A = ∣ 5 0 0 1 5 0 0 1 5 ∣ A=\begin{vmatrix}
5&0&0 \\
1&5&0 \\0&1&5
\end{vmatrix} A = ∣ ∣ 5 1 0 0 5 1 0 0 5 ∣ ∣
you cannot find the column X such that A=cX for c∈ \in ∈ R
since A is 3x3 and cX is 3x1
however, we can find X such that AX=cX for some c∈ \in ∈ A
clearly, 5 is an eign calue & A and the compounding eigen vector is
X = [ 0 0 a ] X=\begin{bmatrix}
0\\0\\a
\end{bmatrix} X = ⎣ ⎡ 0 0 a ⎦ ⎤
foe any a∈ \in ∈ R
we can check that AX=5X
A X = [ 5 0 0 1 5 0 0 1 5 ] AX=\begin{bmatrix}
5 & 0&0 \\1&5&0\\0&1&5
\end{bmatrix} A X = ⎣ ⎡ 5 1 0 0 5 1 0 0 5 ⎦ ⎤ [ 0 0 a ] \begin{bmatrix}
0\\0\\a
\end{bmatrix} ⎣ ⎡ 0 0 a ⎦ ⎤ =X = [ 0 0 5 a ] X=\begin{bmatrix}
0\\0\\5a
\end{bmatrix} X = ⎣ ⎡ 0 0 5 a ⎦ ⎤ =5 [ 0 0 a ] 5\begin{bmatrix}
0\\0\\a
\end{bmatrix} 5 ⎣ ⎡ 0 0 a ⎦ ⎤ =5 X 5X 5 X
as,A = ( − 1 0 2 + 1 0 − 2 + 1 0 0 ) A=\begin{pmatrix}
-1^0 & 2+1^0 \\
-2+1^0 & 0
\end{pmatrix} A = ( − 1 0 − 2 + 1 0 2 + 1 0 0 )
A T = [ − 1 0 − 2 + 1 0 2 + 1 0 0 ] A^T=\begin{bmatrix}
-1^0&-2+1^0\\2+1^0&0
\end{bmatrix} A T = [ − 1 0 2 + 1 0 − 2 + 1 0 0 ]
A T c o n j u g a t e = ( A T ) ∗ = [ 1 0 − 2 + 1 0 2 − 1 0 0 ] A^T conjugate =(A^T)^*=\begin{bmatrix}
1^0&-2+1^0\\2-1^0&0
\end{bmatrix} A T co nj ug a t e = ( A T ) ∗ = [ 1 0 2 − 1 0 − 2 + 1 0 0 ] = = = = [ − 1 0 2 + 1 0 − 2 − 1 0 0 ] =\begin{bmatrix}
-1^0&2+1^0\\-2-1^0&0
\end{bmatrix} = [ − 1 0 − 2 − 1 0 2 + 1 0 0 ]
∴ \therefore ∴ conjugate transpose of A is -A
Therefpre A is skew-hermitian
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