\[A = \begin{bmatrix} 1&2&3&4 \\ -1&0&1&3 \\ 0&1&-2&1 \\ 5&4&-2&-3 \\ \end{bmatrix} \]
## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1So, the rank(A) = 4 because there are 4 linear independent column or using the definition for the book dimension of the column space is 4.
With the assumption that the matrix is non-zero and by the Definition ROM (Rank of Matrix), the maximum rank of a mxn matrix is n since that is the number of columns in the matrix and because the number of linearly independent columns can not be greater than the number of columns of said matrix.
The minimum rank of a non-zero matrix is one because there should be at least one column that is linearly independent.
\[B = \begin{bmatrix} 1&2&1 \\ 3&6&3 \\ 2&4&2 \\ \end{bmatrix} \]
Compute the eigenvalues and eigenvectors of the matrix A. You’ll need to show your work.You’ll need to write out the characteristic polynomial and show your solution.
\[A = \begin{bmatrix} 1&2&3 \\ 0&4&5 \\ 0&0&6 \\ \end{bmatrix} \]
To find the eigenvalues and eigenvectors of this matrix.
First, we need to compute \(det(A-\lambda I)\)
\[ det(A-\lambda I) = \begin{bmatrix} 1-\lambda &2&3 \\ 0&4-\lambda &5 \\ 0&0&6-\lambda \\ \end{bmatrix} \] Notice, we already have upper triangular matrix and recall one of the properties of determinants where the product of the entries on the main diagonal line of said triangular matrix is equal to the determinant.
We have,
\[ det(A-\lambda I) = \begin{bmatrix} 1-\lambda &2&3 \\ 0&4-\lambda &5 \\ 0&0&6-\lambda \\ \end{bmatrix} \Rightarrow (1-\lambda )(4-\lambda )(6-\lambda) \] Thus, we have eigenvalues \(\lambda_1 = 1, \ \lambda_2 = 4 \ and \ \lambda_3 = 6,\)
Next to find the eigen vectors, we need to solve \(A- \lambda I = 0, for \ \lambda_1, \ \lambda_2 \ and \ \lambda_3\)
For \(\lambda_1 = 1\)
\[ (A-(1) I) = 0 \\ \Rightarrow \begin{bmatrix} 1-1 &2&3 \\ 0&4-1 &5 \\ 0&0&6-1 \\ \end{bmatrix} \begin{bmatrix}x \\y\\z\end{bmatrix} = 0 \\ \Rightarrow \begin{bmatrix} 0 &2&3 \\ 0&3 &5 \\ 0&0&5 \\ \end{bmatrix} \begin{bmatrix}x \\y\\z\end{bmatrix} = 0 \\ \Rightarrow \begin{bmatrix}x \\y\\z\end{bmatrix} = \begin{bmatrix}x\\0\\0\end{bmatrix} \ where \ x \in \mathbb{R} \\ \Rightarrow x \begin{bmatrix} 1 \\ 0\\ 0\\ \end{bmatrix} \]
For \(\lambda_2 = 4\)
\[ (A-(4) I) = 0 \\ \Rightarrow \begin{bmatrix} 1-4 &2&3 \\ 0&4-4 &5 \\ 0&0&6-4 \\ \end{bmatrix} \begin{bmatrix}x \\y\\z\end{bmatrix} = 0 \\ \Rightarrow \begin{bmatrix} -3 &2&3 \\ 0&0 &5 \\ 0&0&2 \\ \end{bmatrix} \begin{bmatrix}x \\y\\z\end{bmatrix} = 0 \\ \] Taking the first row, we know z = 0
\[ -3x + 2y+3z = 0 \\ \Rightarrow -3x +2y = 0 \\ \Rightarrow x = 2 \ and \ y = 3 \] So, the eigenvector is \[ \begin{bmatrix} 2\\ 3\\ 0\\ \end{bmatrix} \]
Lastly, for \(\lambda_3 = 6\)
\[ (A-(6) I) = 0 \\ \Rightarrow \begin{bmatrix} 1-6 &2&3 \\ 0&4-6 &5 \\ 0&0&6-6 \\ \end{bmatrix} \begin{bmatrix}x \\y\\z\end{bmatrix} = 0 \\ \Rightarrow \begin{bmatrix} -5 &2&3 \\ 0&-2 &5 \\ 0&0&0 \\ \end{bmatrix} \begin{bmatrix}x \\y\\z\end{bmatrix} = 0 \\ \] We have the following,
\[ -5x + 2y +3z = 0 \\ -2y + 5z = 0 \] Taking the second equation,
\[ -2y + 5z = 0 \Leftrightarrow 5z = 2y \] Plugging that in the first equation. We get,
\[ -5x +5z +3z = 0 \\ \Rightarrow -5x + 8z = 0 \Leftrightarrow 8z = 5x \] Now, for this statement to be true, x = 8 and z = 5. then solving for y using the first equation \[ -5(8) +2y +3(5) = 0 \\ \Rightarrow -40 +2y +15 = 0 \\ \Rightarrow 2y = 25 \\ \Rightarrow y =12.5 \] Then, the eigenvector when \(\lambda = 6\) is \[ \begin{bmatrix}8 \\12.5\\5\end{bmatrix} \]
Therefore, the eigenvectors of matrix A is
\[ (\begin{bmatrix}1 \\0\\0\end{bmatrix}, \ \begin{bmatrix}2 \\3\\0\end{bmatrix}, \ \begin{bmatrix}8 \\12.5\\5\end{bmatrix}) \]