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Matrices are a fundamental concept in mathematics and are used in a variety of fields, including algebra, geometry, calculus, statistics, and computer science. Some of the key concepts related to matrices include:
- Matrix Operations: Addition, Subtraction, Scalar Multiplication, Matrix Multiplication, Transpose, Inverse.
- Determinants: A scalar value associated with a square matrix, used to determine whether a matrix is invertible or not.
- Eigenvalues and Eigenvectors: Eigenvalues are scalar values that represent the amount of stretch or compression in a given direction. Eigenvectors are non-zero vectors that are stretched or compressed by a matrix by a factor equal to the corresponding eigenvalue.
- Rank: The rank of a matrix is the maximum number of linearly independent rows or columns of the matrix.
- Linear Transformations: Matrices can be used to represent linear transformations of vectors in space, and the properties of the matrix can provide information about the transformation.
- Singular Value Decomposition: A method for breaking down a matrix into its constituent parts, which can be used to analyze and manipulate data.
Addition of Matrices
To add two matrices together, you need to add the corresponding elements in each matrix. For example, given two matrices A and B:
$$ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} $$ $$ B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} $$The sum of A and B, denoted A + B, is:
$$ A + B $$ $$ = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} $$ $$ = \begin{bmatrix} 1+5 & 2+6 \\ 3+7 & 4+8 \end{bmatrix} $$ $$ = \begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix} $$Subtracting Matrices
To subtract one matrix from another, you need to subtract the corresponding elements in each matrix. For example, given two matrices A and B:
$$ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} $$ $$ B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} $$The difference of A and B, denoted A - B, is:
$$ A - B $$ $$ = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} - \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} $$ $$ = \begin{bmatrix} 1-5 & 2-6 \\ 3-7 & 4-8 \end{bmatrix} $$ $$ = \begin{bmatrix} -4 & -4 \\ -4 & -4 \end{bmatrix} $$Multiplying Matrices
Scalar Multiplication
Scalar multiplication is an operation that multiplies each element of a matrix by a scalar (a constant value). The result is a matrix with the same dimensions as the original matrix, where each element is the product of the corresponding element in the original matrix and the scalar.
For example, if we have a matrix $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and we want to multiply it by a scalar $k=2$, then the result would be:
$$kA = 2 \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix}$$In this case, each element of matrix $A$ is multiplied by 2 to get the corresponding element of $kA$. So, for example, the element in the first row and second column of $kA$ is equal to 2 times the element in the first row and second column of $A$, which is $2\times2=4$.
Another example would be if we have a matrix $B = \begin{bmatrix} 0 & 1 & -1 \\ 2 & -1 & 3 \end{bmatrix}$ and we want to multiply it by a scalar $k=-3$, then the result would be:
$$kB = -3 \begin{bmatrix} 0 & 1 & -1 \\ 2 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 0 & -3 & 3 \\ -6 & 3 & -9 \end{bmatrix}$$Again, each element of matrix $B$ is multiplied by $-3$ to get the corresponding element of $kB$. So, for example, the element in the second row and third column of $kB$ is equal to $-3$ times the element in the second row and third column of $B$, which is $-3\times3=-9$.
Matrix Multiplication
To multiply two matrices together, you need to multiply the corresponding elements in each row of the first matrix with the corresponding elements in each column of the second matrix, and then sum the products. For example, given two matrices A and B:
$$ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} $$ $$ B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} $$The product of A and B, denoted AB, is:
$$ AB $$ $$ = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$$ $$ = \begin{bmatrix} 1*5+2*7 & 1*6+2*8 \\ 3*5+4*7 & 3*6+4*8 \end{bmatrix} $$ $$ = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix} $$Transpose of a Matrix
In matrix algebra, the transpose of a matrix involves flipping the rows and columns of a matrix. That is, if matrix $A$ has dimensions $m \times n$, then the transpose of matrix A, denoted by $A^T$, has dimensions $n \times m$, where the rows of $A$ become columns of $A^T$, and the columns of $A$ become rows of $A^T$. Mathematically, the $(i, j)th$ element of the transpose of $A$ is equal to the $(j, i)th$ element of $A$.
