row reduced form with element $i$
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0
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find rank(A)
find basis of$ R(L_A)$ consisting of column vectors
find basis of $N(L_A)$
$$A=left[begin{array}{ccc}0&i&-1\1+i&1&1+2i\1-i&2&1+i\-i&1-i&1end{array}right]$$
i figure out the reduce form is
$$A=left[begin{array}{ccc}1&0&1\0&1&i\0&0&0\0&0&0end{array}right]$$, so the rank(A)=2, but what is basis of $R(L_A)$ and$N(L_A)$
linear-algebra matrices linear-transformations
asked Nov 21 at 22:58
DORCT
406
add a comment |
up vote
0
down vote
favorite
find rank(A)
find basis of$ R(L_A)$ consisting of column vectors
find basis of $N(L_A)$
$$A=left[begin{array}{ccc}0&i&-1\1+i&1&1+2i\1-i&2&1+i\-i&1-i&1end{array}right]$$
i figure out the reduce form is
$$A=left[begin{array}{ccc}1&0&1\0&1&i\0&0&0\0&0&0end{array}right]$$, so the rank(A)=2, but what is basis of $R(L_A)$ and$N(L_A)$
linear-algebra matrices linear-transformations
asked Nov 21 at 22:58
DORCT
406
If you reduced right, the same method for getting bases for row space and null space as usual works, since complexes are a field.
– coffeemath
Nov 21 at 23:31
How would you find these bases if you had a $2$ there instead of $i$? Do the same thing.
– amd
Nov 22 at 0:00
for nullspace i got x1=x3 x2=-ix3 x3=x3 so i got basic of N(A)=1 -i 1
– DORCT
Nov 22 at 0:05
for range(A)i find the row of of pivot row which are first and sec row
– DORCT
Nov 22 at 0:07
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
find rank(A)
find basis of$ R(L_A)$ consisting of column vectors
find basis of $N(L_A)$
$$A=left[begin{array}{ccc}0&i&-1\1+i&1&1+2i\1-i&2&1+i\-i&1-i&1end{array}right]$$
i figure out the reduce form is
$$A=left[begin{array}{ccc}1&0&1\0&1&i\0&0&0\0&0&0end{array}right]$$, so the rank(A)=2, but what is basis of $R(L_A)$ and$N(L_A)$
linear-algebra matrices linear-transformations
asked Nov 21 at 22:58
DORCT
406
find rank(A)
find basis of$ R(L_A)$ consisting of column vectors
find basis of $N(L_A)$
$$A=left[begin{array}{ccc}0&i&-1\1+i&1&1+2i\1-i&2&1+i\-i&1-i&1end{array}right]$$
i figure out the reduce form is
$$A=left[begin{array}{ccc}1&0&1\0&1&i\0&0&0\0&0&0end{array}right]$$, so the rank(A)=2, but what is basis of $R(L_A)$ and$N(L_A)$
linear-algebra matrices linear-transformations
linear-algebra matrices linear-transformations
asked Nov 21 at 22:58
DORCT
406
asked Nov 21 at 22:58
DORCT
406
edited Nov 21 at 23:23
asked Nov 21 at 22:58
DORCT
406
asked Nov 21 at 22:58
DORCT
406
asked Nov 21 at 22:58
DORCT
406
406
If you reduced right, the same method for getting bases for row space and null space as usual works, since complexes are a field.
– coffeemath
Nov 21 at 23:31
How would you find these bases if you had a $2$ there instead of $i$? Do the same thing.
– amd
Nov 22 at 0:00
for nullspace i got x1=x3 x2=-ix3 x3=x3 so i got basic of N(A)=1 -i 1
– DORCT
Nov 22 at 0:05
for range(A)i find the row of of pivot row which are first and sec row
– DORCT
Nov 22 at 0:07
add a comment |
If you reduced right, the same method for getting bases for row space and null space as usual works, since complexes are a field.
– coffeemath
Nov 21 at 23:31
How would you find these bases if you had a $2$ there instead of $i$? Do the same thing.
– amd
Nov 22 at 0:00
for nullspace i got x1=x3 x2=-ix3 x3=x3 so i got basic of N(A)=1 -i 1
– DORCT
Nov 22 at 0:05
for range(A)i find the row of of pivot row which are first and sec row
– DORCT
Nov 22 at 0:07
If you reduced right, the same method for getting bases for row space and null space as usual works, since complexes are a field.
– coffeemath
Nov 21 at 23:31
If you reduced right, the same method for getting bases for row space and null space as usual works, since complexes are a field.
– coffeemath
Nov 21 at 23:31
How would you find these bases if you had a $2$ there instead of $i$? Do the same thing.
– amd
Nov 22 at 0:00
How would you find these bases if you had a $2$ there instead of $i$? Do the same thing.
– amd
Nov 22 at 0:00
for nullspace i got x1=x3 x2=-ix3 x3=x3 so i got basic of N(A)=1 -i 1
– DORCT
Nov 22 at 0:05
for nullspace i got x1=x3 x2=-ix3 x3=x3 so i got basic of N(A)=1 -i 1
– DORCT
Nov 22 at 0:05
for range(A)i find the row of of pivot row which are first and sec row
– DORCT
Nov 22 at 0:07
for range(A)i find the row of of pivot row which are first and sec row
– DORCT
Nov 22 at 0:07
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
basic of A is $$N(A)=left[begin{array}{c}1\i\1\end{array}right]$$
and basis of R(A) is $$R(A)=left[begin{array}{cc}0&i\1+i&1\1-i&2\-i&1-iend{array}right]$$, am i doing right
answered Nov 21 at 23:48
DORCT
406
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
basic of A is $$N(A)=left[begin{array}{c}1\i\1\end{array}right]$$
and basis of R(A) is $$R(A)=left[begin{array}{cc}0&i\1+i&1\1-i&2\-i&1-iend{array}right]$$, am i doing right
answered Nov 21 at 23:48
DORCT
406
add a comment |
up vote
0
down vote
basic of A is $$N(A)=left[begin{array}{c}1\i\1\end{array}right]$$
and basis of R(A) is $$R(A)=left[begin{array}{cc}0&i\1+i&1\1-i&2\-i&1-iend{array}right]$$, am i doing right
answered Nov 21 at 23:48
DORCT
406
add a comment |
up vote
0
down vote
up vote
0
down vote
basic of A is $$N(A)=left[begin{array}{c}1\i\1\end{array}right]$$
and basis of R(A) is $$R(A)=left[begin{array}{cc}0&i\1+i&1\1-i&2\-i&1-iend{array}right]$$, am i doing right
answered Nov 21 at 23:48
DORCT
406
basic of A is $$N(A)=left[begin{array}{c}1\i\1\end{array}right]$$
and basis of R(A) is $$R(A)=left[begin{array}{cc}0&i\1+i&1\1-i&2\-i&1-iend{array}right]$$, am i doing right
answered Nov 21 at 23:48
DORCT
406
answered Nov 21 at 23:48
DORCT
406
answered Nov 21 at 23:48
DORCT
406
answered Nov 21 at 23:48
DORCT
406
406
add a comment |
add a comment |
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If you reduced right, the same method for getting bases for row space and null space as usual works, since complexes are a field.
– coffeemath
Nov 21 at 23:31
How would you find these bases if you had a $2$ there instead of $i$? Do the same thing.
– amd
Nov 22 at 0:00
for nullspace i got x1=x3 x2=-ix3 x3=x3 so i got basic of N(A)=1 -i 1
– DORCT
Nov 22 at 0:05
for range(A)i find the row of of pivot row which are first and sec row
– DORCT
Nov 22 at 0:07