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

The orthogonal complement of the column space of $A$ is the span of the set:

Question :

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

the orthogonal complement of the column space of $a$ is the span of the set:

$Let $a = \begin{bmatrix} 1 & 1 & 1 & 0 \ 1 & 2 & 1 & 1 \ 1 & 2 & 1 & -1 \ 1 & | Doubtlet.com$

Solution:

Neetesh Kumar | October 21, 2024

Linear Algebra Homework Help

This is the solution to Math2B Course: Linear Algebra
Final Exam Question Number 21
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Step-by-step solution:

The orthogonal complement of the column space of a matrix $A$ , denoted as $\text{Col}(A)^\perp$ , consists of all vectors $\mathbf{v}$ such that:

Orthogonal Complement Calculator

$A^T \mathbf{v} = 0$

This means we must find the null space (or kernel) of $A^T$ . The null space consists of all vectors $\mathbf{v}$ such that $A^T \mathbf{v} = 0$ .

Step 1: Find $A^T$

First, compute the transpose of matrix $A$ :

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

Step 2: Solve $A^T \mathbf{v} = 0$

We now solve the equation $A^T \mathbf{v} = 0$ , where $\mathbf{v} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}$ .

This gives us the following system of linear equations:

$x_1 + x_2 + x_3 = 0$
$x_1 + 2x_2 + x_3 + x_4 = 0$
$x_1 + 2x_2 + x_3 - x_4 = 0$
$x_2 - x_4 = 0$

Step 3: Solve the system of equations

From equation (4), we have:

$x_2 = x_4$

Substitute $x_2 = x_4$ into equations (2) and (3):

From equation (2):

$x_1 + 2x_4 + x_3 + x_4 = 0 \quad \Rightarrow \quad x_1 + 3x_4 + x_3 = 0$

From equation (3):

$x_1 + 2x_4 + x_3 - x_4 = 0 \quad \Rightarrow \quad x_1 + x_4 + x_3 = 0$

Now subtract the second simplified equation from the first:

$(x_1 + 3x_4 + x_3) - (x_1 + x_4 + x_3) = 0$

This simplifies to:

$2x_4 = 0 \quad \Rightarrow \quad x_4 = 0$

Since $x_4 = 0$ , we have from equation (4) that $x_2 = 0$ . Now, using $x_2 = x_4 = 0$ in equation (1):

$x_1 + x_3 = 0 \quad \Rightarrow \quad x_1 = -x_3$

Step 4: Express the solution

We now have the solution:

$x_1 = -x_3, \quad x_2 = 0, \quad x_4 = 0$

Thus, the vector $\mathbf{v}$ is of the form:

$\mathbf{v} = \begin{bmatrix} -x_3 \\ 0 \\ x_3 \\ 0 \end{bmatrix} = x_2\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix}$

The null space is spanned by the vector $\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$ , $\begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix}$ .

Final Answer:

The correct answer is (e): $(0, 0, 0, 0),(-1, 0, 1, 0)$ .

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Let A=[11101211121−11110]A = \begin{bmatrix} 1 & 1 & 1 & 0 \\ 1 & 2 & 1 & 1 \\ 1 & 2 & 1 & -1 \\ 1 & 1 & 1 & 0 \end{bmatrix}A=​1111​1221​1111​01−10​​ The orthogonal complement of the column space of AAA is the span of the set: