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Find qr decomposition house holder [[1,-1,4],[1,4,-2],[1,4,2],[1,-1,0]]

Solution:
Your problem `->` qr decomposition house holder [[1,-1,4],[1,4,-2],[1,4,2],[1,-1,0]]


Here `A` = 
`1``-1``4`
`1``4``-2`
`1``4``2`
`1``-1``0`


`A_1` = 
`1``-1``4`
`1``4``-2`
`1``4``2`
`1``-1``0`


`a_1` = 
`1`
`1`
`1`
`1`


`||a_1||=sqrt(1^2+1^2+1^2+1^2)=sqrt(4)=2`

`v_1=a_1-sign(a_(11))||a_1||e_1` = 
`1`
`1`
`1`
`1`
 - 2 `xx` 
`1`
`0`
`0`
`0`
 = 
`-1`
`1`
`1`
`1`


`H_1=I - 2 *(v_1*v_1^T)/(v_1^T*v_1)` = 
1000
0100
0010
0001
`-``2/4``*`
`-1`
`1`
`1`
`1`
`*`
[`-1``1``1``1`]
 = 
`1/2``1/2``1/2``1/2`
`1/2``1/2``-1/2``-1/2`
`1/2``-1/2``1/2``-1/2`
`1/2``-1/2``-1/2``1/2`


`H_1 * A_1` = 
`1/2``1/2``1/2``1/2`
`1/2``1/2``-1/2``-1/2`
`1/2``-1/2``1/2``-1/2`
`1/2``-1/2``-1/2``1/2`
 `xx` 
`1``-1``4`
`1``4``-2`
`1``4``2`
`1``-1``0`
 = 
`2``3``2`
`0``0``0`
`0``0``4`
`0``-5``2`




Now removing 1st row and 1st column, we get
`A_2` = 
`0``0`
`0``4`
`-5``2`


`a_2` = 
`0`
`0`
`-5`


`||a_2||=sqrt(0^2+0^2+(-5)^2)=sqrt(25)=5`

`v_2=a_1-sign(a_(11))||a_1||e_1` = 
`0`
`0`
`-5`
 - 5 `xx` 
`1`
`0`
`0`
 = 
`-5`
`0`
`-5`


`H_2=I - 2 *(v_1*v_1^T)/(v_1^T*v_1)` = 
100
010
001
`-``2/50``*`
`-5`
`0`
`-5`
`*`
[`-5``0``-5`]
 = 
`0``0``-1`
`0``1``0`
`-1``0``0`


`H_2 * A_2` = 
`0``0``-1`
`0``1``0`
`-1``0``0`
 `xx` 
`0``0`
`0``4`
`-5``2`
 = 
`5``-2`
`0``4`
`0``0`




Now removing 1st row and 1st column, we get
`A_3` = 
`4`
`0`




Since, `H_2H_1A=R`

`H_2H_1A=`
`1``0``0``0`
`0``0``0``-1`
`0``0``1``0`
`0``-1``0``0`
 `xx` 
`1/2``1/2``1/2``1/2`
`1/2``1/2``-1/2``-1/2`
`1/2``-1/2``1/2``-1/2`
`1/2``-1/2``-1/2``1/2`
 `xx` 
`1``-1``4`
`1``4``-2`
`1``4``2`
`1``-1``0`
 = 
`2``3``2`
`0``5``-2`
`0``0``4`
`0``0``0`
 = R


Also `A=H_1H_2R` and `A=QR`, `:.Q=H_1H_2`

`Q=H_1H_2`=
`1/2``1/2``1/2``1/2`
`1/2``1/2``-1/2``-1/2`
`1/2``-1/2``1/2``-1/2`
`1/2``-1/2``-1/2``1/2`
 `xx` 
`1``0``0``0`
`0``0``0``-1`
`0``0``1``0`
`0``-1``0``0`
 = 
`1/2``-1/2``1/2``-1/2`
`1/2``1/2``-1/2``-1/2`
`1/2``1/2``1/2``1/2`
`1/2``-1/2``-1/2``1/2`




checking `Q xx R = A?`

`Q xx R` = 
`1/2``-1/2``1/2``-1/2`
`1/2``1/2``-1/2``-1/2`
`1/2``1/2``1/2``1/2`
`1/2``-1/2``-1/2``1/2`
 `xx` 
`2``3``2`
`0``5``-2`
`0``0``4`
`0``0``0`
 = 
`1``-1``4`
`1``4``-2`
`1``4``2`
`1``-1``0`


and `A` = 
`1``-1``4`
`1``4``-2`
`1``4``2`
`1``-1``0`







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