How do I compute an adjunct matrix?
David Scott
Here's an example from my Linear Algebra book:
A = [1 2 3
2 3 4
3 4 5]
From the answers in the back of the book, the adjunct matrix is,
adj A = [-1 2 -1
2 -4 2
-1 2 -1]
I've been using det(A) in Matlab to compute the determinant and check
my work. Is there a similar function to compute the adjunct matrix?
Greg von Winckel
notation adj refers to the matrix adjoint.
In my experience, I have never actually needed the adjoint explicitly
for anything. It is computed from the determinants of all the minors
and is used in defining the matrix inverse (an object I have needed
explicitly only very rarely).
What are you trying to do that requires the adjoint?
-Greg
David Scott
unable to find any functions in Matlab that would compute the adjoint
matrix.
I'm trying to check my Linear Algebra homework. ^_-
Greg von Winckel
would really only be useful for checking linear algebra homework.
Since I'm feeling generous, here's an adjoint function:
function B=adj(A)
n=size(A,1); d=1:n-1;
B=zeros(n);
AA=[A,A;A,A]';
for j=1:n
for k=1:n
B(j,k)=(-1)^(j+k)*det(AA(j+d,k+d));
end
end
Don't use this for big matrices!
HTH,
Greg
Roger Stafford
<greg...@mathDEL.unmDEL.edu> wrote:
> I see. I'd guess the reason there is not a Matlab function is that it
> would really only be useful for checking linear algebra homework.
>
> Since I'm feeling generous, here's an adjoint function:
>
> function B=adj(A)
>
> n=size(A,1); d=1:n-1;
> B=zeros(n);
>
> AA=[A,A;A,A]';
>
> for j=1:n
> for k=1:n
> B(j,k)=(-1)^(j+k)*det(AA(j+d,k+d));
> end
> end
>
> Don't use this for big matrices!
>
> HTH,
>
> Greg
-----------------------
Greg, if you leave out the factor (-1)^(j+k), you'll have it right.
Roger Stafford
Greg von Winckel
the correction:
function B=adj(A)
n=size(A,1); d=1:n-1;
B=zeros(n); AA=[A,A;A,A]';
for j=1:n
for k=1:n
B(j,k)=det(AA(j+d,k+d));
end
end
Still, let me re-emphasize that this is only useful as an educational
aid.
Jos
The word adjoint is ambiguous. The above is also known as the
adjugate of matrix A. Formally, the adjoint is the same as the
conjugate transpose of A.
B = A.' ; % adjoint
B = det(A) * inv(A) ; % adjugate, for square invertible matrices
hth
Jos
Greg von Winckel
classes. "Adjoint" is ambiguous by itself, but usually pretty clear
from context. In most of my classes the conjugate transpose was
called "Hermitian conjugate" whereas the adjugate was only ever
mention in control systems classes, never in math classes.
Jos
<SNIP ... adjoint adjugate
> B = A.' ; % adjoint
should be B = A' of course ...
Jos
David Scott
> adjugate of matrix A. Formally, the adjoint is the same as the
> conjugate transpose of A.
>
> B = A.' ; % adjoint
>
> B = det(A) * inv(A) ; % adjugate, for square invertible matrices
>
> hth
> Jos
I should have thought of this myself. In the very chapter I'm
studying I learned that,
inv(A) = 1/det(A) * adj(A)
(in fact, this is why I'm computing adj(A) in the first place, but
I'm doing it by hand and want to be able to check it with Matlab)
I guess it makes sense that,
adj(A) = det(A) * inv(A)
Thanks.
Greg von Winckel
exist even if inv(A) does not.
-Greg
Roger Stafford
<greg...@mathDEL.unmDEL.edu> wrote:
> That is the case when A is nonsingular. However, adj(A) will always
> exist even if inv(A) does not.
>
> -Greg
---------------------
I believe the 'svd' function can be used to find the adjoint (or
adjugate) of a square matrix A even if it is not invertible. This method
avoids having to tediously compute n^2 different n-1 x n-1 cofactor
determinants.
n = size(A,1);
[u,s,v] = svd(A);
s = diag(s);
ix = toeplitz(ones(n-1,1),[1 zeros(1,n-1)])+repmat((1:n-1)',1,n);
B = v*diag(prod(reshape(s(ix),n-1,n)))*u'/det(v*u'); % B = adj(A)
(To ease the understanding of index ix, I point out that each of the terms
used in the diagonal matrix for obtaining B in the last line is the
product of all but one of the diagonal terms in s.)