If Matrix $A$ is: $$ A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} $$ then Transpose of Matrix $A^T$ is: $$ A^T = \begin{bmatrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{bmatrix} $$Example
$$ \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}^T \implies \begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{bmatrix} $$Inverse of a Matrix or Dividing Matrices
Matrix division is not a well-defined operation in general, but we can talk about the inverse of a matrix. Given a matrix A, its inverse A^{-1} is a matrix such that AA^{-1} = I, where I is the identity matrix. The inverse of a matrix is a matrix that when multiplied with the original matrix yields the identity matrix. The inverse of a square matrix A is denoted as $A^{-1}$ and is defined as:
$$AA^{-1} = A^{-1}A = I$$where I is the identity matrix.
The formula to find the inverse of a 2x2 matrix A is:
$$ A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} $$where a, b, c and d are the elements of matrix A.
For a square matrix A of higher dimensions, the inverse can be found using various methods such as row reduction or the adjugate matrix method.
Let $A$ be an $n \times n$ matrix. If $A^{-1}$ exists, then:
$$ A A^{-1} = A^{-1} A = I_n $$where $I_n$ is the $n x n$ identity matrix.
Here's an example of how to find the inverse of a matrix:
$$ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} $$We can find the inverse of A by solving the equation:
$$ A A^{-1} = I $$Multiplying A by its inverse gives:
$$ \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$ $$ = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$This gives us the following system of equations:
$$ a + 2c = 1 $$ $$ b + 2d = 0 $$ $$ 3a + 4c = 0 $$ $$ 3b + 4d = 1 $$Solving this system of equations, we get:
$$ a = -2/5, $$ $$ b = 1/5,$$ $$ c = 3/10, $$ $$d = -1/10 $$Therefore, the inverse of A is:
$$ A^{-1} = \begin{bmatrix} \frac{-2}{5} & \frac{1}{5} \\ \frac{3}{10} & \frac{-1}{10} \end{bmatrix} $$Note that not all matrices have inverses. If a matrix does not have an inverse, it is said to be singular or non-invertible.
Determinants
Consider the following 3x3 matrix:
$$ A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} $$To compute the determinant of A using the row method, we first choose a row or column of the matrix to expand along. For simplicity, let's expand along the first row:
$$ \begin{aligned} \det(A) &= a_{11} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12} \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13} \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix} \\ &= a_{11} (a_{22}a_{33} - a_{23}a_{32}) - a_{12} (a_{21}a_{33} - a_{23}a_{31}) + a_{13} (a_{21}a_{32} - a_{22}a_{31}) \end{aligned} $$Next, we compute the determinants of the 2x2 matrices in each term. For example, the first term becomes:
$$ \begin{aligned} a_{11} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} &= a_{11} (a_{22}a_{33} - a_{23}a_{32}) \end{aligned} $$Similarly, we can compute the other two terms:
$$ \begin{aligned} - a_{12} \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} &= - a_{12} (a_{21}a_{33} - a_{23}a_{31}) \\ + a_{13} \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix} &= a_{13} (a_{21}a_{32} - a_{22}a_{31}) \end{aligned} $$Finally, we add up these three terms to obtain the determinant of A:
$$ \det(A) = a_{11} (a_{22}a_{33} - a_{23}a_{32}) - a_{12} (a_{21}a_{33} - a_{23}a_{31}) + a_{13} (a_{21}a_{32} - a_{22}a_{31}) $$Example
Lets look by an example: Consider the following 3x3 matrix:
$$A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$$Method 1
To find the determinant of A, we can use the following formula:
$$\det(A) = a_{11} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12} \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13} \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}$$where $a_{ij}$ denotes the element in the $i^{th}$ row and $j^{th}$ column of $A$, and $|A|$ denotes the determinant of $A$.
Using this formula, we can compute the determinant of $A$ as follows:
$$\begin{aligned} \det(A) &= 1 \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} - 2 \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} + 3 \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} \\ &= 1 (45 - 48) - 2 (36 - 42) + 3 (32 - 35) \\ &= -3 \end{aligned}$$Therefore, the determinant of $A$ is $-3$.
