Linear algebra, Gram-Schmidt say I have a matrix A with independent columns: I do Gram-Schmidt to this matrix A so that I get an orthonormal matrix named Q out of it. I understand that matrix Q will have the same collumn space as A. But does Q have the same nullspace as A ? And if not then, why is Q useful to us ?
Let me ask you a question
What's a basis for our familiar coordinate grid (xy plane) ?
isn't is [0 1] , [1 0] ?
it *
Yes, are they orthonormal ?
yeah
Why ? can't we work with some other basis that is not orthonormal ?
this is also called the standard basis
Why do we so much want our axes to orthogonal ?
its as simple as it can get I think
Yes, all the calculations become simple when you choose an orthonormal basis. So why are you asking the usefulness of Q ?
because the procedure of finding Q is more complex than finding a standard basis
I could just find the rref instead of doing Gram - Schmidt
I think ..
Haha its an one time thing
Once you have a good basis for a vectorspace, you can use it to work all the problems
but will it have the same nullspace as the starting A ?
It will right ?
You've said that A has independent column vectors; this means nullspace of A is {0}
I guess what I am really trying to pin down is : say we have a rectangular A , and I want to find its rref. Instead of row operations , I do collumn operations , or I do both row and column operations . Am I allowed to do such a thing and still arive at rref form >
cause Gram - Schmidt is basically col operations
Gram-Schmidt is about constructing an orthonormal basis from a given set of independent vectors
rref = reduced row echelon form you can't mess with columns here; doing so will disturb the rowspace
it will disturb the row space, but will it disturb the nullspace ?
Exactly! messing with columns does change the nullspace
Consider below matrix \[A=\begin{bmatrix}1&2\\2&4 \end{bmatrix}\]
so for the matrix you just showed ( A ) , if I want to find Q , ill just subtruct from the second col a multiple of the first col
ofc its more complicated than that but essentially thats what I do
so we do a col operation to this matrix
* oh and dived by the length
so there is another col manipulation
You can't find a Q; there exists no Q, because its columns are not independent.
Look at that matrix again, can you tell me its nullspace ?
[ -2,1 ] ?
Yep, next add first column to second column
same result
\[B=\begin{bmatrix}1&\color{red}{3}\\2&\color{red}{6} \end{bmatrix}\] whats the null space of B ?
[ -2,1 ]
sure ?
[ -3,1 ] ?
Yes, so the nullspace does change with column operations
row operations preserve rowspace and nullspace column operations preserve columnspace
i see
row operations preserve rowspace and nullspace, but disturb columnspace column operations preserve columnspace, but disturb rowspace and nullspace
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