ 2&4 &6 \end{pmatrix}$

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To determine the matrix, we turn the matrix into reduced row echelon form

By adding times the first row to the second we get

$\begin{pmatrix} 1&2 &3 \ 2+(-2)&4+(-4) &6+(-6) \end{pmatrix}=\begin{pmatrix} 1&2 &3 \ 0&0 &0 \end{pmatrix}$

And find that the row rank is

Determine the row rank of the matrix

$\begin{pmatrix} 1&4 &5 \ 2&8 &10 \ 3&6 &4 \end{pmatrix}$

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To determine the row rank of the matrix we reduce the matrix into reduced echelon form.

First we add times the 1st row to the 2nd row

$\begin{pmatrix} 1&4 &5 \ 2+(-2)&8+(-8) &10+(-10) \ 3&6 &4 \end{pmatrix}=\begin{pmatrix} 1&4 &5 \ 0&0 &0 \ 3&6 &4 \end{pmatrix}$

add times the 1st row to the 3rd row

$\begin{pmatrix} 1&4 &5 \ 0&0 &0 \ 3+(-3)&6+(-12) &4+(-15) \end{pmatrix}=\begin{pmatrix} 1&4 &5 \ 0&0 &0 \ 0&-6 &-11 \end{pmatrix}$

Switch the 2nd row and the 3rd row

$\begin{pmatrix} 1&4 &5 \ 0&-6 &-11 \ 0&0 &0 \end{pmatrix}$

multiply the 2nd row by $-$\frac{1}{6}$$

$\begin{pmatrix} 1&4 &5 \ 0&-6(-$\frac{1}{6}$) &-11(-$\frac{1}{6}$) \ 0&0 &0 \end{pmatrix}=\begin{pmatrix} 1& 4& 5\ 0&1 &$\frac{11}{6}$ \ 0&0 &0 \end{pmatrix}$

add times the 2nd row to the 1st row

$\begin{pmatrix} 1& 4+(-4)& 5+(-$\frac{44}{6}$)\ 0&1 &$\frac{11}{6}$ \ 0&0 &0 \end{pmatrix} = \begin{pmatrix} 1&0 &-$\frac{7}{3}$ \ 0&1 &$\frac{11}{6}$ \ 0&0 & 0 \end{pmatrix}$

And we find that the row rank is

Consider the following set of vectors

$\b{v}_1= (1,0,0)$

$\b{v}_2= (0,1,0)$

$\b{v}_3= (0,0,1)$

Is the the set linearly independent?

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Yes, the set is linearly independent. There are multiple ways to see this

Way 1) Put the vectors into matrix form,

$\begin{bmatrix} 1 &0 &0 \0 &1 &0 \0 &0 &1 \end{bmatrix}$

The matrix is already in reduced echelon form. Notice there are three rows that have a nonzero number in them and we started with 3 vectors. Thus the set is linearly independent.

Way 2) Consider the equation $\alpha \b{v}_1 + \beta \b{v}_2 + \gamma \b{v}_3 = 0$

If when we solve the equation, we get $\alpha = \beta = \gamma = 0$ then it is linearly independent. Let's solve the equation and see what we get.

$\alpha \b{v}_1 + \beta \b{v}_2 + \gamma \b{v}_3 =\b{0}$

$\alpha (1, 0, 0) +\beta (0,1,0) +\gamma(0,0,1) = (0,0,0)$

Distribute the scalar constants to get

$(\alpha, 0, 0) + (0,\beta,0) +(0,0,\gamma) = (0,0,0)$

Thus we get a system of 3 equations

$\alpha = 0$

$\beta = 0$

$\gamma = 0$

Since $\alpha = \beta = \gamma = 0$ the vectors are linearly independent.

Consider the following set of vectors

$\b{v}_1= (1,0,0)$

$\b{v}_2= (0,1,0)$

$\b{v}_3= (0,0,1)$

$\v_4 = (2, 1, 5)$ $\b{v}_4 = (2,1,5)$

Is the the set linearly independent?

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The vectors have dimension 3. Therefore the largest possible size for a linearly independent set is 3. But there are 4 vectors given. Thus, the set cannot be linearly independent and must be linearly dependent

Another way to see this is by noticing that $\b{v}_4$ can be written as a linear combination of the other vectors:

$\b{v}_4 = 2 \b{v}_1 + 1 \b{v}_2 + 5 \b{v}_3$

In a vector space with dimension 5, what is the maximum number of vectors that can be in a linearly independent set?

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The dimension of a vector space is the maximum number of vectors possible in a linearly independent set. (notice you can have linearly independent sets with 5 or less, but never more than 5)

In a vector space of dimension 5, can you have a linearly independent set of 3 vectors?

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The dimension of the vector space is the maximum number of vectors in a linearly independent set. It is possible to have linearly independent sets with less vectors than the dimension.

So for this example it is possible to have linear independent sets with

1 vector, or 2 vectors, or 3 vectors, all the way up to 5 vectors.