Method 2: Sarrus Rule
Sarrus' rule is a mnemonic technique to compute the determinant of a 3x3 matrix. It involves duplicating the first two columns of the matrix on the right-hand side, and then adding the products of the diagonals going downwards and subtracting the products of the diagonals going upwards. Here is the formula:
$$ \det(A) = a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{31}a_{22}a_{13} - a_{32}a_{23}a_{11} - a_{33}a_{21}a_{12} $$Using the same matrix $A$ as before, we can compute the determinant of $A$ using Sarrus' rule as follows:
$$ \begin{aligned} \det(A) &= 1 \times 5 \times 9 + 2 \times 6 \times 7 + 3 \times 4 \times 8 \\ &\quad - 7 \times 5 \times 3 - 8 \times 6 \times 1 - 9 \times 4 \times 2 \\ &= 45 + 84 + 96 - 105 - 48 - 72 \\ &= -3 \end{aligned} $$Method 3: Cofactor Expansion
Cofactor expansion is a method to compute the determinant of a matrix by expanding it along any row or column. It involves computing the determinants of the (n-1)x(n-1) submatrices obtained by deleting the row and column containing each element, and multiplying them by the corresponding cofactors. Here is the formula:
$$\det(A) = \sum_{j=1}^{n} (-1)^{i+j} a_{ij} C_{ij}$$where $C_{ij}$ is the cofactor of $a_{ij}$, defined as $C_{ij} = (-1)^{i+j} \det(A_{ij})$, where $A_{ij}$ is the submatrix obtained by deleting the $i^{th}$ row and $j^{th}$ column from $A$.
Using the same matrix $A$ as before, we can compute the determinant of $A$ by expanding it along the first row as follows:
$$ \begin{aligned} \det(A) &= 1 \times C_{11} - 2 \times C_{12} + 3 \times C_{13} \\ &= 1 \times \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} - 2 \times \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} + 3 \times \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} \\ &= 1 \times (45 - 48) - 2 \times (36 - 42) + 3 \times (32 - 35) \\ &= -3 \end{aligned} $$Note that the result is the same as before, as expected.
EigenValues
Let us find the eigenvalues of a 3x3 matrix $A$:
First, we need to find the determinant of the matrix $A - \lambda I$, where $\lambda$ is an eigenvalue and $I$ is the 3x3 identity matrix:
$$\det(A - \lambda I) = \begin{vmatrix} a_{11} - \lambda & a_{12} & a_{13} \\ a_{21} & a_{22} - \lambda & a_{23} \\ a_{31} & a_{32} & a_{33} - \lambda \end{vmatrix}$$We then expand the determinant using the cofactor expansion method along the first row or column:
$$\det(A - \lambda I) = (a_{11} - \lambda) \begin{vmatrix} a_{22} - \lambda & a_{23} \\ a_{32} & a_{33} - \lambda \end{vmatrix} - a_{12} \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} - \lambda \end{vmatrix} + a_{13} \begin{vmatrix} a_{21} & a_{22} - \lambda \\ a_{31} & a_{32} \end{vmatrix}$$Simplifying this expression gives us a polynomial equation in $\lambda$:
$$\lambda^3 + c_2\lambda^2 + c_1\lambda + c_0 = 0$$where $c_2$, $c_1$, and $c_0$ are constants determined by the entries of the matrix $A$. We can solve this equation to find the three eigenvalues of $A$. Following are the steps to solve the polynomial equation $\lambda^3 + c_2\lambda^2 + c_1\lambda + c_0 = 0$ and find the eigenvalues of a 3x3 matrix $A$:
- Write the characteristic polynomial: $\det(A - \lambda I) = \lambda^3 + c_2\lambda^2 + c_1\lambda + c_0 = 0$
- Use the cubic formula or other numerical methods to solve the polynomial equation and find the three roots $\lambda_1$, $\lambda_2$, and $\lambda_3$.
- Verify that each of the three roots is an eigenvalue of the matrix $A$ by substituting it into the equation $(A - \lambda I)\mathbf{v} = \mathbf{0}$, where $\mathbf{v}$ is a nonzero eigenvector associated with the eigenvalue $\lambda$. If the equation is satisfied, then the corresponding $\lambda$ is an eigenvalue of $A$.
- Repeat the above steps to find the eigenvectors associated with each eigenvalue. To find the eigenvector $\mathbf{v}_i$ associated with eigenvalue $\lambda_i$, solve the equation $(A - \lambda_i I)\mathbf{v}_i = \mathbf{0}$. Note that there may be more than one linearly independent eigenvector associated with each eigenvalue.
- Normalize each eigenvector by dividing by its Euclidean norm, so that $|\mathbf{v}_i| = 1$.
