Linear Algebra - Linear Independence and Rank

Find the rank of the following matrix.

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

Explanation

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.

True or False: If a $5 \times 5$ matrix has linearly independent columns, then .

True

False

Explanation

Since is a $5 \times 5$ matrix, . Since has three linearly independent columns, it must have a column space (and hence row space) of dimension , causing by the definition of rank. Hence.

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?

Yes.

No.

Not enough information

Explanation

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.

The rank and the nullity of a matrix with four rows and six columns are the same. What number do they share?

Explanation

The sum of the rank and the nullity of a matrix is equal to the number of columns in the matrix. Since the matrix in question has six columns, for the rank and the nullity to be equal, they must each be 3.

Calculate the Rank of the following matrix

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

Explanation

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.

The nullity of a matrix is 3; the nullity of $A^{T }$ is also 3.

True, false or undetermined: is a square matrix.

True

False

Explanation

The sum of the rank and the nullity of a matrix is equal to the number of columns in the matrix. Therefore, the number of columns in is equal to $\textup{Rnk}(A) + 3$ , and the number of columns in $A^{T}$ - the number of rows in - is $\textup{Rnk}(A^{T}) + 3$ .

Furthermore, the rank of a matrix is equal to its transpose, so

$\textup{Rnk}(A) = \textup{Rnk}(A^{T})$ ,

and

$\textup{Rnk}(A) + 3 = \textup{Rnk}(A^{T}) + 3$

This makes a matrix with the same number of rows as columns. It is square, and the statement is true.

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?

Yes

Not enough information

Explanation

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:

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

The rank and nullity of a matrix are 4 and 2, respectively. The nullity of $A^{T }$ is 3. What are the dimensions of ?

is a $7 \times 6$ matrix.

is a $6 \times 7$ matrix.

is a $4 \times 5$ matrix.

is a $5 \times 4$ matrix.

Insufficient information is given to determine the dimensions of the matrix.

Explanation

The sum of the rank and the nullity of any matrix is the number of columns in the matrix, so , whose rank and nullity are 4 and 2, respectively, must have 6 columns. Also, the rank of the transpose $A^{T }$ is always equal to that of itself. $A^{T }$ therefore has rank 4 and nullity 3, so it has 7 columns - the number of rows in . This makes a matrix with seven rows and six columns - a $7 \times 6$ matrix.

Determine if the following matrix is linearly independent or not.

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

Linearly Dependent

Linearly Independent

Explanation

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.

$A = \begin{bmatrix} 1 & 0 & 2 \4 & 3 & 11 \ 2 & 2 & 6 \-3 & 1 & -5 \ 3 & -2 & 4\end{bmatrix}$

What are the rank and nullity of ?

Rank 2, nullity 1

Rank 2; nullity 3

Rank 3; nullity 0

Rank 3; nullity 2

Rank 5; nullity 0

Explanation

The rank of a matrix can be found by performing row operations until it is reduced to row-echelon form. On matrix , we want to have leading 1's in each nonzero row, we want the leading 1's to go from upper left to lower right, and we want zero rows gathered at the bottom. Beginning with :

$A = \begin{bmatrix} 1 & 0 & 2 \4 & 3 & 11 \ 2 & 2 & 6 \-3 & 1 & -5 \ 3 & -2 & 4\end{bmatrix}$

The following four operations can be performed simultaneously:

$-4R1 + R2 \rightarrow R2$

$-2R1 + R 3 \rightarrow R3$

$3R1 + R4 \rightarrow R4$

$- 3R1 + R5 \rightarrow R5$

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

$\frac{1}{3}R2 \rightarrow R2$

$A = \begin{bmatrix} 1 & 0 & 2 \0 & 1 & 1 \ 0& 2 & 2 \0 & 1 & 1 \ 0 & -2 & -2\end{bmatrix}$

The next three can be performed simultaneously:

$-2R2 + R3 \rightarrow R3$

$- R2 + R4 \rightarrow R4$

$2R2 + R5 \rightarrow R5$

$A = \begin{bmatrix} 1 & 0 & 2 \0 & 1 & 1 \ 0& 0 &0 \0& 0 &0 \ 0& 0 &0 \end{bmatrix}$

The matrix is now in reduced row-echelon form. There are two nonzero rows, so the rank is 2. The sum of the rank and the nullity is equal to the number of columns, of which there are 3, so the nullity is 1.

Linear Independence and Rank

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