Question #218165
i. Let A= [ 5 0 0]
[ 1 5 0]
[ 0 1 5]
Find a column vector X for which A= cX, for some c belong to R.

ii. Give an example with justification, of a skew-hermitian operator on C^2
1
Expert's answer
2021-07-20T09:22:05-0400

A=500150015A=\begin{vmatrix} 5&0&0 \\ 1&5&0 \\0&1&5 \end{vmatrix}

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=[00a]X=\begin{bmatrix} 0\\0\\a \end{bmatrix}

foe any a\in R

we can check that AX=5X

AX=[500150015]AX=\begin{bmatrix} 5 & 0&0 \\1&5&0\\0&1&5 \end{bmatrix} [00a]\begin{bmatrix} 0\\0\\a \end{bmatrix} =X=[005a]X=\begin{bmatrix} 0\\0\\5a \end{bmatrix} =5[00a]5\begin{bmatrix} 0\\0\\a \end{bmatrix} =5X5X

as,A=(102+102+100)A=\begin{pmatrix} -1^0 & 2+1^0 \\ -2+1^0 & 0 \end{pmatrix}

AT=[102+102+100]A^T=\begin{bmatrix} -1^0&-2+1^0\\2+1^0&0 \end{bmatrix}

ATconjugate=(AT)=[102+102100]A^T conjugate =(A^T)^*=\begin{bmatrix} 1^0&-2+1^0\\2-1^0&0 \end{bmatrix} == =[102+102100]=\begin{bmatrix} -1^0&2+1^0\\-2-1^0&0 \end{bmatrix}


\therefore conjugate transpose of A is -A

Therefpre A is skew-hermitian




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