- The three eigenvalue-eigenvector pairs $(\lambda_1, \mathbf{v}_1)$, $(\lambda_2, \mathbf{v}_2)$, and $(\lambda_3, \mathbf{v}_3)$ form a complete set of eigenvalue-eigenvector pairs that diagonalize the matrix $A$. Specifically, we can write $A$ as $A = PDP^{-1}$, where $P$ is the matrix whose columns are the eigenvectors $\mathbf{v}_1$, $\mathbf{v}_2$, and $\mathbf{v}_3$, and $D$ is the diagonal matrix whose diagonal entries are the eigenvalues $\lambda_1$, $\lambda_2$, and $\lambda_3$.
Example
Let's say we have the matrix:
$$ A = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix} $$To find the eigenvalues of $A$, we need to solve the characteristic polynomial equation:
$$ \det(A - \lambda I) = 0 $$where $I$ is the identity matrix and $\lambda$ is the eigenvalue we are trying to find. Expanding the determinant, we get:
$$ \begin{vmatrix} 2 - \lambda & -1 & 0 \\ -1 & 2 - \lambda & -1 \\ 0 & -1 & 2 - \lambda \end{vmatrix} $$ $$= (2 - \lambda) \begin{vmatrix} 2 - \lambda & -1 \\ -1 & 2 - \lambda \end{vmatrix} - (-1) \begin{vmatrix} -1 & -1 \\ 0 & 2 - \lambda \end{vmatrix} + 0 \begin{vmatrix} -1 & 2 - \lambda \\ 0 & -1 \end{vmatrix} $$Simplifying this expression, we get:
$$ (2 - \lambda)^3 - 3(2 - \lambda) = 0 $$which is a cubic polynomial equation in $\lambda$. To solve for the eigenvalues, we can use various methods such as factoring, synthetic division, or the cubic formula. For simplicity, let's use factoring and synthetic division. We can factor out $(2 - \lambda)$ from the equation and obtain:
$$ (2 - \lambda) [(2 - \lambda)^2 - 3] = 0 $$The roots of this equation are $\lambda_1 = 2$, $\lambda_2 = 2 - \sqrt{3}$, and $\lambda_3 = 2 + \sqrt{3}$. Therefore, the eigenvalues of $A$ are:
$$ \lambda_1 = 2, \quad \lambda_2 = 2 - \sqrt{3}, \quad \lambda_3 = 2 + \sqrt{3} $$EigenVectors
Eigenvalues and eigenvectors are important concepts in linear algebra that are used in various fields of mathematics, science, and engineering. Given a square matrix $A$, an eigenvector $\mathbf{v}$ and its corresponding eigenvalue $\lambda$ are defined as follows:
$$A\mathbf{v} = \lambda\mathbf{v}$$In other words, when the matrix $A$ is multiplied by its eigenvector $\mathbf{v}$, the result is a scalar multiple of $\mathbf{v}$, where the scalar is the eigenvalue $\lambda$.
To find the eigenvectors of a matrix $A$, we first need to find its eigenvalues. We do this by solving the characteristic equation:
$$\det(A-\lambda I) = 0$$where $I$ is the identity matrix of the same size as $A$. The solutions to this equation are the eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$ of $A$.
Once we have found the eigenvalues of $A$, we can find the eigenvectors by solving the system of linear equations:
$$(A-\lambda_i I)\mathbf{v}_i = \mathbf{0}$$where $\mathbf{v}_i$ is the eigenvector corresponding to the eigenvalue $\lambda_i$.
To solve this system, we first compute the matrix $A-\lambda_i I$ and row reduce it to its reduced row echelon form. We then solve the resulting system of equations for the unknowns $v_{i1}, v_{i2}, \ldots, v_{in}$, where $n$ is the size of the matrix $A$. The solution to this system is the eigenvector $\mathbf{v}_i$.
It is important to note that a matrix may have multiple eigenvectors corresponding to the same eigenvalue, and that eigenvectors are only defined for square matrices. Additionally, not all matrices have eigenvectors or eigenvalues.
