\[A=\begin{bmatrix} 1&2&3&4\\-1&0&1&3\\0&1&-2&1\\5&4&-2&-3 \end{bmatrix} \]
\[dim(C(A)) = 4\]
\[rank(A) = 4\]
A <- matrix(c(1,-1,0,5,2,0,1,4,3,1,-2,-2,4,3,1,-3), ncol = 4, byrow = F)
pracma::rref(A)## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1pracma::Rank(A)## [1] 4\[m>n\rightarrow max(rank(A))=n \] \[min(rank(NonZeroA))=1\]
\[ B=\begin{bmatrix} 1&2&1\\3&6&3\\2&4&2 \end{bmatrix} \]
\[ \begin{bmatrix} 1&2&1\\3&6&3\\2&4&2 \end{bmatrix} \xrightarrow[R3-2R1]{R2-3R1} \begin{bmatrix} 1&2&1\\0&0&0\\0&0&0 \end{bmatrix} \]
\[dim(C(B)) = 1\] \[rank(B) = 1\]
B <- matrix(c(1,3,2,2,6,4,1,3,2), ncol = 3, byrow = F)
pracma::rref(B)## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 0 0 0pracma::Rank(B)## [1] 1\[ A= \begin{bmatrix} 1&2&3\\0&4&5\\0&0&6 \end{bmatrix} \]
\[ \det(A-\lambda I_n)=0 \]
\[ det( \begin{bmatrix} 1&2&3\\0&4&5\\0&0&6 \end{bmatrix} \begin{bmatrix} \lambda&0&0\\0&\lambda&0\\0&0&\lambda \end{bmatrix} ) \]
\[ \begin{bmatrix} 1-\lambda &2&3\\ 0&4-\lambda &5\\0&0&6-\lambda \end{bmatrix} \]
\[ Rule\:of\:Sarrus = det(M) = det\begin{bmatrix}a11&a12&a13\\a21&a22&a23\\a31&a32&a33\end{bmatrix}=a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-+a_{31}a_{22}a_{13}-+a_{32}a_{23}a_{11}+a_{33}a_{21}a_{12} \]
\[ \begin{bmatrix} 1-\lambda &2&3\\ 0&4-\lambda &5\\0&0&6-\lambda \end{bmatrix} \begin{bmatrix} 1-\lambda &2\\0&4-\lambda\\0&0& \end{bmatrix} \]
\[ (1-\lambda)(4-\lambda)(6-\lambda) \]
\[ 4-5\lambda+\lambda^2(6-\lambda) \]
\[ 4(6-\lambda)-5\lambda(6-\lambda)+\lambda^2(6-\lambda) \]
\[ 24-4\lambda-30\lambda+5\lambda^2+6\lambda^2-\lambda^3 \]
\[p_A(\lambda)=24-34\lambda+11\lambda^2-\lambda^3\]
\[ 24-34\lambda+11\lambda^2-\lambda^3\implies(1-\lambda)(4-\lambda)(6-\lambda) \]
\[\lambda=\{1,4,6\}\]
\[ (\lambda I_n-A)\vec{v}=\vec{0} \]
\[ \begin{bmatrix} 1-\lambda &2&3\\ 0&4-\lambda &5\\0&0&6-\lambda \end{bmatrix},\lambda=1 \implies \begin{bmatrix} 1-1 &2&3\\ 0&4-1 &5\\0&0&6-1 \end{bmatrix} \rightarrow \begin{bmatrix} 0 &2&3\\ 0&3&5\\0&0&5 \end{bmatrix} \]
\[ \begin{bmatrix} 0 &2&3\\ 0&3&5\\0&0&5 \end{bmatrix} \xrightarrow[]{1/2R1} \begin{bmatrix} 0 &1&3/2\\ 0&3&5\\0&0&5 \end{bmatrix} \xrightarrow[]{R2+-3R1} \begin{bmatrix} 0 &1&3/2\\ 0&0&1/2\\0&0&5 \end{bmatrix} \]
\[ \begin{bmatrix} 0 &1&3/2\\ 0&0&1/2\\0&0&5 \end{bmatrix} \xrightarrow[]{2R2} \begin{bmatrix} 0 &1&3/2\\ 0&0&1\\0&0&5 \end{bmatrix} \xrightarrow[R3+-5R2]{} \begin{bmatrix} 0 &1&3/2\\ 0&0&1\\0&0&0 \end{bmatrix} \]
