# Does this lose the benefit of QR decomposition?
**URL:** https://discourse.julialang.org/t/does-this-lose-the-benefit-of-qr-decomposition/58324
**Category:** Performance
**Tags:** linearalgebra, qr
**Created:** [March 31, 2021, 7:37pm UTC](https://discourse.julialang.org/t/does-this-lose-the-benefit-of-qr-decomposition/58324 "2021-03-31T19:37:30Z")
**Posts on this page:** 3
**Page:** 1
### Author:  [@Ross\_Boylan](https://discourse.julialang.org/u/Ross_Boylan)
#### Post date: [March 31, 2021, 7:37pm UTC](https://discourse.julialang.org/t/does-this-lose-the-benefit-of-qr-decomposition/58324/1 "2021-03-31T19:37:30Z")
My understanding is that one can do fast multiplies by the Q matrix of the QR decomposition because it is represented as a sequence of Householder reflections. I want to compute
```julia
deltai = 3:size(T, 1)
δ = Q[:, deltai] * fp[deltai]
```
(not self-containted; `fp` is one dimensional) where `Q` is the result of a call to `qr().Q`. I’m concerned that, since I subset `Q` explicitly this loses the benefit.
Should I be concerned? If so, what can I do about it?
I don’t think `(Q*fp)[deltai]` yields the same result.
I guess I could force the first 2 values of `fp` to 0; then I think `Q*fp` does give the result I want.
---
### Author:  [@simonbyrne](https://discourse.julialang.org/u/simonbyrne)
#### Post date: [March 31, 2021, 9:32pm UTC](https://discourse.julialang.org/t/does-this-lose-the-benefit-of-qr-decomposition/58324/2 "2021-03-31T21:32:50Z")
> [@Ross\_Boylan](#):
>
> I guess I could force the first 2 values of `fp` to 0; then I think `Q*fp` does give the result I want.
I think padding the first two entries to be zero would be the most efficient option here: otherwise it has to form `Q`, which requires applying the reflectors of `Q` to an identity matrix:
>
---
### Author:  [@Ross\_Boylan](https://discourse.julialang.org/u/Ross_Boylan)
#### Post date: [April 1, 2021, 11:49pm UTC](https://discourse.julialang.org/t/does-this-lose-the-benefit-of-qr-decomposition/58324/3 "2021-04-01T23:49:29Z")
Thank you. After overcoming the fact that my code became [so optimized it wouldn’t run at all](https://discourse.julialang.org/t/forwarddiff-and-lapack-dont-get-along/58397), I got it to work. The difference was very large, ~8x faster based on the means, considering that is only one part of the calculations. Memory use shows similar improvements.
# Before
```julia-auto
julia> @benchmark RBSilly.fit(yx, 2)
BenchmarkTools.Trial:
memory estimate: 3.17 MiB
allocs estimate: 602
--------------
minimum time: 71.913 μs (0.00% GC)
median time: 168.027 μs (0.00% GC)
mean time: 241.267 μs (41.50% GC)
maximum time: 2.101 ms (89.42% GC)
--------------
samples: 10000
evals/sample: 1
```
# After
```julia-auto
julia> @benchmark RBSilly.fit(yx, 2)
BenchmarkTools.Trial:
memory estimate: 59.86 KiB
allocs estimate: 190
--------------
minimum time: 23.335 μs (0.00% GC)
median time: 24.966 μs (0.00% GC)
mean time: 29.430 μs (11.89% GC)
maximum time: 3.025 ms (94.90% GC)
--------------
samples: 10000
evals/sample: 1
```