Example
Let's find the eigenvectors of the following 3x3 matrix:
$$A = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{bmatrix}$$To find the eigenvalues of $A$, we need to solve the characteristic equation:
$$\det(A - \lambda I) = \begin{vmatrix} 2-\lambda & 1 & 0 \\ 1 & 2-\lambda & 1 \\ 0 & 1 & 2-\lambda \end{vmatrix} = 0$$Expanding the determinant using the first row, we get:
$$(2-\lambda) \begin{vmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 0 & 2-\lambda \end{vmatrix} + 0 \begin{vmatrix} 1 & 2-\lambda \\ 0 & 1 \end{vmatrix} = 0$$Simplifying this expression, we get:
$$(2-\lambda)^3 - 3(2-\lambda) + 1 = 0$$Expanding and simplifying, we get the cubic equation:
$$\lambda^3 - 6\lambda^2 + 11\lambda - 6 = 0$$Using synthetic division or the rational root theorem, we can find that the roots of this equation are $\lambda_1 = 1$, $\lambda_2 = 2$, and $\lambda_3 = 3$. These are the eigenvalues of the matrix $A$.
Next, we need to find the eigenvectors corresponding to each eigenvalue. Let's start with the eigenvalue $\lambda_1 = 1$. To find the eigenvector $\mathbf{v}_1$, we need to solve the system of linear equations:
$$(A - \lambda_1 I) \mathbf{v}1 = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} v_{11} \\ v_{12} \\ v_{13} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$Row reducing the augmented matrix $\left[\begin{array}{ccc|c} 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right]$, we get:
$$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_{11} \\ v_{12} \\ v_{13} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$From the last row of the row-reduced matrix, we see that $v_{13}$ is a free variable. Setting $v_{13} = t$ for some arbitrary scalar $t$, we can solve for the other variables:
$$v_{12} = -v_{13} = -t$$ $$v_{11} = v_{13} - v_{12} = 2t$$Therefore, the eigenvector corresponding to $\lambda_1 = 1$ is $\mathbf{v}_1 = \begin{bmatrix} 2t \\ -t \\ t \end{bmatrix}$, where $t$ is any nonzero scalar.
Now let's find the eigenvectors corresponding to $\lambda_2 = 2$. We need to solve the system of linear equations:
$$(A - \lambda_2 I) \mathbf{v}2 = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} v{21} \\ v_{22} \\ v_{23} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$Row reducing the augmented matrix $\left[\begin{array}{ccc|c} 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right]$, we get:
$$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_{21} \\ v_{22} \\ v_{23} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$From the second row of the row-reduced matrix, we see that $v_{22}$ is a free variable. Setting $v_{22} = t$ for some arbitrary scalar $t$, we can solve for the other variables:
$$v_{21} = -v_{23} = -t$$ $$v_{23} = v_{21} = -t$$Therefore, the eigenvector corresponding to $\lambda_2 = 2$ is $\mathbf{v}_2 = \begin{bmatrix} -t \\ t \\ -t \end{bmatrix}$, where $t$ is any nonzero scalar.
Finally, let's find the eigenvectors corresponding to $\lambda_3 = 3$. We need to solve the system of linear equations:
$$(A - \lambda_3 I) \mathbf{v}3 = \begin{bmatrix} -1 & 1 & 0 \\ 1 & -1 & 1 \\ 0 & 1 & -1 \end{bmatrix} \begin{bmatrix} v{31} \\ v_{32} \\ v_{33} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$Row reducing the augmented matrix $\left[\begin{array}{ccc|c} -1 & 1 & 0 & 0 \\ 1 & -1 & 1 & 0 \\ 0 & 1 & -1 & 0 \end{array}\right]$, we get:
$$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_{31} \\ v_{32} \\ v_{33} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$From the last row of the row-reduced matrix, we see that $v_{13}$ is a free variable. Setting $v_{13} = t$ for some arbitrary scalar $t$, we can solve for the other variables:
$$v_{12} = -v_{13} = -t$$ $$v_{11} = v_{13} - v_{12} = 2t$$Therefore, the eigenvector corresponding to the eigenvalue $\lambda_1 = 1$ is given by:
$$\mathbf{v}_1 = \begin{bmatrix} 2t \\ -t \\ t \end{bmatrix} = t \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$$Note that since the eigenvectors are only defined up to a scalar multiple, we can choose any nonzero value for $t$. Therefore, we can express the eigenvector $\mathbf{v}_1$ as:
$$\mathbf{v}_1 = c \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$$where $c$ is any nonzero scalar.