\[ \begin{bmatrix} 0 &1&3/2\\ 0&0&1\\0&0&0 \end{bmatrix} \xrightarrow[]{R1+-3/2R2} \begin{bmatrix} 0 &1&0\\ 0&0&1\\0&0&0 \end{bmatrix} \]
\[ \begin{bmatrix} 0 &1&0\\ 0&0&1\\0&0&0 \end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix} 0\\0\\0 \end{bmatrix} \]
\[Let\: x_1 = 1,\:therefore\]
\[x_1 = 1\]
\[x_2 = 0 \]
\[x_3 = 0\]
\[E_{\lambda=1}=span(\begin{bmatrix}1\\0\\0\end{bmatrix})\]
\[ \begin{bmatrix} 1-\lambda &2&3\\ 0&4-\lambda &5\\0&0&6-\lambda \end{bmatrix},\lambda=4 \implies \begin{bmatrix} 1-4 &2&3\\ 0&4-4 &5\\0&0&6-4 \end{bmatrix} \rightarrow \begin{bmatrix} -3 &2&3\\ 0&0&5\\0&0&2 \end{bmatrix} \]
\[ \begin{bmatrix} -3 &2&3\\ 0&0&5\\0&0&2 \end{bmatrix} \xrightarrow[1/5R2]{-1/3R1} \begin{bmatrix} 1 &-2/3&-1\\ 0&0&1\\0&0&2 \end{bmatrix} \xrightarrow[R3+-2R2]{R1+R2} \begin{bmatrix} 1 &-2/3&0\\ 0&0&1\\0&0&0 \end{bmatrix} \]
\[ \begin{bmatrix} 1 &-2/3&0\\ 0&0&1\\0&0&0 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\x_3 \end{bmatrix} = \begin{bmatrix} 0\\0\\0 \end{bmatrix} \]
\[x_1 + \frac{-2}{3}x_2 = 0\implies x_1=\frac{2}{3}x_2\]
\[Let\: x_2 = 1,\:therefore\]
\[x_1 = \frac{2}{3}\]
\[x_2 = 1 \]
\[x_3 = 0\]
\[E_{\lambda=4}=span(\begin{bmatrix}2/3\\1\\0\end{bmatrix})\]
\[ \begin{bmatrix} 1-\lambda &2&3\\ 0&4-\lambda &5\\0&0&6-\lambda \end{bmatrix},\lambda=6 \implies \begin{bmatrix} 1-6 &2&3\\ 0&4-6 &5\\0&0&6-6 \end{bmatrix} \rightarrow \begin{bmatrix} -5 &2&3\\ 0&-2&5\\0&0&0 \end{bmatrix} \]
\[ \begin{bmatrix} -5 &2&3\\ 0&-2&5\\0&0&0 \end{bmatrix} \xrightarrow[-1/2R2]{-1/5R1} \begin{bmatrix} 1&-2/5&-3/5\\0&1&-5/2\\0&0&0 \end{bmatrix} \xrightarrow[]{R1+2/5R2} \begin{bmatrix} 1&0&-8/5\\0&1&-5/2\\0&0&0 \end{bmatrix} \]
\[x_1+(-\frac{8}{5})x_3=0\] \[x_1=\frac{8}{5}x_3\]
\[x_2+(-\frac{5}{2})x_3=0\] \[x_2=\frac{5}{2}x_3\]
\[Let\: x_3 = 1,\:therefore\]
\[x_1 = \frac{8}{5}\] \[x_2 =\frac{5}{2}\] \[x_3 = 1\]
\[E_{\lambda=6}=span(\begin{bmatrix}8/5\\5/2\\1\end{bmatrix})\]
A <- matrix(data = c(1,0,0,2,4,0,3,5,6), ncol = 3, byrow = FALSE)
(pracma::charpoly(A))## [1] 1 -11 34 -24e <- eigen(A)
# Column order for associated Lambdas is 6,4,1
(e)## eigen() decomposition
## $values
## [1] 6 4 1
##
## $vectors
## [,1] [,2] [,3]
## [1,] 0.5108407 0.5547002 1
## [2,] 0.7981886 0.8320503 0
## [3,] 0.3192754 0.0000000 0