Roger Stafford
Roger Stafford
<ellieandrogerxyzz...@dialup-4.232.57.48.dial1.losangeles1.level3.net>,
ellieandr...@mindspring.com.invalid (Roger Stafford) wrote:
> I believe the 'svd' function can be used to find the adjoint (or
> adjugate) of a square matrix A even if it is not invertible. This method
> avoids having to tediously compute n^2 different n-1 x n-1 cofactor
> determinants.
>
> n = size(A,1);
> [u,s,v] = svd(A);
> s = diag(s);
> ix = toeplitz(ones(n-1,1),[1 zeros(1,n-1)])+repmat((1:n-1)',1,n);
> B = v*diag(prod(reshape(s(ix),n-1,n)))*u'/det(v*u'); % B = adj(A)
>
> (To ease the understanding of index ix, I point out that each of the terms
> used in the diagonal matrix for obtaining B in the last line is the
> product of all but one of the diagonal terms in s.)
>
> Roger Stafford
------------------
I think it is conceptually better to write that last line as:
B = det(u*v')*v*diag(prod(reshape(s(ix),n-1,n)))*u'; % B = adj(A)
though the answer is the same. This way shows its derivation better from
det(A)*inv(A) = det(u*s*v') * v*inv(s)*u'
= det(u*v') * v*(inv(s)*det(s))*u'
as a limiting case as A approaches singularity. After cancellation, the
diagonal terms in inv(s)*det(s) each have all but one of the terms in s,
and a zero-divided-by-zero situation is avoided.
Roger Stafford
Roger Stafford
<greg...@mathDEL.unmDEL.edu> wrote:
------------------
Hello Greg,
You were too quick to accept the correction concerning the factor
(-1)^(j+k) which I made yesterday to your "educational" adj function. It
turns out you need it in case n = size(A,1) is even but not if n is odd.
The OP's case was odd and the factor would then be incorrect, so I jumped
to the erroneous conclusion that it should never be used. Doh! for me
too.
Roger Stafford
Greg von Winckel
function B=adj(A)
n=size(A,1); d=1:n-1;
B=zeros(n); c=2*mod(size(A,1),2)-1;
AA=[A,A;A,A]';
for j=1:n
for k=1:n
B(j,k)=c^(j+k)*det(AA(j+d,k+d));
end
end
Joe
adjunct_A=inv(A)*det(A)
Reasoning:
inv(A)=adjunct(A)/det(A)
enjoy.
Joe
"David Scott" <ded...@gmail.com> wrote in message <ef43d...@webcrossing.raydaftYaTP>...
Nasser M. Abbasi
> It is a one liner!!
> adjunct_A=inv(A)*det(A)
>
> Reasoning:
> inv(A)=adjunct(A)/det(A)
>
> enjoy.
> Joe
>
-------------------
A=[2 2;
2 2]
inv(A)*det(A)
Warning: Matrix is singular to working precision.
ans =
NaN NaN
NaN NaN
---------------------
However, the Adjunct of A is defined even for singular matix as
this, and it is
2 -2
-2 2
fyi, Mathematica has a Cofactor function that allows
finding the cofactor for each matrix element. Then one
simply transposes the result to obtain the adjunct matrix
----------------------------
<<"Combinatorica`"
mat={{2,2},{2,2}}
cof=Table[Cofactor[mat,{i,j}],{i,2},{j,2}];
adjunct=Transpose[cof]
Out[9]= {{2, -2},
{-2, 2}}
------------------
It can be used on symbolic matrices ofcourse
-------------------
mat ={ {a, b},
{c, d} };
cof = Table[Cofactor[mat,{i,j}],{i,2},{j,2}];
adjunct = Transpose[cof]
Out[18]= {{d, -b},
{-c, a}}
------------------------
There has been some discussion on this sometime ago here I
remember. I think something on Matlab file exchange can do this.
may be OP can google it.
--Nasser
Matt J
> This above does not work if det(A)=0
>
> -------------------
> A=[2 2;
> 2 2]
How about this? Too painful for large matrices? I wonder if that can be avoided, and if one would even want the adjunct for large A.
>> c=poly(A); adjunct_A=-polyvalm(c(1:end-1),A)
adjunct_A =
2 -2
-2 2