# Linear algebra bug?
**URL:** https://discourse.julialang.org/t/linear-algebra-bug/44513
**Category:** General Usage
**Tags:** matrices
**Created:** [August 7, 2020, 4:47pm UTC](https://discourse.julialang.org/t/linear-algebra-bug/44513 "2020-08-07T16:47:05Z")
**Posts on this page:** 5
**Page:** 1
### Author:  [@jensen](https://discourse.julialang.org/u/jensen)
#### Post date: [August 7, 2020, 4:47pm UTC](https://discourse.julialang.org/t/linear-algebra-bug/44513/1 "2020-08-07T16:47:05Z")
I translating some code from octave/matlab to Julia. However, the code does not work. I have isolated the problem in the enclosed octave/julia files. The octave code produces the correct result.
The problem is in the calculation of :
G = ((A’\*A)^-1)\*A’ # Octave
G = inv(transpose(A)\*A)\*transpose(A) # Julia
My system is :
uname --all
Linux optiplex 5.4.0-42-generic #46~18.04.1-Ubuntu SMP Fri Jul 10 07:21:24 UTC 2020 x86\_64 x86\_64 x86\_64 GNU/Linux
The problems are observed on julia-1.5.0, however it is also observed in 1.4.2.
Please advise. I assume that I have misunderstood the Julia syntax.
## Octave code
#!/usr/bin/octave
A = [ 30 20 10 ;
40 30 20 ;
50 40 30 ;
60 50 40 ;
70 60 50 ]
size(A)
Q = (A’\*A)^-1
G = ((A’\*A)^-1)\*A’
A
=\> G =
3.8940e-02 2.5146e-02 1.0376e-02 -6.3477e-03 -2.1118e-02
7.3242e-04 1.4648e-03 2.4414e-04 2.9297e-03 -2.1973e-03
-6.0059e-02 -3.5156e-02 -1.0742e-02 1.5625e-02 3.9062e-02
## julia code
#!/opt/julia-1.5.0/bin/julia
using LinearAlgebra
A = [ 30 20 10 ;
40 30 20 ;
50 40 30 ;
60 50 40 ;
70 60 50 ]
println(size(A))
print("A = ")
display(A)
println()
Q = inv(transpose(A)\*A)
print("Q = ")
display(Q)
println()
G = inv(transpose(A)\*A)\*transpose(A)
print("G = ")
display(G)
println()
print("A = ")
display(A)
println()
===\> G = 3×5 Array{Float64,2}:
0.0078125 0.00390625 0.00195312 -0.00195312 -0.00390625
0.0615234 0.0371094 0.0136719 -0.0117188 -0.0390625
-0.0908203 -0.0527344 -0.0166016 0.0214844 0.0566406
---
### Author:  [@AndiMD](https://discourse.julialang.org/u/AndiMD)
#### Post date: [August 7, 2020, 5:09pm UTC](https://discourse.julialang.org/t/linear-algebra-bug/44513/2 "2020-08-07T17:09:19Z")
the condition number of `A'*A` seems to be ~infinite? That would mean, both the Julia and the Octave results are wrong.
By the way, you can use the same syntax in Julia:
`G = ((A'*A)^-1)*A'`
Check condition number:
`using LinearAlgebra`
Use more precision:
`B = BigFloat.(A)`
`cond(B'*B)`
---
### Author:  [@Tamas\_Papp](https://discourse.julialang.org/u/Tamas_Papp)
#### Post date: [August 7, 2020, 6:27pm UTC](https://discourse.julialang.org/t/linear-algebra-bug/44513/3 "2020-08-07T18:27:03Z")
To complement @AndiMD’s answer: `A` is singular, eg
```julia
julia> svd(A).S
3-element Array{Float64,1}:
167.01031102734288
10.370921393501915
1.1114435990051559e-14
```
so this is a badly conditioned problem.
Do you need this for linear regression? Consider some form of regularization.
---
### Author:  [@Seif\_Shebl](https://discourse.julialang.org/u/Seif_Shebl)
#### Post date: [August 7, 2020, 9:36pm UTC](https://discourse.julialang.org/t/linear-algebra-bug/44513/4 "2020-08-07T21:36:17Z")
As others have observed, the matrix is badly conditioned. MATLAB actually tells you this explicitely in the warning message:
```julia
>> G = ((A'*A)^-1)*A'
Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 1.015773e-18.
```
Also, `inv` is really rarely needed, and is not always more accurate than the back-solve `\` operator. So, you may write this elegantly in Julia like this:
```julia
julia> (A'A)\A'
3×5 Array{Float64,2}:
0.00820313 0.0053125 0.004375 -0.00046875 -0.0053125
0.0635937 0.039375 0.01125 -0.0090625 -0.029375
-0.0917969 -0.0546875 -0.015625 0.0195312 0.0546875
```
The results are not accurate as I and others have said, though.
---
### Author:  [@jensen](https://discourse.julialang.org/u/jensen)
#### Post date: [August 7, 2020, 9:55pm UTC](https://discourse.julialang.org/t/linear-algebra-bug/44513/5 "2020-08-07T21:55:29Z")
Many thanks. You are both right !!
When I constructed the unit test for the function I just picked a very “regular” patter that turned out to be poorly conditioned. I assumed that the octave results were correct as I have used the function for many years without problems ( with proper well-conditioned data).
You saved me a lot of time.
Again. Many thanks.
Peter